If the length of the diagonal of a square is (a + b), then the area of the square will be:

$(a + b)^{2}$

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$\frac{1}{2} (a + b)^{2}$

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$(a^{2} + b^{2})$

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$\frac{1}{2} (a^{2} + b^{2})$

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Solution

The correct option is B
$\frac{1}{2} (a + b)^{2}$

Diagonal = $\sqrt{2}$ side

or side $= \frac{a + b}{\sqrt{2}}$

Area = side $^{2} = {(\frac{a + b}{\sqrt{2}})}^{2} = \frac{(a + b)^{2}}{2}$ .

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If the length of the diagonal of a square is (a + b), then the area of the square will be:

The correct option is B 12(a+b)2 Diagonal = √2 side or side =a+b√2 Area = side2=(a+b√2)2=(a+b)22 .

The correct option is B
$\frac{1}{2} (a + b)^{2}$

Diagonal = $\sqrt{2}$ side

or side $= \frac{a + b}{\sqrt{2}}$

Area = side $^{2} = {(\frac{a + b}{\sqrt{2}})}^{2} = \frac{(a + b)^{2}}{2}$ .