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

We bring here the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IVB). This part is a continuation of ‘Triangles: NCERT Exemplar Problems and Solutions (Part-IVB)’ and contains solutions to Q. No. 7-12 from Exercise 6.4 that consists only of the Long Answer Type Questions. 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 DPQR, 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, ÐA + ÐD = 90°. Prove that\

AC2 + BD2 = AD2 + BC2.

Solution.

Given: In quadrilateral ABCD, ÐA + ÐD = 90°

To prove: AC2 + BD2 = AD2 + BC2

Construct: Produce AB and CD to meet at E.

Also, join AC and BD.  CBSE Class 10 NCERT Textbooks & NCERT Solutions

NCERT Solutions for CBSE Class 10 Maths

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