, cbse worksheets for class 10 maths.
Download free printable worksheets for CBSE Class 10 Maths with important chapter wise questions as per Latest NCERT Syllabus. These Worksheets help Grade 10 students practice Maths Important Questions and exercises on various topics like Coordinate Geometry, Probability, Quadratic Equations, Statistics, Surface Area, Arithmetic Progression, Polynomials, Linear Equation in two Variables, Real Numbers. These free PDF download of Class 10 Maths worksheets consist of visual simulations to help your child visualize concepts being taught and reinforce their learning.
Get free Kendriya Vidyalaya Class 10 Maths Worksheets shared by teachers, parents & students to understand the concepts. All the necessary topics are covered in these 10th grade worksheets. These class 10 Maths worksheets provide skills and experience necessary to ace in Exams
CBSE Class 10 Maths Syllabus
UNIT I: NUMBER SYSTEMS
1. REAL NUMBER
Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of irrationality.
UNIT II: ALGEBRA
UNIT III: COORDINATE GEOMETRY Coordinate Geometry Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division).
UNIT IV: GEOMETRY
UNIT V: TRIGONOMETRY
UNIT VI: MENSURATION
UNIT VII: STATISTICS AND PROBABILITY
PRESCRIBED BOOKS:
Structure of CBSE Maths Sample Paper for Class 10 is
For Preparation of board exams students can also check out other resource material
CBSE Class 10 Maths Sample Papers
Important Questions for Class 10 Maths Chapter Wise
Maths Revision Notes for class 10
Previous Year Question Paper CBSE Class 10 Maths
Why do one Children need Worksheets for Practice ?
It is very old saying that one can build a large building if the foundation is strong and sturdy. This holds true for studies also. Worksheets are essential and help students in the in-depth understanding of fundamental concepts. Practicing solving a lot of worksheets, solving numerous types of questions on each topic holds the key for success. Once basic concepts and fundamentals have been learnt, the next thing is to learn their applications by practicing problems. Practicing the problems helps us immensely to gauge how well we have understood the concepts.
There are times when students just run through any particular topic with casual awareness there by missing out on a few imperative “between the lines” concepts. Such things are the major causes of weak fundamental understandings of students. So in such cases Worksheets act as a boon and critical helpful tool which gauges the in-depth understanding of children highlighting doubts and misconceptions, if any.
Worksheets classifies the important aspects of any topic or chapter taught in the class in a very easy manner and increases the awareness amongst students.When students try to solve a worksheet they get to understand what are the key important factors which needs the main focus.Sometimes it happens that due to shortage of time all the major points of any particular topic gets skipped in the class or teacher rushes through , due to shortage of time. A worksheet thus provides a framework for the entire chapter and can help covering those important aspects which were rushed in the class and ensure that students record and understand all key items.
In a class of its say 40 students howsoever teacher tries to be active and work towards making each student understand whatever she has to teach in the class but there are always some students who tend to be in their own world and they wander in their thoughts.Worksheets which are provided timely to all the students, causes them to focus on the material at hand. it’s simply the difference between passive and active learning. Worksheets of this type can be used to introduce new material, particularly material with many new definitions and terms.
Worksheets help students be focussed and attentive in the class because they know after the class is over they will be assigned a worksheet which they need to solve so if they miss or skip any point in the class they may not be able to solve the worksheet completely and thereby lose reputation in the class.
Often students revise the chapter at home reading their respective textbooks. Thus more often than not they do miss many important points. Worksheets thus can be used intentionally to help guide student’s to consult textbooks. Having students write out responses encourages their engagement with the textbooks, the questions chosen indicate areas on which to focus. Explicitly discussing the worksheets and why particular questions are asked helps students reflect on what is important.
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Refer to Case Study Chapter 6 Triangles Mathematics, these class 10 maths case study based questions have been designed as per the latest examination guidelines issued for the current academic year by CBSE, NCERT, KVS. Students should go through these solves case studies so that they are able to understand the pattern of questions expected in exams and get good marks.
I. Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles.The height of Vijay’s house if 20m when Vijay’s house casts a shadow 10m long on the ground. At the same time, the tower casts a shadow 50m long on the ground and the house of Ajay casts 20m shadow on the ground.
Question. What is the height of Ajay’s house? (a) 30m (b) 40m (c) 50m (d) 20m
Question. When the tower casts a shadow of 40m, same time what will be the length of the shadow of Vijay’s house? (a) 15m (b) 32m (c) 16m (d) 8m
Question. What is the height of the tower? (a) 20m (b) 50m (c) 100m (d) 200m
Question. When the tower casts a shadow of 40m, same time what will be the length of the shadow of Ajay’s house? (a) 16m (b) 32m (c) 20m (d) 8m
Question. What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12m? (a) 75m (b) 50m (c) 45m (d) 60m
l. Read the following and answer Two trees are standing parallel to each other. The bigger tree 8 m high, casts a shadow of 6 m.
Question. If the ratio of the height of two trees is 3 : 1, then the shadow of the smaller tree is (a) 2 m (b) 6 m (c) 8/3 m (d) 8 m
Question. If , ΔABC ∼ ΔPQR , ar (ΔABC)/ ar (ΔPQR) = 4/25 , PQ = 10 cm, then AB is equal to (a) 4 cm (b) 2 cm (c) 5 cm (d) 5 8 cm
Question. If AB and CD are the two trees and AE is the shadow of the longer tree, then (a) ΔAEB ∼ ΔCED (b) ΔABE ∼ ΔCED (c) ΔAEB ∼ ΔDEC (d) ΔBEA ∼ ΔDEC
Question. Since AB ll CD , so by basic proportionality theorem, we have (a) AE/CE = BD/DE (b) AC/AE = DE/BE (c) AE/CE = AB/CD (d) AE/CE = BE/DE
Question. The distance of point B from E is (a) 10 m (b) 8 m (c) 18 m (d) 10/3 m
ll. Read the following and answer A ladder was placed against a wall such that it touches a point 4 m above the ground. The distance of the foot of the ladder from the bottom of the ground was 3 m. Keeping its foot at the same point, Akshay turns the ladder to the opposite side so that it reached the window of his house.
Question. In an isosceles right triangle PQR, right angled at P, then (a) QR 2 = 2PQ 2 (b) QP 2 = 2PR 2 (c) QP 2 = 2QR 2 (d) PR 2 = 2QR 2
Question. If OA 2 = OB 2 + AB 2 , then (a) ΔOBA is an equilateral triangle. (b) ΔOAB is an isosceles right triangle. (c) ΔOAB is a right triangle right angled at O. (d) ΔOAB is a right triangle right angled at B.
Question. The theorem which can be used for find the length of the ladder is (a) Thales Theorem (b) Converse of Thales Theorem (c) Pythagoras Theorem (d) Converse of Pythagoras Theorem
Question. The length of the ladder, in metre is (a) 4 m (b) 5 m (c) 9 m (d) 2 m
Question. If the window of the house is 3 m above the ground, then the distance of the point C from D is (a) 3 m (b) 4 m (c) 5 m (d) 3.5 m
lll. Read the following and answer Two buildings (say A and B) are located 12 m apart. The height of the two buildings are 32 m and 41 m.
Question. The distance DF is equal to (a) 15 m (b) 12 m (c) 9 m (d) 21 m
Question. In a triangle PQR, PQ = 7 cm, QR = 25 cm, RP = 24 cm, then the triangle is right angled at (a) P (b) Q (c) R (d) can’t say
Question. ABC is an equilateral triangle of side ‘2a’ units. The length of each of its altitude is (a) a units (b) 2a units (c) √2 a units (d) √3 a units
Question. The distance between the top of the two buildings can be calculated using (a) Thales Theorem (b) Pythagoras Theorem (c) Converse of Thales Theorem (d) Converse of Pythagoras Theorem
Question. The length EF in the figure is (a) 32 m (b) 41 m (c) 41 m/2 (d) 9 m
lV. Read the following and answer A farmer had a triangular piece of land. He put a fence, parallel to one of the sides of the field as shown in the figure.
