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Question 58
The diagonal of a square is 14 cm. What will be the length of the diagonal of the square whose area is double of the area of first square?
Solution
Diagonal of first square = 14 cm
$$\sqrt{2}a = 14 => a = \dfrac{14}{\sqrt{2}}$$ where a = side of first square
Area of first square $$= (Â \dfrac{14}{\sqrt{2}})^2 = \dfrac{196}{2} = 98 cm^2$$
Area of second square $$= 2\times98 = 196 cm^2$$
Side of second square $$ = \sqrt{196} = 14 cm$$
Diagonal of second square $$=14\times\sqrt{2}$$ = $$14\sqrt{2} cm$$
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