How much greater, in square inches, is the area of a square with a dia

Question

How much greater, in square inches, is the area of a square with a diagonal of 8 inches than the area of a square with a diagonal of 4 inches?

A. 4
B. 24
C. 32
D. 48
E. 96

hibobotamuss · Answer

Area of a square D^2/2

So, Area of Big square - Area of small square = 32-8 = 24

B

pushpitkc · Answer

Area of a square:  s^2  | Diagonal =  s * \sqrt{2}  -> s =  \frac{Diagonal}{\sqrt{2}}  where s - side of the square

The area of a square with a diagonal of 8 inches is  (\frac{8}{\sqrt{2}})^2 = \frac{64}{2} = 32

Also, the area of a square with a diagonal of 4 inches is  (\frac{4}{\sqrt{2}})^2 = \frac{16}{2} = 8

Therefore, the area of a square(diagonal: 8 inches) is  24(Option B)  square inches greater than a square(diagonal: 4 inches)

KarishmaB · Answer

The area of a square in terms of diagonal is  \frac{Diagonal^2}{2}

Difference in area =  \frac{8^2}{2} - \frac{4^2}{2} = \frac{48}{2} = 24

Answer (B)

Archit3110 · Answer

For square ratio is in 45:45:90
so side would x:x:sqrt 2x
or sqrt 2 x: 8 ; x= 4sqrt 2
and for other square x= 2 sqrt 2

difference in area : (4sqrt 2)^2-(2 sqrt 2)^2= 24 option B

Archit3110 · Answer

side of 8 inches diagonal square would be sqrt 32 and of diagonal 4 inches 2 sqrt2
difference = 24 option B

Kinshook · Answer

Asked: How much greater, in square inches, is the area of a square with a diagonal of 8 inches than the area of a square with a diagonal of 4 inches?

Area of a square with diagonal d =  (\frac{d}{\sqrt{2}})^2 = \frac{d^2}{2}

Difference in areas =  \frac{8^2}{2} - \frac{4^2}{2} = 32 - 8 = 24

IMO B

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