Question. If AD = x + 1, DB = 3x – 1, AE = x + 3, EC = 3x + 4, then (a) x = 5 (b) x = 7 (c) x = 8 (d) x = 4
Question. If the point D is 20 m away from A, where as AB and AC are 80 m and 100 m respectively, then (a) AE = 20 m (b) EC = 25 cm (c) AE = 25 cm (d) EC = 60 cm
Question. Which of the following is not true? (a) AD/AB = AE/AC (b) AD/AE = AB/AC (c) AB/BD = AC/EC (d) BD/AD = AE/EC
Question. Which of the following statements is true? (a) AD/DB = AE/EC , using Thales Theorem (b) AD/DB = AE/EC , using Pythagoras Theorem (c) AD/DB = AE/EC , using Pythagoras Theorem (d) AD/DB = AE/EC , using Thales Theorem
Question. If P and Q are the mid points of sides YZ and XZ respectively, then (a) PQ ll XY (b) PQ ll YZ (c) PQ ll ZX (d) None of these
V. Read the following and answer The ratio of two corresponding sides in similar figures is called scale factor. Scale factor = Length of image / Actual length of object
Question. Two similar triangles have a scale factor of 1 : 2. Then their corresponding altitudes have a ratio (a) 2 : 1 (b) 4 : 1 (c) 1 : 2 (d) 1 : 1
Question. If two similar triangles have a scale factor of 2 : 5, then which of the following statements is true ? (a) The ratio of their medians is 2 : 5. (b) The ratio of their altitudes is 5 : 2. (c) The ratio of their perimeters is 2 × 3 : 5. (d) The ratio of their altitudes is 22 : 52.
Question. The shadow of a statue 8 m long has length 5 m. At the same time the shadow of a pole 5.6 m high is (a) 3 m (b) 3.5 m (c) 4 cm (d) 4.5 m
Question. For two similar polygons which of the following is not true? (a) They are not flipped horizontally. (b) They are dilated by a scale factor. (c) They cannot be translated down. (d) They are mirror images of each other.
Question. A model of a car is made on the scale 1 : 8. The model is 40 cm long and 20 cm wide. The actual length of car is (a) 320 cm (b) 160 cm (c) 5 cm (d) 2.5 cm
Worksheets for class 10, worksheets for class 10 mathematics circles.
Ncert solutions for class 10 maths chapter 6 – download free pdf.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here, which is considered to be one of the most important study materials for the students studying in CBSE Class 10. Chapter 6 of NCERT Solutions for Class 10 Maths is well structured in accordance with the CBSE Syllabus for 2023-24. It covers a vast topic, including a number of rules and theorems. Students often tend to get confused about which theorem to use while solving a variety of questions.
Download most important questions for class 10 maths chapter – 6 triangles.
The solutions provided at BYJU’S are designed in such a way that every step is explained clearly and in detail. The Solutions for NCERT Class 10 Maths are prepared by the subject experts to help students prepare better for their board exams. These solutions will be helpful not only for exam preparations, but also in solving homework and assignments.
The CBSE Class 10 examination often asks questions, either directly or indirectly, from the NCERT textbooks. Thus, the NCERT Solutions for Chapter 6 Triangles of Class 10 Maths is one of the best resources to prepare, and equip oneself to solve any type of questions in the exam, from the chapter. It is highly recommended that the students practise these NCERT Solutions on a regular basis to excel in the Class 10 board examinations.
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Exercise 6.1 page: 122.
1. Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
Answer: Similar
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral) Answer: Equilateral
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
Answer: (a) Equal
(b) Proportional
2. Give two different examples of pair of (i) Similar figures (ii) Non-similar figures
3. State whether the following quadrilaterals are similar or not:
From the given two figures, we can see their corresponding angles are different or unequal. Therefore, they are not similar.
1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
(i) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒1.5/3 = 1/EC
⇒EC = 3/1.5
EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.
(ii) Given, in △ ABC, DE∥BC
⇒ AD/7.2 = 1.8 / 5.4
⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10
Hence, AD = 2.4 cm.
2. E and F are points on the sides PQ and PR, respectively of a ΔPQR. For each of the following cases, state whether EF || QR. (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm
Given, in ΔPQR, E and F are two points on side PQ and PR, respectively. See the figure below;
(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm
Therefore, by using Basic proportionality theorem, we get,
PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, we get, PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
PE/QE = 4/4.5 = 40/45 = 8/9
And, PF/RF = 8/9
So, we get here,
PE/QE = PF/RF
Hence, EF is parallel to QR.
(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
From the figure,
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55 …………. (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55 ………… (ii)
PE/EQ = PF/FR
3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD
In the given figure, we can see, LM || CB,
By using basic proportionality theorem, we get,
AM/AB = AL/AC ……………………..(i)
Similarly, given, LN || CD and using basic proportionality theorem,
∴AN/AD = AL/AC ……………………………(ii)
From equation (i) and (ii) , we get,
AM/AB = AN/AD
Hence, proved.
4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC
In ΔABC, given as, DE || AC
Thus, by using Basic Proportionality Theorem, we get,
∴BD/DA = BE/EC ……………………………………………… (i)
In ΔBAE, given as, DF || AE
∴BD/DA = BF/FE ……………………………………………… (ii)
From equation (i) and (ii) , we get
BE/EC = BF/FE
5. In the figure, DE||OQ and DF||OR, show that EF||QR.
In ΔPQO, DE || OQ
So by using Basic Proportionality Theorem,
PD/DO = PE/EQ……………… ..(i)
Again given, in ΔPOR, DF || OR,
PD/DO = PF/FR………………… (ii)
Therefore, by converse of Basic Proportionality Theorem,
EF || QR, in ΔPQR.
6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Given here,
In ΔOPQ, AB || PQ
By using Basic Proportionality Theorem,
OA/AP = OB/BQ……………. (i)
Also given,
In ΔOPR, AC || PR
By using Basic Proportionality Theorem
∴ OA/AP = OC/CR……………(ii)
OB/BQ = OC/CR
In ΔOQR, BC || QR.
7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Given, in ΔABC, D is the midpoint of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.
We have to prove that E is the mid point of AC.
Since, D is the mid-point of AB.
⇒AD/DB = 1 …………………………. (i)
In ΔABC, DE || BC,
Therefore, AD/DB = AE/EC
From equation (i), we can write,
⇒ 1 = AE/EC
Hence, proved, E is the midpoint of AC.
8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Given, in ΔABC, D and E are the mid points of AB and AC, respectively, such that,
AD=BD and AE=EC.
We have to prove that: DE || BC.
Since, D is the midpoint of AB
⇒AD/BD = 1……………………………….. (i)
Also given, E is the mid-point of AC.
⇒ AE/EC = 1
AD/BD = AE/EC
By converse of Basic Proportionality Theorem,
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.
We have to prove, AO/BO = CO/DO
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔADC, we have OE || DC
Therefore, by using Basic Proportionality Theorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE || AB
DE/EA = DO/BO …………….(ii)
AO/CO = BO/DO
⇒AO/BO = CO/DO
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.
Given, Quadrilateral ABCD where AC and BD intersect each other at O such that,
AO/BO = CO/DO.
We have to prove here, ABCD is a trapezium
In ΔDAB, EO || AB
DE/EA = DO/OB ……………………(i)
Also, given,
AO/BO = CO/DO
⇒ AO/CO = BO/DO
⇒ CO/AO = DO/BO
⇒DO/OB = CO/AO …………………………..(ii)
DE/EA = CO/AO
Therefore, by using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
⇒ AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD.
1. State which pairs of triangles in the figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
(i) Given, in ΔABC and ΔPQR,
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
Therefore, by AAA similarity criterion,
∴ ΔABC ~ ΔPQR
(ii) Given, in ΔABC and ΔPQR,
AB/QR = 2/4 = 1/2,
BC/RP = 2.5/5 = 1/2,
CA/PA = 3/6 = 1/2
By SSS similarity criterion,
ΔABC ~ ΔQRP
(iii) Given, in ΔLMP and ΔDEF,
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF = 2.7/5 = 27/50
Here , MP/DE = PL/DF ≠ LM/EF
Therefore, ΔLMP and ΔDEF are not similar.
(iv) In ΔMNL and ΔQPR, it is given,
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
Therefore, by SAS similarity criterion
∴ ΔMNL ~ ΔQPR
(v) In ΔABC and ΔDEF, given that,
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here , AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
Hence, ΔABC and ΔDEF are not similar.
(vi) In ΔDEF, by sum of angles of triangles, we know that,
∠D + ∠E + ∠F = 180°
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
Similarly, In ΔPQR,
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
Now, comparing both the triangles, ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, ΔDEF ~ ΔPQR
2. In figure 6.35, ΔODC ~ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
As we can see from the figure, DOB is a straight line.
Therefore, ∠DOC + ∠ COB = 180°
⇒ ∠DOC = 180° – 125° (Given, ∠ BOC = 125°)
In ΔDOC, sum of the measures of the angles of a triangle is 180º
Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180°
⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°)
⇒ ∠DCO = 55°
It is given that, ΔODC ~ ΔOBA,
Therefore, ΔODC ~ ΔOBA.
Hence, corresponding angles are equal in similar triangles
∠OAB = ∠OCD
⇒ ∠ OAB = 55°
⇒ ∠OAB = 55°
3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD
In ΔDOC and ΔBOA,
AB || CD, thus alternate interior angles will be equal,
∴∠CDO = ∠ABO
∠DCO = ∠BAO
Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;
∴∠DOC = ∠BOA
Hence, by AAA similarity criterion,
ΔDOC ~ ΔBOA
Thus, the corresponding sides are proportional.
DO/BO = OC/OA
⇒OA/OC = OB/OD
4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
∠PQR = ∠PRQ
∴ PQ = PR ……………………… (i)
QR/QS = QT/PRUsing equation (i) , we get
QR/QS = QT/QP ……………….(ii)
In ΔPQS and ΔTQR, by equation (ii),
QR/QS = QT/QP
∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]
5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Given, S and T are point on sides PR and QR of ΔPQR
And ∠P = ∠RTS.
In ΔRPQ and ΔRTS,
∠RTS = ∠QPS (Given)
∠R = ∠R (Common angle)
∴ ΔRPQ ~ ΔRTS (AA similarity criterion)
6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
Given, ΔABE ≅ ΔACD.
∴ AB = AC [By CPCT] ………………………………. (i)
And, AD = AE [By CPCT] …………………………… (ii)
In ΔADE and ΔABC, dividing eq.(ii) by eq(i),
AD/AB = AE/AC
∠A = ∠A [Common angle]
∴ ΔADE ~ ΔABC [SAS similarity criterion]
7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC
Given, altitudes AD and CE of ΔABC intersect each other at the point P.
(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (90° each)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
ΔAEP ~ ΔCDP
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB ( 90° each)
∠ABD = ∠CBE (Common Angles)
ΔABD ~ ΔCBE
(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (90° each)
∠PAE = ∠DAB (Common Angles)
ΔAEP ~ ΔADB
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (90° each)
∠PCD = ∠BCE (Common angles)
ΔPDC ~ ΔBEC
8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,
In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (AA similarity criterion)
9. In the figure, ABC and AMP are two right triangles, right angled at B and M, respectively, prove that:
(i) ΔABC ~ ΔAMP
(ii) CA/PA = BC/MP
Given, ABC and AMP are two right triangles, right angled at B and M, respectively.
(i) In ΔABC and ΔAMP, we have,
∠CAB = ∠MAP (common angles)
∠ABC = ∠AMP = 90° (each 90°)
∴ ΔABC ~ ΔAMP (AA similarity criterion)
(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)
If two triangles are similar then the corresponding sides are always equal,
Hence, CA/PA = BC/MP
10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG (ii) ΔDCB ~ ΔHGE (iii) ΔDCA ~ ΔHGF
Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG, respectively.
(i) From the given condition,
ΔABC ~ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
Since, ∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠ACD = ∠FGH
∴ ΔACD ~ ΔFGH (AA similarity criterion)
⇒CD/GH = AC/FG
(ii) In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Already proved)
∠B = ∠E (Already proved)
∴ ΔDCB ~ ΔHGE (AA similarity criterion)
(iii) In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Already proved)
∠A = ∠F (Already proved)
∴ ΔDCA ~ ΔHGF (AA similarity criterion)
11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
Given, ABC is an isosceles triangle.
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∠BAD = ∠CEF (Already proved)
∴ ΔABD ~ ΔECF (using AA similarity criterion)
12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.
Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR
i.e. AB/PQ = BC/QR = AD/PM
We have to prove: ΔABC ~ ΔPQR
As we know here,
AB/PQ = BC/QR = AD/PM
⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)
⇒ ΔABD ~ ΔPQM [SSS similarity criterion]
∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]
⇒ ∠ABC = ∠PQR
In ΔABC and ΔPQR
AB/PQ = BC/QR …………………………. (i)
∠ABC = ∠PQR …………………………… (ii)
ΔABC ~ ΔPQR [SAS similarity criterion]
13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA 2 = CB.CD
Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.
In ΔADC and ΔBAC,
∠ADC = ∠BAC (Already given)
∠ACD = ∠BCA (Common angles)
∴ ΔADC ~ ΔBAC (AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
∴ CA/CB = CD/CA
⇒ CA 2 = CB.CD.
14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.
Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;
AB/PQ = AC/PR = AD/PM
We have to prove, ΔABC ~ ΔPQR
Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.
In ΔABD and ΔCDE, we have
AD = DE [By Construction.]
BD = DC [Since, AP is the median]
and, ∠ADB = ∠CDE [Vertically opposite angles]
∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]
⇒ AB = CE [By CPCT] ………………………….. (i)
Also, in ΔPQM and ΔMNR,
PM = MN [By Construction.]
QM = MR [Since, PM is the median]
and, ∠PMQ = ∠NMR [Vertically opposite angles]
∴ ΔPQM = ΔMNR [SAS criterion of congruence]
⇒ PQ = RN [CPCT] ……………………………… (ii)
Now, AB/PQ = AC/PR = AD/PM
From equation (i) and (ii) ,
⇒CE/RN = AC/PR = AD/PM
⇒ CE/RN = AC/PR = 2AD/2PM
⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]
∴ ΔACE ~ ΔPRN [SSS similarity criterion]
Therefore, ∠2 = ∠4
Similarly, ∠1 = ∠3
∴ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P ……………………………………………. (iii)
Now, in ΔABC and ΔPQR, we have
AB/PQ = AC/PR (Already given)
From equation (iii),
∴ ΔABC ~ ΔPQR [ SAS similarity criterion]
15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Given, Length of the vertical pole = 6m
Shadow of the pole = 4 m
Let Height of tower = h m
Length of shadow of the tower = 28 m
In ΔABC and ΔDEF,
∠C = ∠E (angular elevation of sum)
∠B = ∠F = 90°
∴ ΔABC ~ ΔDEF (AA similarity criterion)
∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)
∴ 6/h = 4/28
⇒h = (6×28)/4
⇒ h = 6 × 7
⇒ h = 42 m
Hence, the height of the tower is 42 m.
16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.
Given, ΔABC ~ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.
∴AB/PQ = AC/PR = BC/QR ……………………………(i )
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….….. (ii)
Since AD and PM are medians, they will divide their opposite sides.
∴ BD = BC/2 and QM = QR/2 ……………..…………. (iii)
From equations (i) and (iii) , we get
AB/PQ = BD/QM ……………………….(iv)
In ΔABD and ΔPQM,
From equation (ii), we have
From equation (iv), we have,
AB/PQ = BD/QM
∴ ΔABD ~ ΔPQM (SAS similarity criterion)
⇒AB/PQ = BD/QM = AD/PM
Exercise 6.4 Page: 143
1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm 2 and 121 cm 2 . If EF = 15.4 cm, find BC.
Solution: Given, ΔABC ~ ΔDEF,
Area of ΔABC = 64 cm 2
Area of ΔDEF = 121 cm 2
EF = 15.4 cm
As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
= AC 2 /DF 2 = BC 2 /EF 2
∴ 64/121 = BC 2 /EF 2
⇒ (8/11) 2 = (BC/15.4) 2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm
2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.
In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [AAA similarity criterion]
As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,
Area of (ΔAOB)/Area of (ΔCOD) = AB 2 /CD 2
= (2CD) 2 /CD 2 [∴ AB = 2CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD 2 /CD 2 = 4/1
Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1
3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.
Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.
We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO
Let us draw two perpendiculars AP and DM on line BC.
We know that area of a triangle = 1/2 × Base × Height
In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (AA similarity criterion)
∴ AP/DM = AO/DO
⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO.
4. If the areas of two similar triangles are equal, prove that they are congruent.
Say ΔABC and ΔPQR are two similar triangles and equal in area
Now let us prove ΔABC ≅ ΔPQR.
Since, ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC 2 /QR 2
⇒ BC 2 /QR 2 =1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC 2 /QR 2
Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]
5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.
We have to prove: Area(ΔABC)/Area(ΔDEF) = AM 2 /DN 2
Since, ΔABC ~ ΔDEF (Given)
∴ Area(ΔABC)/Area(ΔDEF) = (AB 2 /DE 2 ) …………………………… (i)
and, AB/DE = BC/EF = CA/FD ……………………………………… (ii)
In ΔABM and ΔDEN,
Since ΔABC ~ ΔDEF
AB/DE = BM/EN [Already Proved in equation (i) ]
∴ ΔABC ~ ΔDEF [SAS similarity criterion]
⇒ AB/DE = AM/DN ………………………………………………….. (iii)
∴ ΔABM ~ ΔDEN
As the areas of two similar triangles are proportional to the squares of the corresponding sides.
