Question

the diagonal of a square A is (a+b) units.Find the diagonal of a square B whose area is twice the area of A.

Answer 1

Okay here is your answer

The diagonal of Square A = a+b
Hence All angles of the square are of 90°

Hence we can apply the Pythagoras Theorem

(a + b)² = x² + x²
All sides of the Square are equal and hence let's take it as x

( a+ b)² = 2x²
x² = ( a + b )²/2

Hence we get the area of Square A , because as side is x hence the area will be side ² = x²

Now it is given that the area of Square B is twice the Area of Square A

Hence Square B Area = (a+b)²/2 + ( a+b)²/2
(a+b)² + (a+b)². /2
2( a+b)² / 2
( a+ b)²

Hence the area of Square B is (a+b)²

So the side of Square B will be , we can find it through area of Square

Area of Square= side²
(a+b)² = side²
Side = ( a+b)

Hence now we have found out the the side of Square B now let's find the diagonal with the help of Pythagoras Theorem as All angles are Right angles

According to Pythagoras Theorem

Diagonal ² = y² + y²
All sides of the square are equal hence let's take the sides of Square B as y , y = ( a+b)

Diagonal ² = 2y²
Diagonal ² = 2( a+b)²
Diagonal = √2 ( a+b)

Hence the diagonal of Square B is √2 ( a+b)

Hope it helps

Answer 2

Answer:

Area of the square A = \frac{\left ( diagonal \right )^{2}}{2}= \frac{\left ( a+b \right )^{2}}{2}

Area of the new square

= \frac{\left ( a+b \right )^{2}}{2}\times 2 = \left ( a+b \right )^{2}

=> side = (a+b)

\therefore Diagonal = \sqrt{2}\times side

= \sqrt{2}(a+b)

Hence option [B] is the right answer

Step-by-step explanation: