Here you get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IB). This part of the chapter includes solutions of Question Number 7 to 12 from Exercise 6.1 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Triangles. This exercise comprises only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed solution.

## NCERT Exemplar Solution for CBSE Class 10 Mathematics: Triangles (Part-IA)

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**CBSE Class 10 Mathematics Syllabus 2017-2018**

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Triangles:**

**Exercise 6.1**

**Multiple Choice Questions (Q. No. 7-12)**

**Question. 7** In Δ*ABC *and Δ*DEF, *∠*B = **∠E,* ∠F = *∠C* and *AB* = 3*DE* then, the two triangles are

(a) congruent but not similar

(b) similar but not** **congruent

(c) neither congruent nor similar

(d) congruent as well as similar

**Answer. (b) **

**Explanation: **

Here, two triangles D*ABC* and D*DEF* do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.

**Answer. (a)*** *

**Explanation: **

**Question. 9** If Δ*ABC* ~ Δ*DFE, ∠**A = *30°, ∠*C *= 50°, *AB* = 5cm, *AC *= 8cm and *DF *= 7.5cm. Then, which of the following is true?

(a) *DE *=12cm, ∠*F* =50°

(b) *DE* =12cm, ∠*F* =100°

(*c*) *EF *=12cm, ∠*D* =100°

(d) *EF *=12cm, ∠*D *=30°

**Answer. (b) **

**Explanation: **

**Answer. (c) **

**Explanation: **

And, angle formed by DE and FD is ∠D.

So, ∠B = ∠D

⟹ DABC ~ DEDF (By SAS criterion)

**Answer. (a) **

**Explanation: **

As the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.

Therefore, we get:

**Question. 12 **If *S* is a point on side *PQ* of a Δ*PQR *such that *PS *= *QS* = *RS*, then

(a) *PR .* QR = *RS*^{2}

(b) *QS*^{2} + *RS*^{2} *=QR*^{2}

(c) *PP*^{2} + *QP*^{2 }= *PQ*^{2}

(d) *PS*^{2} + *RS*^{2} = *PR*^{2}

**Answer. (c) **

**Explanation: **

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