∴ area(ΔABC)/area(ΔDEF) = AB 2 /DE 2 = AM 2 /DN 2
7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Area(ΔBQC) = ½ Area(ΔAPC)
Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC 2 /BC 2 ) = AC 2 /BC 2
Since, Diagonal = √2 side = √2 BC = AC
⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)
Tick the correct answer and justify:
8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4
Given , ΔABC and ΔBDE are two equilateral triangle. D is the midpoint of BC.
∴ BD = DC = 1/2BC
Let each side of triangle is 2 a .
As, ΔABC ~ ΔBDE
∴ Area(ΔABC)/Area(ΔBDE) = AB 2 /BD 2 = (2 a ) 2 /( a ) 2 = 4 a 2 / a 2 = 4/1 = 4:1
Hence, the correct answer is (C).
9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81
Given, Sides of two similar triangles are in the ratio 4 : 9.
Let ABC and DEF are two similar triangles, such that,
ΔABC ~ ΔDEF
And AB/DE = AC/DF = BC/EF = 4/9
As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
∴ Area(ΔABC)/Area(ΔDEF) = AB 2 /DE 2
∴ Area(ΔABC)/Area(ΔDEF) = (4/9) 2 = 16/81 = 16:81
Hence, the correct answer is (D).
1. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm
(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of the sides of the, we will get 49, 576, and 625.
49 + 576 = 625
(7) 2 + (24) 2 = (25) 2
Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm
(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
Clearly, 9 + 36 ≠ 64
Or, 3 2 + 6 2 ≠ 8 2
Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.
Hence, the given triangle does not satisfies Pythagoras theorem.
(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 50 2 + 80 2 ≠ 100 2
As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle does not satisfies Pythagoras theorem.
Hence, it is not a right triangle.
(iv) Given, sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Thus, 144 +25 = 169
Or, 12 2 + 5 2 = 13 2
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Hence, length of the hypotenuse of this triangle is 13 cm.
2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM 2 = QM × MR.
Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR
We have to prove, PM 2 = QM × MR
In ΔPQM, by Pythagoras theorem
PQ 2 = PM 2 + QM 2
Or, PM 2 = PQ 2 – QM 2 …………………………….. (i)
In ΔPMR, by Pythagoras theorem
PR 2 = PM 2 + MR 2
Or, PM 2 = PR 2 – MR 2 ……………………………………….. (ii)
Adding equation, (i) and (ii) , we get,
2PM 2 = (PQ 2 + PM 2 ) – (QM 2 + MR 2 )
= QR 2 – QM 2 – MR 2 [∴ QR 2 = PQ 2 + PR 2 ]
= (QM + MR) 2 – QM 2 – MR 2
∴ PM 2 = QM × MR
3. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB 2 = BC × BD (ii) AC 2 = BC × DC (iii) AD 2 = BD × CD
(i) In ΔADB and ΔCAB,
∠DAB = ∠ACB (Each 90°)
∠ABD = ∠CBA (Common angles)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB 2 = CB × BD
(ii) Let ∠CAB = x
∠CBA = 180° – 90° – x
∠CBA = 90° – x
Similarly, in ΔCAD
∠CAD = 90° – ∠CBA
= 90° – x
∠CDA = 180° – 90° – (90° – x)
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each 90°)
∴ ΔCBA ~ ΔCAD [AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC 2 = DC × BC
(iii) In ΔDCA and ΔDAB,
∠DCA = ∠DAB (Each 90°)
∠CDA = ∠ADB (common angles)
∴ ΔDCA ~ ΔDAB [AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD 2 = BD × CD
4. ABC is an isosceles triangle right angled at C. Prove that AB 2 = 2AC 2 .
Given, ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (By isosceles triangle property)
AB 2 = AC 2 + BC 2 [By Pythagoras theorem]
= AC 2 + AC 2 [Since, AC = BC]
AB 2 = 2AC 2
5. ABC is an isosceles triangle with AC = BC. If AB 2 = 2AC 2 , prove that ABC is a right triangle.
Given, ΔABC is an isosceles triangle having AC = BC and AB 2 = 2AC 2
AB 2 = AC 2 + AC 2
= AC 2 + BC 2 [Since, AC = BC]
Hence, by Pythagoras theorem ΔABC is right angle triangle.
6. ABC is an equilateral triangle of side 2a. Find each of its altitudes .
Given, ABC is an equilateral triangle of side 2a.
Draw, AD ⊥ BC
In ΔADB and ΔADC,
∠ADB = ∠ADC [Both are 90°]
Therefore, ΔADB ≅ ΔADC by RHS congruence.
Hence, BD = DC [by CPCT]
In right angled ΔADB,
AB 2 = AD 2 + BD 2
(2 a ) 2 = AD 2 + a 2
⇒ AD 2 = 4 a 2 – a 2
⇒ AD 2 = 3 a 2
⇒ AD = √3a
7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.
We have to prove, as per the question,
AB 2 + BC 2 + CD 2 + AD 2 = AC 2 + BD 2
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
AB 2 = AO 2 + BO 2 …………………….. (i) [By Pythagoras theorem]
AD 2 = AO 2 + DO 2 …………………….. (ii)
DC 2 = DO 2 + CO 2 …………………….. (iii)
BC 2 = CO 2 + BO 2 …………………….. (iv)
Adding equations (i) + (ii) + (iii) + (iv) , we get,
AB 2 + AD 2 + DC 2 + BC 2 = 2(AO 2 + BO 2 + DO 2 + CO 2 )
= 4AO 2 + 4BO 2 [Since, AO = CO and BO =DO]
= (2AO) 2 + (2BO) 2 = AC 2 + BD 2
AB 2 + AD 2 + DC 2 + BC 2 = AC 2 + BD 2
8. In Fig. 6.54, O is a point in the interior of a triangle.
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that: (i) OA 2 + OB 2 + OC 2 – OD 2 – OE 2 – OF 2 = AF 2 + BD 2 + CE 2 , (ii) AF 2 + BD 2 + CE 2 = AE 2 + CD 2 + BF 2 .
Given, in ΔABC, O is a point in the interior of a triangle.
And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
Join OA, OB and OC
(i) By Pythagoras theorem in ΔAOF, we have
OA 2 = OF 2 + AF 2
Similarly, in ΔBOD
OB 2 = OD 2 + BD 2
Similarly, in ΔCOE
OC 2 = OE 2 + EC 2
Adding these equations,
OA 2 + OB 2 + OC 2 = OF 2 + AF 2 + OD 2 + BD 2 + OE 2 + EC 2
OA 2 + OB 2 + OC 2 – OD 2 – OE 2 – OF 2 = AF 2 + BD 2 + CE 2 .
(ii) AF 2 + BD 2 + EC 2 = (OA 2 – OE 2 ) + (OC 2 – OD 2 ) + (OB 2 – OF 2 )
∴ AF 2 + BD 2 + CE 2 = AE 2 + CD 2 + BF 2 .
9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Given, a ladder 10 m long reaches a window 8 m above the ground.
Let BA be the wall and AC be the ladder,
Therefore, by Pythagoras theorem,
AC 2 = AB 2 + BC 2
10 2 = 8 2 + BC 2
BC 2 = 100 – 64
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.
10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Given, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.
Let AB be the pole and AC be the wire.
By Pythagoras theorem,
24 2 = 18 2 + BC 2
BC 2 = 576 – 324
BC 2 = 252
BC = 6√7m
Therefore, the distance from the base is 6√7m.
Speed of first aeroplane = 1000 km/hr
Speed of second aeroplane = 1200 km/hr
In right angle ΔAOB, by Pythagoras Theorem,
AB 2 = AO 2 + OB 2
⇒ AB 2 = (1500) 2 + (1800) 2
⇒ AB = √(2250000 + 3240000)
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.
12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Given, Two poles of heights 6 m and 11 m stand on a plane ground.
And distance between the feet of the poles is 12 m.
Let AB and CD be the poles of height 6m and 11m.
Therefore, CP = 11 – 6 = 5m
From the figure, it can be observed that AP = 12m
By Pythagoras theorem for ΔAPC, we get,
AP 2 = PC 2 + AC 2
(12m) 2 + (5m) 2 = (AC) 2
AC 2 = (144+25) m 2 = 169 m 2
Therefore, the distance between their tops is 13 m.
13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE 2 + BD 2 = AB 2 + DE 2 .
Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
By Pythagoras theorem in ΔACE, we get
AC 2 + CE 2 = AE 2 …………………………………………. (i)
In ΔBCD, by Pythagoras theorem, we get
BC 2 + CD 2 = BD 2 ……………………………….. (ii)
From equations (i) and (ii) , we get,
AC 2 + CE 2 + BC 2 + CD 2 = AE 2 + BD 2 ………….. (iii)
In ΔCDE, by Pythagoras theorem, we get
DE 2 = CD 2 + CE 2
In ΔABC, by Pythagoras theorem, we get
AB 2 = AC 2 + CB 2
Putting the above two values in equation (iii) , we get
DE 2 + AB 2 = AE 2 + BD 2 .
14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB 2 = 2AC 2 + BC 2 .
Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;
AD ⊥BC and BD = 3CD
In right angle triangle, ADB and ADC, by Pythagoras theorem,
AB 2 = AD 2 + BD 2 ………………………. (i)
AC 2 = AD 2 + DC 2 …………………………….. (ii)
Subtracting equation (ii) from equation (i) , we get
AB 2 – AC 2 = BD 2 – DC 2
= 9CD 2 – CD 2 [Since, BD = 3CD]
= 8(BC/4) 2 [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB 2 – AC 2 = BC 2 /2
⇒ 2(AB 2 – AC 2 ) = BC 2
⇒ 2AB 2 – 2AC 2 = BC 2
∴ 2AB 2 = 2AC 2 + BC 2 .
15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD 2 = 7AB 2 .
Given, ABC is an equilateral triangle.
And D is a point on side BC such that BD = 1/3BC
Let the side of the equilateral triangle be a , and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given, BD = 1/3BC
DE = BE – BD = a/2 – a/3 = a/6
In ΔADE, by Pythagoras theorem,
AD 2 = AE 2 + DE 2
⇒ 9 AD 2 = 7 AB 2
16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Given, an equilateral triangle say ABC,
Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
In ΔABE, by Pythagoras Theorem, we get
AB 2 = AE 2 + BE 2
4AE 2 = 3a 2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is: (A) 120°
(B) 60° (C) 90°
Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
We can observe that,
AB 2 = 108
AC 2 = 144
And, BC 2 = 36
AB 2 + BC 2 = AC 2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
1. In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/PQ = SR/PR
Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
Given, PS is the angle bisector of ∠QPR. Therefore,
∠QPS = ∠SPR………………………………..(i)
As per the constructed figure,
∠SPR=∠PRT(Since, PS||TR)……………(ii)
∠QPS = ∠QRT(Since, PS||TR) …………..(iii)
From the above equations, we get,
In △QTR, by basic proportionality theorem,
QS/SR = QP/PT
Since, PT=TR
QS/SR = PQ/PR
BD ⊥AC, DM ⊥ BC and DN ⊥ AB
Now from the figure we have,
DN || CB, DM || AB and ∠B = 90 °
Therefore, DMBN is a rectangle.
So, DN = MB and DM = NB
The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.
∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (i)
In ∆CDM, ∠1 + ∠2 + ∠DMC = 180°
⇒ ∠1 + ∠2 = 90° …………………………………….. (ii)
In ∆DMB, ∠3 + ∠DMB + ∠4 = 180°
⇒ ∠3 + ∠4 = 90° …………………………………….. (iii)
From equation (i) and (ii), we get
From equation (i) and (iii), we get
In ∆DCM and ∆BDM,
∠1 = ∠3 (Already Proved)
∠2 = ∠4 (Already Proved)
∴ ∆DCM ∼ ∆BDM (AA similarity criterion)
BM/DM = DM/MC
DN/DM = DM/MC (BM = DN)
⇒ DM 2 = DN × MC
(ii) In right triangle DBN,
∠5 + ∠7 = 90° ……………….. (iv)
In right triangle DAN,
∠6 + ∠8 = 90° ………………… (v)
D is the point in triangle, which is foot of the perpendicular drawn from B to AC.
∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)
From equation (iv) and (vi), we get,
From equation (v) and (vi), we get,
In ∆DNA and ∆BND,
∠6 = ∠7 (Already proved)
∠8 = ∠5 (Already proved)
∴ ∆DNA ∼ ∆BND (AA similarity criterion)
AN/DN = DN/NB
⇒ DN 2 = AN × NB
⇒ DN 2 = AN × DM (Since, NB = DM)
3. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that
AC 2 = AB 2 + BC 2 + 2 BC.BD.
By applying Pythagoras Theorem in ∆ADB, we get,
AB 2 = AD 2 + DB 2 ……………………… (i)
Again, by applying Pythagoras Theorem in ∆ACD, we get,
AC 2 = AD 2 + DC 2
AC 2 = AD 2 + (DB + BC) 2
AC 2 = AD 2 + DB 2 + BC 2 + 2DB × BC
AC 2 = AB 2 + BC 2 + 2DB × BC
4. In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that
AC 2 = AB 2 + BC 2 – 2 BC.BD.
AB 2 = AD 2 + DB 2
We can write it as;
⇒ AD 2 = AB 2 − DB 2 ……………….. (i)
By applying Pythagoras Theorem in ∆ADC, we get,
AD 2 + DC 2 = AC 2
From equation (i),
AB 2 − BD 2 + DC 2 = AC 2
AB 2 − BD 2 + (BC − BD) 2 = AC 2
AC 2 = AB 2 − BD 2 + BC 2 + BD 2 −2BC × BD
AC 2 = AB 2 + BC 2 − 2BC × BD
5. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) AC 2 = AD 2 + BC.DM + 2 (BC/2) 2
(ii) AB 2 = AD 2 – BC.DM + 2 (BC/2) 2
(iii) AC 2 + AB 2 = 2 AD 2 + ½ BC 2
(i) By applying Pythagoras Theorem in ∆AMD, we get,
AM 2 + MD 2 = AD 2 ………………. (i)
Again, by applying Pythagoras Theorem in ∆AMC, we get,
AM 2 + MC 2 = AC 2
AM 2 + (MD + DC) 2 = AC 2
(AM 2 + MD 2 ) + DC 2 + 2MD.DC = AC 2
From equation(i), we get,
AD 2 + DC 2 + 2MD.DC = AC 2
Since, DC=BC/2, thus, we get,
AD 2 + (BC/2) 2 + 2MD.(BC/2) 2 = AC 2
AD 2 + (BC/2) 2 + 2MD × BC = AC 2
(ii) By applying Pythagoras Theorem in ∆ABM, we get;
AB 2 = AM 2 + MB 2
= (AD 2 − DM 2 ) + MB 2
= (AD 2 − DM 2 ) + (BD − MD) 2
= AD 2 − DM 2 + BD 2 + MD 2 − 2BD × MD
= AD 2 + BD 2 − 2BD × MD
= AD 2 + (BC/2) 2 – 2(BC/2) MD
= AD 2 + (BC/2) 2 – BC MD
(iii) By applying Pythagoras Theorem in ∆ABM, we get,
AM 2 + MB 2 = AB 2 ………………….… (i)
By applying Pythagoras Theorem in ∆AMC, we get,
AM 2 + MC 2 = AC 2 …………………..… (ii)
Adding both the equations (i) and (ii), we get,
2AM 2 + MB 2 + MC 2 = AB 2 + AC 2
2AM 2 + (BD − DM) 2 + (MD + DC) 2 = AB 2 + AC 2
2AM 2 +BD 2 + DM 2 − 2BD.DM + MD 2 + DC 2 + 2MD.DC = AB 2 + AC 2
2AM 2 + 2MD 2 + BD 2 + DC 2 + 2MD (− BD + DC) = AB 2 + AC 2
2(AM 2 + MD 2 ) + (BC/2) 2 + (BC/2) 2 + 2MD (-BC/2 + BC/2) 2 = AB 2 + AC 2
2AD 2 + BC 2 /2 = AB 2 + AC 2
6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.
By applying Pythagoras Theorem in ∆DEA, we get,
DE 2 + EA 2 = DA 2 ……………….… (i)
By applying Pythagoras Theorem in ∆DEB, we get,
DE 2 + EB 2 = DB 2
DE 2 + (EA + AB) 2 = DB 2
(DE 2 + EA 2 ) + AB 2 + 2EA × AB = DB 2
DA 2 + AB 2 + 2EA × AB = DB 2 ……………. (ii)
By applying Pythagoras Theorem in ∆ADF, we get,
AD 2 = AF 2 + FD 2
Again, applying Pythagoras theorem in ∆AFC, we get,
AC 2 = AF 2 + FC 2 = AF 2 + (DC − FD) 2
= AF 2 + DC 2 + FD 2 − 2DC × FD
= (AF 2 + FD 2 ) + DC 2 − 2DC × FD AC 2
AC 2 = AD 2 + DC 2 − 2DC × FD ………………… (iii)
Since ABCD is a parallelogram,
AB = CD ………………….…(iv)
And BC = AD ………………. (v)
In ∆DEA and ∆ADF,
∠DEA = ∠AFD (Each 90°)
∠EAD = ∠ADF (EA || DF)
AD = AD (Common Angles)
∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)
⇒ EA = DF ……………… (vi)
Adding equations (i) and (iii), we get,
DA 2 + AB 2 + 2EA × AB + AD 2 + DC 2 − 2DC × FD = DB 2 + AC 2
DA 2 + AB 2 + AD 2 + DC 2 + 2EA × AB − 2DC × FD = DB 2 + AC 2
From equation (iv) and (vi),
BC 2 + AB 2 + AD 2 + DC 2 + 2EA × AB − 2AB × EA = DB 2 + AC 2
AB 2 + BC 2 + CD 2 + DA 2 = AC 2 + BD 2
7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) ∆APC ~ ∆ DPB
(ii) AP . PB = CP . DP
Firstly, let us join CB, in the given figure.
