Question

THE DIAGONAL OF A SQUARE 'A' IS (a+b). FIND THE DIAGONAL OF SQUARE 'B' IF AREA OF SQUARE 'B' IS HALF OF THE AREA OF SQUARE 'A'.

Answer 1

Required length of diagonal is (a+b) /root 2

Answer 2

AnswEr:

GivEn:

• Diagonal of 'A' (square) - ( a+b )
• Area of B = Area of A/2

To Find:

• Area of the square 'B'
• Side of the square 'B'
• Diagonal of the square 'B'

Formula Used:

• $\bullet{\boxed{\sf{ Diagonal_{(square)} = a \sqrt{2} }}}$

• $\bullet{\boxed{\sf{ Area_{(square)} = a^2 }}}$

Solution:

Let the side of square 'A' is x.

then,

Diagonal of square 'A' = x√2 = a + b

$\implies{\sf{ x = \dfrac{a+b}{ \sqrt{2}} }}$

Area of Square A =

${\sf{ \left ( \dfrac{a+b}{ \sqrt{2}} \left ) ^2 }} \\ \\ \implies{\sf{ \dfrac{ (a+b)^2}{2} }}$

So,

2nd Situation:

Area of square 'B' =

$\implies{\sf{ \dfrac{1}{2} \times (area\:of\: square\:A) }} \\ \\ \implies {\sf{ \dfrac{1}{4} (a+b)^2 }}$

Now,

we can find the side of the square 'B';

$\implies\huge{\sf{ \sqrt{ \dfrac{1}{4} (a+b)^2 } }} \\ \\ \implies{\sf{ \dfrac{a+b}{2} }}$

Again,

Diagonal of square 'B' =

$\implies{\sf{ \dfrac{a+b}{2} . \sqrt{2} }} \\ \\ \implies{\sf{ \sqrt{2} \dfrac{(a+b)}{2} }}$