Question

The ratio of the areas of two squares, one having its diagonal double than the other, is

(a) 1 : 2
(b) 2 : 3
(c) 3 : 1
(d) 4 : 1

Answer 1

(d) 4 : 1

Let the two squares be ABCD and PQRS. Further, the diagonal of square PQRS is twice the diagonal of square ABCD.

PR = 2 AC

Now, area of the square = $\frac{{\left(Diagonal\right)}^2}{2}$

Area of PQRS = $\frac{{\left(PR\right)}^2}{2}$

Similarly, area of ABCD = $\frac{{\left(AC\right)}^2}{2}$

According to the question:

If AC = x units, then, PR = 2x units

Therefore, $\frac{AreaofPQRS}{AreaofABCD}$ = $\frac{{\left(PR\right)}^2\times 2}{2\times {\left(AC\right)}^2}$ = $\frac{{\left(2x\right)}^2\times 2}{2\times {\left(x\right)}^2}$ = $\frac{4}{1}$

Thus, the ratio of the areas of squares PQRS and ABCD = 4 : 1