(i) In ∆APC and ∆DPB,
∠APC = ∠DPB (Vertically opposite angles)
∠CAP = ∠BDP (Angles in the same segment for chord CB)
∆APC ∼ ∆DPB (AA similarity criterion)
(ii) In the above, we have proved that ∆APC ∼ ∆DPB
We know that the corresponding sides of similar triangles are proportional.
∴ AP/DP = PC/PB = CA/BD
⇒AP/DP = PC/PB
∴AP. PB = PC. DP
8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:
(i) ∆ PAC ~ ∆ PDB
(ii) PA . PB = PC . PD.
(i) In ∆PAC and ∆PDB,
∠P = ∠P (Common Angles)
As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.
∠PAC = ∠PDB
Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)
(ii) We have already proved above,
∆APC ∼ ∆DPB
AP/DP = PC/PB = CA/BD
AP/DP = PC/PB
∴ AP. PB = PC. DP
9. In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC.
In the given figure, let us extend BA to P such that;
Now join PC.
Given, BD/CD = AB/AC
⇒ BD/CD = AP/AC
By using the converse of basic proportionality theorem, we get,
∠BAD = ∠APC (Corresponding angles) ……………….. (i)
And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)
By the new figure, we have;
⇒ ∠APC = ∠ACP ……………………. (iii)
On comparing equations (i), (ii), and (iii), we get,
∠BAD = ∠APC
Therefore, AD is the bisector of the angle BAC.
10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the
horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.
To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;
AC 2 = AB 2 + BC 2
AB 2 = (1.8 m) 2 + (2.4 m) 2
AB 2 = (3.24 + 5.76) m 2
AB 2 = 9.00 m 2
⟹ AB = √9 m = 3m
Thus, the length of the string out is 3 m.
As its given, she pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m
Let us say now, the fly is at point D after 12 seconds.
Length of string out after 12 seconds is AD.
AD = AC − String pulled by Nazima in 12 seconds
= (3.00 − 0.6) m
In ∆ADB, by Pythagoras Theorem,
AB 2 + BD 2 = AD 2
(1.8 m) 2 + BD 2 = (2.4 m) 2
BD 2 = (5.76 − 3.24) m 2 = 2.52 m 2
BD = 1.587 m
Horizontal distance of fly = BD + 1.2 m
= (1.587 + 1.2) m = 2.787 m
NCERT Solutions Class 10 Maths Chapter 6 , Triangles, is part of the Unit Geometry, which constitutes 15 marks of the total marks of 80. On the basis of the updated CBSE Class 10 Syllabus for 2023-24, this chapter belongs to the Unit-Geometry and has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and problem-solving methods in this chapter is mandatory to score well in the board examination of Class 10 Maths.
6.1 introduction.
From your earlier classes, you are familiar with triangles and many of their properties. In Class 9, you have studied congruence of triangles in detail. In this chapter, we shall study about those figures which have the same shape, but not necessarily the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.
In Class 9, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. The topic explains similarity of figures by performing the relevant activity. Similar figures are two figures having the same shape, but not necessarily the same size.
The topic recalls triangles and its similarities. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). It explains Basic Proportionality Theorem and different theorems are discussed performing various activities.
In the previous section, we stated that two triangles are similar (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). The topic discusses the criteria for similarity of triangles referring to the topics we have studied in earlier classes. It also contains different theorems explained with proper examples.
You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. The topic Areas of Similar Triangles consists of theorem and relatable examples to prove the theorem.
You are already familiar with the Pythagoras Theorem from your earlier classes. You have also seen proof of this theorem in Class 9. Now, we shall prove this theorem using the concept of similarity of triangles. Hence, the theorem is verified through some activities, and you can make use of it while solving certain problems.
The summary contains the points you have studied in the chapter. Going through the points mentioned in the summary will help you to recollect all the important concepts and theorems of the chapter.
Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)
Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)
Exercise 6.3 Solutions 16 Questions (1 main question with 6 sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)
Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)
Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)
Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)
Triangle is one of the most interesting and exciting chapters of the unit Geometry as it takes us through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles, including a detailed explanation of similar figures, different theorems related to the similarities of triangles with proof, and the areas of similar triangles. The chapter concludes by explaining the Pythagoras theorem and the ways to use it in solving problems. Read and learn Chapter 6 of the Class 10 Maths NCERT textbook to learn more about Triangles and the concepts covered in it. Ensure to learn the NCERT Solutions for Class 10 effectively to score high in the board examinations.
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By QB365 on 22 May, 2021
QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams
10th Standard CBSE
Final Semester - June 2015
Case Study Questions
(ii) If m, n and r are the sides of right triangle ABJ, then which of the following can be correct?
(iii) If \(\Delta\) ABJ ~ \(\Delta\) ADH, then which similarity criterion is used here?
(iv) If \(\angle\) ABJ = 90° and B, J are mid points of sides AD and AH respectively and BJ || DH, then which of the following option is false?
(ii) Distance travelled by aeroplane towards west after \(1 \frac{1}{2}\) hr is
(iii) In the given figure, \(\angle\) POQ is
(iv) Distance between aeroplanes after \(1 \frac{1}{2}\) hr is
(v) Area of \(\Delta\) POQ is
(ii) What will be the length of shadow of tower when Meenal's house casts a shadow of 15 m?
(iii) Height of Aruns house is
(iv) If tower casts a shadow of 40 rn, then find the length of shadow of Arun's house
(v) If tower casts a shadow of 40 m, then what will be the length of shadow of Meenal's house?
(ii) The value of x is
(iii) The value of PR is
(iv) The value of RQ is
(v) How much distance will be saved in reaching city Q after the construction of highway?
(ii) In if AB || CD, and DO = 3x - 19, OB = x - 5, OC = x - 3 and AO = 3, then the value of x can be
(iii) In if OD = 3x - 1, OB = 5x - 3, OC = 2x + 1 and AO = 6x - 5, then the value of x is
(iv) In \(\Delta\) ABC, if PQ || BC and AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, then AB + PQ is equal to
(v) In \(\Delta\) DEF, if RS || EF, DR = 4x - 3, DS = 8x - 7, ER = 3x - 1 and FS = 5x - 3, then the value of x is
Cbse 10th standard maths subject triangles case study questions 2021 answer keys.
