Class 10 Maths MCQs of Chapter 6- Triangle with Solutions

Class 10 Maths MCQs of Chapter 6- Triangle with Solutions are created for helping class 10 students in their preparation for the CBSE Board exam 2022-23. These solutions of MCQs based on chapter 6 will give an idea of the MCQ questions which are going to be asked in Term 1 CBSE Board exam 2021-22. The practice of these MCQ questions is needed to every student of class 10 for achieving excellent marks. Here you can study all NCERT solutions of maths and science chapter-wise free of cost. If you like our work, you can donate any amount of money mentioned here at the end of the post.

Class 10 Maths MCQs of Chapter 6- Triangle with Solutions

Q1. O is a point on side PQ of a ΔPQR such that PO = QO = RO, then

(a) QO² = PR² + QR² (b) PR² = QR²+PQ² (c) QR² = QO² + RO² (d) PO² +RO²=PR²

Ans. (b) PR² = QR²+PQ²

We are given that PO = RO = OQ

In ΔPOQ

∠QPO =∠PQO= x (angle of opposite equal sides)

In triangle QOP

∠QRO =∠OQR= y (angle of opposite equal sides)

Applying angle sum property of the triangle

x + (x +y) +y = 180°

2x +2y = 180°

2( x + y) = 180°

x + y = 90°

Therefore ΔPQR is a right triangle in which ∠Q is right angle

H² = P² + B²

PR² = QR²+PQ²

See the video-Solutions for class 10 maths exercise 6.1 Triangle

Q2.D and E are the mid points of AB and AC of a triangle ABC ,respectively and BC =6 cm, if DE∥ BC,then the length of DE is

(a) 2.5 cm (b) 3 cm (c) 5 cm (d) 6 cm

Ans. (b) 3 cm

In triangle ABC, DE ∥ BC and D and E are the mid point of AB and AC respectively

Applying mid point theorem

DE = 1/2(BC) = 1/2(6) = 3 m

Class 10 Maths MCQ’s of Chapter 6- Triangle with Solutions

Q3.The diagonals of a rhombus are 16 cm and 12 cm,in length,the side of the rhombus is in length is:

(a) 20 cm (b) 8 cm (c) 10 cm (d) 9 cm

Ans.(c) 10 cm

Let the rhombus is ABCD and diagonal AC and BD bisect each other at O,in which AC = 12 cm and BD = 16 cm

The diagonal of the rhombus bisect each other at right angle, therefore ΔAOB is a right triangle

OA = AC/2 =12/2 = 6 cm and OB = BD/2 = 16/2 = 8 cm

Applying Pythagoras theorem

AB² = OA² +OB² = 6² + 8² = 36 + 64 =100

AB = √100 = 10 cm

See the video-Class 10 Maths NCERT Solurions Exercise 6.2 chapter 6 Triangles

Q4.The diagonals of rhombus are 24 cm and 10 cm ,in length ,the altitude of rhombus is :

(a) 120 cm (b) (120/13) cm (c) (110/13) cm (d) (240/13) cm

Let the rhombus is ABCD,height is h and diagonal AC and BD bisect each other at O,in which AC = 24 cm and BD = 10 cm

The diagonal of the rhombus bisects each other at right angle, therefore ΔAOB is a right triangle

OA = AC/2 =24/2 = 12 cm and OB = BD/2 = 10/2 = 5 cm

Applying pythogorus

Applying pythogorus theorem

AB² = OA² +OB² = 12² + 5² = 144 + 25 =169

AB = √169 = 13 cm

The area of rhombus is = (product of diagonal)/2 = (24×10)/2 = 120 cm²

The area of rhombus = side × corresponding altitude =13 h

13 h =120 ⇒ h = (120/13) cm

h = (120/13) cm

Q5. If in two triangles, ΔABC and ΔDEF , AB/DF = BC/FE = CA/ED ,then

(a) ΔABC ∼ΔDEF (b) ΔABC ∼ΔEDF (c) ΔABC ∼ΔEFD (d) ΔABC ∼ΔDFE

Ans. (d) ΔABC ∼ΔDFE

In ΔABC and ΔDEF, we are given

AB/DF = BC/FE = CA/ED

Opposite angle of AB , ∠C = Opposite angle of DF,∠E

Opposite angle of BC , ∠A = Opposite angle of DF,∠D

Opposite angle of CA , ∠B = Opposite angle of ED,∠F

∴ΔABC ∼ΔDFE

Q6. It is given that ΔABC ∼ ΔDEF and BC/EF = 2/3, then ratio of the areas of ΔDEF and ΔABC is:

(a) 2: 3 (b) 9: 4 (c) 4 : 9 (d) 3 : 2

Ans.(c) 4 : 9

We know ,if ΔABC ∼ ΔDEF then the ratio of their areas is equal to the ratio of the square of their corresponding sides

ar ΔABC /arΔDEF = (BC/EF)² = (2/3)² = 4/9

Hence ar ΔABC : arΔDEF = 4 :9

Q7. In ΔABC, ∠BAC = 90° and AD ⊥ BC,then:

(a) BD.CD = BC² (b) AB.AC = BC² (c) BD.CD = AD² (d)AB.AC = AD²

Ans. (c) AD² = BD.CD

In ΔABC, ∠BAC = 90° and AD ⊥ BC are given

According to the given condition

ΔABC ∼ ΔDBA , ΔABC ∼ΔDAC and ΔDBA ∼DAC

AB/BD =BC/AB =AC/AD, AB/AD =BC/AC =AC/CD and BD/AD =AB/AC =AD/CD

AB ² = BC.BD, AC² = BC.CD and AD² = BD.CD

In the given alternative AD² = BD.CD is the match

Q8. If in triangles ABC and DEF ,AB/DE =BC/FD,then they will be similar ,if

(a) ∠B = ∠E (B) ∠A = ∠D (a) ∠B = ∠D (a) ∠A = ∠F

Ans.(a)∠B = ∠D

In triangles ABC and DEF, we are given

AB/DE = BC/FD

In AB and BC ,the ∠B is common and in DE and FD the ∠D is common

Therefore for SAS criteria of similar triangle,∠B =∠D

Q9. If ΔABC ∼ ΔDEF, the ratio between area of the triangles ABC and DEF is 25 : 36, find the ratio between AB and DE.

(a) 5 : 6 (b) 9: 4 (c) 4 : 9 (d) 3 : 2

Ans.(a) 5 : 6

We know ,if ΔABC ∼ ΔDEF then the ratio of their areas is equal to the ratio of the square of their corresponding sides

ar ΔABC /arΔDEF = (AB/DE)²

25/36 = (AB/DE)²

AB/DE = √(25/36) = 5/6

Class 10 Maths MCQs of Chapter 6- Triangle with Solutions

Q10. The area of equilateral triangle whose side is ‘a’

(a) a²√3/4 (b) a√3/4 (c) a³√3/4 (d) a²√5/4

Ans.(a) a²√3/4

MCQs questions with solutions published here definitely help you since the pattern of term 1 CBSE exam is totally MCQs based, so you are needed to study all the MCQs based on other chapters, the link for them we have mentioned at the starting of the post.

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NCERT Solutions of class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry