CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IIIB) is available here. This part of the chapter includes solutions for question number 9-15 from Exercise 6.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Triangles. This exercise comprises only the Short Answer Type Questions 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-IIIA)**

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.

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

**Exercise 6.3**

**Short Answer Type Questions (Q. NO. 9-15):**

**Question. 9 ***ABCD *is a trapezium in which *AB *||*DC *and *P,Q *are points on *AD *and *BC*

respectively, such that* PQ *||* DC, *if *PD = *18cm, *BQ *= 35cm and QC = 15cm, find *AD*.

**Solution.**

**Question. 10 **Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of the smaller triangle is 48cm^{2}, then find the area of the larger triangle.

**Solution.**

Let the two triangles be *ABC *and *PQR*, where Δ*ABC* is smaller of the two.

Given, Δ*ABC~ *Δ*DEF*

AB : DE = 2:3

Area of smaller Δ*ABC* = 48cm^{2}

Since Δ*ABC~ *Δ*DEF*

Using the property, “The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides,” we have:

**Question. 11 **In a Δ*PQR, N *is a point on *PR, *such that *QN* ^ *PR. *If *PN.NR = QN*^{2}*, *then prove that ∠*PQR = *90°.

**Solution.**

**Question. 12 **Areas of two similar triangles are 36cm^{2} and 100cm^{2}. If the length of a side of the larger triangle is 20cm. Find the length of the corresponding side of the smaller triangle.

**Solution.**

Let the two triangles be, Δ*ABC* and Δ*DEF *where Δ*ABC* is the smaller of the two and

Area of Δ*ABC* = 36cm^{2}

And, Area Δ*DEF* = 100cm^{2}

Also, DE = 20cm

Let length of the corresponding side of Δ*ABC*, AB = *x *cm

As Δ*ABC* ~ Δ*DEF*

Therefore, using property, “The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides,” we have:

Hence, the length of corresponding side of the smaller triangle is 12cm.

**Question. 13 **In given figure, if Ð*ACB* =* *Ð*CDA, AC *= 8cm and *AD* = 3cm, then find *BD.*

**Question. 14 **A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

**Solution.**

Let BC = 15m be the tower and its shadow *AB *is 24m. Again, let *EF *= *h *be telephone pole and *DE = *16m be its shadow.

**Question. 15 **Foot of a 10m long ladder leaning against a vertical wall is 6m away from the base of the wall Find the height of the point on the wall where the top of the Ladder reaches.

**Solution.**

Let *BC* be the vertical wall and *AC* = 10m is a ladder.

Distance of ladder from the base of the wall *BC* is 6m

Hence, the height of the point on the wall where the top of the ladder reaches is 8cm.

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

**CBSE Class 10 NCERT Textbooks & NCERT Solutions**

**NCERT Solutions for CBSE Class 10 Maths**

**NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters**

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