(i) (a): Speed = 1200 km/hr \(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\) \(\therefore\) Required distance = Speed x Time \(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\) (ii) (c): Speed = 1500 km/hr Time = \(\frac{3}{2}\) hr. \(\therefore\) Required distance = Speed x Time \(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\) (iii) (b): Clearly, directions are always perpendicular to each other. \(\therefore \quad \angle P O Q=90^{\circ}\) (iv) (a): Distance between aeroplanes after \(1\frac{1}{2}\) hour \(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\) (v) (d): Area of \(\Delta\) POQ= \(\frac{1}{2}\) x base x height \(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)
(i) (b) (ii) (c): Using Pythagoras theorem, we have PQ 2 = PR 2 + RQ 2 \(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \) \(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\) \(\Rightarrow x^{2}+12 x-5 x-60=0 \) \(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \) \(\Rightarrow x=5, x=-12\) \(\therefore \quad x=5\) [Since length can't be negative] (iii) (a) : PR = 2x = 2 x 5 = 10 km (iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km (v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km
(i) (c) (ii) (b): Since \(\Delta\) AOB ~ \(\Delta\) COD [ByAA similarity criterion] \(\therefore \frac{A O}{O C}=\frac{B O}{O D} \Rightarrow \frac{3}{x-3}=\frac{x-5}{3 x-19}\) \(\Rightarrow 3(3 x-19)=(x-5)(x-3) \) \(\Rightarrow 9 x-57=x^{2}-3 x-5 x+15 \Rightarrow x^{2}-17 x+72=0 \) \(\Rightarrow(x-8)(x-9)=0 \Rightarrow x=8 \text { or } 9\) (iii) (c) : Since, \(\Delta\) AOB ~ \(\Delta\) COD [ByAA similarity criterion] \(\therefore \frac{A O}{O C}=\frac{B O}{O D} \Rightarrow \frac{6 x-5}{2 x+1}=\frac{5 x-3}{3 x-1}\) \(\Rightarrow(6 x-5)(3 x=1)=(5 x-3)(2 x+1) \) \(\Rightarrow \quad 18 x^{2}-6 x-15 x+5=10 x^{2}+5 x-6 x-3 \) \(\Rightarrow \quad 8 x^{2}-20 x+8=0 \Rightarrow 2 x^{2}-5 x+2=0\) From options, x = 2 is the only value that satisfies this equation. (iv) (d): Since, \(\Delta\) APQ ~ \(\Delta\) ABC [ByAA similarity criterion] \(\therefore \quad \frac{A P^{\circ}}{A B}=\frac{A Q}{A C}=\frac{P Q}{B C} \Rightarrow \frac{2.4}{A B}=\frac{2}{5}=\frac{P Q}{6} \) \(\therefore \quad A B=\frac{2.4 \times 5}{2}=6 \mathrm{~cm} \text { and } P Q=\frac{2 \times 6}{5}=2.4 \mathrm{~cm} \) \(\therefore \quad A B+P Q=6+2.4=8.4 \mathrm{~cm}\) (v) (a): Since, \(\Delta\) DRS ~ \(\Delta\) DEF [ByAA similarity criterion] \(\therefore \quad \frac{D E}{D R}=\frac{D F}{D S} \Rightarrow \frac{D E}{D R}-1=\frac{D F}{D S}-1 \) \(\Rightarrow \frac{D E-D R}{D R}=\frac{D F-D S}{D S} \Rightarrow \frac{E R}{D R}=\frac{F S}{D S} \) \(\Rightarrow \quad \frac{D R}{E R}=\frac{D S}{F S} \Rightarrow \frac{4 x-3}{3 x-1}=\frac{8 x-7}{5 x-3} \) \(\Rightarrow \quad 20 x^{2}-12 x-15 x+9=24 x^{2}-8 x-21 x+7 \) \(\Rightarrow \quad 4 x^{2}-2 x-2=0 \Rightarrow 2 x^{2}-x-1=0\) Only option (a) i.e., x = 1 satisfies this equation.
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Tn state board / cbse, 3000+ q&a's per subject, score high marks.
CBSE Board has introduced the case study questions for the ongoing academic session 2021-22. The board will ask the paper on the basis of a different exam pattern which has been introduced this year where 50% syllabus is occupied for MCQ for Term 1 exam. Selfstudys has provided below the chapter-wise questions for CBSE Class 10 Maths. Students must solve these case study based problems as soon as they are done with their syllabus.
These case studies are in the form of Multiple Choice Questions where students need to answer them as asked in the exam. The MCQs are not that difficult but having a deep and thorough understanding of NCERT Maths textbooks are required to answer these. Furthermore, we have provided the PDF File of CBSE Class 10 maths case study 2021-2022.
In order to score good marks in the term 1 exam students must be aware of the Important formulas, Case Based Questions, MCQ and Assertion Reasons with solutions. Solving these types of questions is important because the board will ask them in the Term 1 exam as per the changed exam pattern of CBSE Class 10th.
Important formulas should be necessarily learned by the students because the case studies are solved with the help of important formulas. Apart from that there are assertion reason based questions that are important too.
Assertion Reasoning is a kind of question in which one statement (Assertion) is given and its reason is given (Explanation of statement). Students need to decide whether both the statement and reason are correct or not. If both are correct then they have to decide whether the given reason supports the statement or not. In such ways, assertion reasoning questions are being solved. However, for doing so and getting rid of confusions while solving. Students are advised to practice these as much as possible.
For doing so we have given the PDF that has a bunch of MCQs questions based on case based, assertion, important formulas, etc. All the Multiple Choice problems are given with detailed explanations.
Recently CBSE Board has the exam pattern and included case study questions to make the final paper a little easier. However, Many students are nervous after hearing about the case based questions. They should not be nervous because case study are easy and given in the board papers to ease the Class 10th board exam papers. However to answer them a thorough understanding of the basic concepts are important. For which students can refer to the NCERT textbook.
Basically, case study are the types of questions which are developed from the given data. In these types of problems, a paragraph or passage is given followed by the 5 questions that are given to answer . These types of problems are generally easy to answer because the data are given in the passage and students have to just analyse and find those data to answer the questions.
These types of questions are solved by reading the statement, and given reason. Sometimes these types of problems can make students confused. To understand the assertion and reason, students need to know that there will be one statement that is known as assertion and another one will be the reason, which is supposed to be the reason for the given statement. However, it is students duty to determine whether the statement and reason are correct or not. If both are correct then it becomes important to check, does reason support the statement?
Moreover, to solve the problem they need to look at the given options and then answer them.
CBSE Class 10 Maths Case Based MCQ are either Multiple Choice Questions or assertion reasons. To solve such types of problems it is ideal to use elimination methods. Doing so will save time and answering the questions will be much easier. Students preparing for the board exams should definitely solve these types of problems on a daily basis.
Also, the CBSE Class 10 Maths MCQ Based Questions are provided to us to download in PDF file format. All are developed as per the latest syllabus of CBSE Class Xth.
Class 10 Mathematics Multiple Choice Questions for all the chapters helps students to quickly revise their learnings, and complete their syllabus multiple times. MCQs are in the form of objective types of questions whose 4 different options are given and one of them is a true answer to that problem. Such types of problems also aid in self assessment.
Case Study Based Questions of class 10th Maths are in the form of passage. In these types of questions the paragraphs are given and students need to find out the given data from the paragraph to answer the questions. The problems are generally in Multiple Choice Questions.
The Best Class 10 Maths Case Study Questions are available on Selfstudys.com. Click here to download for free.
To solve Class 10 Maths Case Studies Questions you need to read the passage and questions very carefully. Once you are done with reading you can begin to solve the questions one by one. While solving the problems you have to look at the data and clues mentioned in the passage.
In Class 10 Mathematics the assertion and reasoning questions are a kind of Multiple Choice Questions where a statement is given and a reason is given for that individual statement. Now, to answer the questions you need to verify the statement (assertion) and reason too. If both are true then the last step is to see whether the given reason support=rts the statement or not.
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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.
Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.
We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.
Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .
First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.
Class 10 Maths has the following chapters.
CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.
Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.
Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .
In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.
In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.
The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.
We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .
As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.
Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.
Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.
Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.
CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.
CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.
You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.
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Important Questions for Class 10 Maths Chapter 6 Triangles - Explore detailed answers and step by step solutions which will help you ace your Class 10 Math examination.
January 18, 2024
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Important Questions for Class 10 Maths Chapter 6: Triangles play a crucial role in the exam preparations for students, aligning with the updated CBSE pattern for the 2023-2024 academic session. As students gear up for the board exams, it is strongly recommended that they dedicate focused practice to the important questions related to Triangles.
Scoring full marks in this chapter is attainable through diligent preparation, and these questions serve as a comprehensive resource for achieving that goal. With the potential inclusion of similar questions in the actual exam, students are urged not to overlook any aspect of their preparation.
CBSE Date Sheet 2024 Revised
Additional support for comprehensive exam readiness is available through important questions covering all chapters in 10th-grade math at BYJU’S.
The Triangles chapter consists of a myriad of topics, including criteria for similarity, congruency, areas of similar triangles, and the Pythagorean theorem. For a thorough understanding and effective exam performance, students can refer to the detailed solutions provided for the important questions in the NCERT solutions for Class 10 Maths.
Not only are answers available, but the chapter also offers extra practice questions without solutions, encouraging students to reinforce their grasp on the Triangles topic. By diligently solving these additional problems, students can solidify their command over the subject and enhance their confidence for the upcoming board exams.
Triangles are essential for students gearing up for their board exams. This chapter covers diverse topics, including the criteria for the similarity of triangles, the Basic Proportionality Theorem (Thales Theorem), proving triangles similar, areas of similar triangles, and the application of the Pythagorean theorem. Carefully created, these important questions offer a strategic approach to exam preparation, ensuring comprehensive coverage of the syllabus and alignment with the CBSE exam pattern.
