Here you get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IVB) is available here. This part of the chapter includes solutions for question number 1-8from Exercise 6.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Triangles. This exercise comprises only the Long 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-IVA)**

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 6.4**

**Long Answer Type Questions (Q. NO. 7-12):**

**Question. 7 **A** **flag pole 18m high casts a shadow 9.6m long. Find the distance of the top of the pole from the far end of the shadow.

**Solution. **

Let *BC* = 18m be the flag pole and *AB *be its shadow which is 9.6m long.

**Question.**** 8 **A street light bulb is fixed on a pole 6m above the level of the street. If a woman of height 1.5m casts a shadow of 3m, then find how far she is away from the base of the pole.

**Solution.**

Let the bulb be fixed at top *A* of pole* AB* and *CD *= 1.5m be the height of a woman and *ED = *3m be the length of her shadow.

Hence, the required distance between base of pole and woman is 9m.

**Question.**** 9 **In given figure, *ABC *is a triangle right angled at *B *and *BD *⏊* AC. *If *AD *= 4cm and *CD = *5cm, then find *BD *and *AB*.

**Question. ****10** In given figure *PQR *is a right triangle, right angled at *Q *and *QS *^* PR. *If *PQ *= 6cm and *PS* = 4cm, then find *QS, RS *and *QR*.

**Question.**** 11** In D*PQR, PD *^*QR *such that *D *Lies on *QR*, if *PQ = a, PR = b, QD = c *and *DR* = *d, *then prove that *(a + b)(a *− *b) = (c + d) *(c − *d).*

**Solution. **

Hence proved.

**Question.**** 12** In a quadrilateral *ABCD*, Ð*A *+ Ð*D = *90°. Prove that\

** ***AC*^{2}* + BD*^{2}* = AD*^{2}* + BC*^{2}*.*

**Solution.**

**Given: **In quadrilateral *ABCD, *Ð*A + *ÐD = 90°

**To prove: ***AC*^{2}* *+ *BD*^{2}* *= *AD*^{2} *+ BC*^{2}

**Construct: **Produce *AB *and *CD *to meet at *E. *

Also, join *AC *and *BD.*

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