NCERT Solutions 10th Maths Chapter 6 Triangles Exercise 6.4

NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 we will restart our exploration of the world of Triangles. Thus, we are providing you Chapter 6 Triangles NCERT Class 10 Maths Solutions that will help in achieving more marks. You don't have to wander and waste your precious time in finding best CBSE NCERT Solutions.

Exercise 6.4
1. 1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Answer
It is given that,
Area of △ABC = 64 cm2
Area of △DEF = 121 cm2
EF = 15.4 cm
and, △ABC ~ △DEF
∴ Area of △ABC/Area of △DEF = AB2/DE2
= AC2/DF2 = BC2/EF2 ...(i)
[If two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides]
∴ 64/121 = BC2/EF2
⇒ (8/11)2 = (BC/15.4)2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm
2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Answer

ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.
In △AOB and ?COD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ △AOB ~ △COD [By AAA similarity criterion]
Now, Area of (△AOB)/Area of (△COD)
= AB2/CD2 [If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]
= (2CD)2/CD2 [? AB = CD]
∴ Area of (△AOB)/Area of (△COD)
= 4CD2/CD = 4/1
Hence, the required ratio of the area of △AOB and △COD = 4:1