CBSE Important Questions for Class 10 Maths
By solving these questions, students can strengthen their conceptual clarity, master the application of criteria for similarity, and enhance their problem-solving skills in the context of triangles. Accompanied by detailed solutions, these questions facilitate self-assessment, providing a valuable tool for gauging proficiency and rectifying any misconceptions. Utilizing these important questions systematically in their study routine will contribute to a confident and prepared performance in the Class 10 Maths board exams.
CBSE Class 10 Notes
Physics Wallah teachers are known for providing important questions and valuable resources for Maths subjects. These questions are designed to complement the learning material and aid students in their preparation. Wallah’s approach involves imparting not only theoretical knowledge but also practical problem-solving skills. Students are encouraged to solving with these important questions as part of their study routine to reinforce their understanding, practice application of concepts, and prepare effectively for their physics exams.
Q.1: In the figure, DE // AC and DF // AE. Prove that BF/FE = BE/EC.
Given that,
In triangle ABC, DE // AC.
By Basic Proportionality Theorem,
BD/DA = BE/EC……….(i)
Also, given that DF // AE.
Again by Basic Proportionality Theorem,
BD/DA = BF/FE……….(ii)
From (i) and (ii),
BE/EC = BF/FE
Hence proved.
Q.2: In ∆DEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB. (2015)
Q.3: In the given figure, PS/SQ = PT/TR and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle.
PS/SQ = PT/TR
We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Therefore, ST // QR
And ∠ PST = ∠ PQR (Corresponding angles) ……..(i)
Also, given,
∠ PST = ∠ PRQ………(ii)
∠ PRQ = ∠ PQR
Therefore, PQ = PR (sides opposite the equal angles)
Hence, PQR is an isosceles triangle.
Solution: In ∆ADE and ∆ABC, ∠DAE = ∠BAC …Common ∠ADE – ∠ABC … [Corresponding angles ∆ADE – ∆ΑΒC …[AA corollary
Q.5: In the given figure, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that:
(i) ∆AEP ~ ∆ CDP
(ii) ∆ABD ~ ∆ CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆ PDC ~ ∆ BEC
Given that AD and CE are the altitudes of triangle ABC and these altitudes intersect each other at P.
(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (90° each)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by AA similarity criterion,
ΔAEP ~ ΔCDP
(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB ( 90° each)
∠ABD = ∠CBE (Common Angles)
ΔABD ~ ΔCBE
(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (90° each)
∠PAE = ∠DAB (Common Angles)
ΔAEP ~ ΔADB
(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (90° each)
∠PCD = ∠BCE (Common angles)
ΔPDC ~ ΔBEC
Q.6: The lengths of the diagonals of a rhombus are 24 cm and 32 cm. Calculate the length of the altitude of the rhombus. (2013)
Solution: Diagonals of a rhombus are ⊥ bisectors of each other. ∴ AC ⊥ BD, OA = OC = A C 2 ⇒ 24 2 = 12 cm OB = OD = B D 2 ⇒ 32 2 = 16 cm In rt. ∆BOC,
Q.7: A boy walks 12 m due east and 5 m due south. How far is he from the starting point ?
(a). 31 m (b). 26 m (c). 62 m (d). 13 m
Solution: (b)
Applying Pythagoras’ Theorem
AC 2 =AB 2 +BC 2 ⇒ AC 2 =12 2 +5 2 =144+25=169=13 2 ⇒ AC=13m
Q.8: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Length of the vertical pole = 6 m
Shadow of the pole = 4 m
Let the height of the tower be h m.
Length of the shadow of the tower = 28 m
In ΔABC and ΔDFE,
∠C = ∠E (angle of elevation)
∠B = ∠F = 90°
By AA similarity criterion,
ΔABC ~ ΔDFE
We know that the corresponding sides of two similar triangles are proportional.
AB/DF = BC/EF
h = (6 ×28)/4
Hence, the height of the tower = 42 m.
Q.9: The sides AB and AC and the perimeter P, of ∆ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of ∆DEF. Are the two triangles similar? If yes, find a r ( △ A B C ) a r ( △ D E F ) (2012).
Q.10: If the areas of two similar triangles are equal, prove that they are congruent.
Let ΔABC and ΔPQR be the two similar triangles with equal area.
To prove ΔABC ≅ ΔPQR.
ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC 2 /QR 2 = AB 2 /PQ 2 = AC 2 /PR 2 ⇒ BC 2 /QR 2 = AB 2 /PQ 2 = AC 2 /PR 2 = 1 [Since, ar (ΔABC) = ar (ΔPQR)] ⇒ BC 2 /QR 2 = 1 ⇒ AB 2 /PQ 2 = 1 ⇒ AC 2 /PR 2 = 1 BC = QR AB = PQ AC = PR Therefore, ΔABC ≅ ΔPQR [SSS criterion of congruence]
Q.11: O is any point inside a rectangle ABCD as shown in the figure. Prove that OB 2 + OD 2 = OA 2 + OC 2 .
Through O, draw PQ || BC so that P lies on AB and Q lies on DC.
Therefore, PQ ⊥ AB and PQ ⊥ DC (∠ B = 90° and ∠ C = 90°)
So, ∠ BPQ = 90° and ∠ CQP = 90°
Hence, BPQC and APQD are both rectangles.
By Pythagoras theorem,
OB 2 = BP 2 + OP 2 …..(1)
OD 2 = OQ 2 + DQ 2 …..(2)
OC 2 = OQ 2 + CQ 2 …..(3)
OA 2 = AP 2 + OP 2 …..(4)
Adding (1) and (2),
OB 2 + OD 2 = BP 2 + OP 2 + OQ 2 + DQ 2
= CQ 2 + OP 2 + OQ 2 + AP 2
(since BP = CQ and DQ = AP)
= CQ 2 + OQ 2 + OP 2 + AP 2
= OC 2 + OA 2 [From (3) and (4)]
Hence proved that OB 2 + OD 2 = OA 2 + OC 2 .
Q.12: In the figure, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.
AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm
By basic proportionality theorem,
AD/DB = AE/EC
AD/7.2 = 1.8/5.4
AD = (1.8 × 7.2)/5.4
Therefore, AD = 2.4 cm.
Q.14: The sides of two similar triangles are in the ratio 7 : 10. Find the ratio of areas of these triangles.
The ratio of sides of two similar triangles = 7 : 10
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
The ratio of areas of these triangles = (Ratio of sides of two similar triangles) 2
= (7) 2 : (10) 2
Therefore, the ratio of areas of the given similar triangles is 49 : 100.
Q.15: In an equilateral ΔABC, D is a point on side BC such that BD = (⅓) BC. Prove that 9(AD) 2 = 7(AB) 2 .
Given, ABC is an equilateral triangle.
And D is a point on side BC such that BD = (1/3)BC.
Let a be the side of the equilateral triangle and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given, BD = 1/3BC
DE = BE – BD = a/2 – a/3 = a/6
In ΔADE, by Pythagoras theorem,
AD 2 = AE 2 + DE 2
= [(a√3)/2] 2 + (a/6) 2
= (3a 2 /4) + (a 2 /36)
= (37a 2 + a 2 )/36
= (28a 2 )/36
= (7/9) (AB) 2
Therefore, 9(AD) 2 = 7(AB) 2 .
Engaging with important questions for Class 10 Maths Chapter 6 – Triangles offers a range of benefits for students preparing for their board exams:
Conceptual Reinforcement:
Application of Criteria:
Problem-Solving Skills:
Self-Assessment:
Thorough Revision:
The Side-Side-Side (SSS) Similarity Criterion asserts that if the corresponding sides of any two triangles are in the same ratio, then their corresponding angles will be equal, and the triangles will be considered similar triangles.
The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, then the segments formed on those sides are divided in the same ratio. In other words, if we have a triangle ABC with a line PQ parallel to BC, intersecting AB at P and AC at Q, then the theorem asserts that AP/PB = AQ/QC.
The six main types of triangles in mathematics are:
Scalene Triangle: All three sides and angles are different. Isosceles Triangle: Two sides and two angles are equal. Equilateral Triangle: All three sides and angles are equal (each angle is 60 degrees). Acute Triangle: All three angles are less than 90 degrees. Obtuse Triangle: One angle is greater than 90 degrees. Right Triangle: One angle is precisely equal to 90 degrees.
Congruence in triangles refers to the condition where two triangles have the same shape and size, and all their corresponding angles and sides are equal. When two triangles are congruent, they can be superimposed on each other, and their corresponding parts will coincide perfectly. This concept is denoted by the symbol "≅" and is used to establish the equality of triangles in various geometric proofs and constructions.
The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
Important Questions for Class 10 Maths Chapter 5 Arithmetic Progression
NIOS Class 10 Syllabus 2024, Subject Wise Class 10 Syllabus
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