Question

square and a rhombus are of the same side a centimetre if area of square is twice the area of rhombus then find the sum of diagonals of rhombus

Answer 1

Given: Side of square=Side of rhombus= a cm

  →     Area of Square = 2 × Area of rhombus

→   (Side )² = 1/2 × ( Product of diagonals)

→   a²= 1/2 × $D_{1}\times D_{2}$

→ 2 a²= $D_{1}\times D_{2}$

→ 4 a² = 2 $D_{1}\times D_{2}$

Now , $[D_{1}+ D_{2}]^{2} =[D_{1}] ^{2}+ [D_{2}]^{2}+2 \times D_{1}\times D_{2}$

but, $[\frac{D_{1}}{2}] ^{2}+ [\frac{D_{2}}{2}]^{2}=a^{2},$, where a is the side of rhombus as diagonals of rhombus bisect each other at right angles.

$[D_{1}] ^{2}+ [D_{2}]^{2}= 4a^{2}$

Substituting the values in above equation,

$[D_{1}+ D_{2}]^{2}=4 a^{2} + 4 a^{2}$

$[D_{1}+ D_{2}]^{2}=8a^{2}$

$D_{1}+ D_{2} =2\sqrt {2}a$ cm

Answer 2

Answer:

Credit @CarlynBronk

Step-by-step explanation: