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.

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.