Question

The diagonal AC of the given square is 4 cm. Then, its perimeter and area will be equal to ______ and _______ respectively.

• A. $8\sqrt{2}cm,4c{m}^2$
• B. $4\sqrt{2}cm,8c{m}^2$
• C. $8\sqrt{2}cm,8c{m}^2$
• D. $4\sqrt{2}cm,4c{m}^2$

Answer 1

The correct option is C $8\sqrt{2}cm,8c{m}^2$

Angles of the triangle ABC are ${45}^{\circ },{45}^{\circ },{90}^{\circ }$

If we take the sine of all these angles, the ratio of these angles will be
$\Rightarrow \sin \left({45}^{\circ }\right):\sin \left({45}^{\circ }\right):\sin \left({90}^{\circ }\right)$

We know that the value sin ${45}^{\circ }=\frac{1}{\sqrt{2}}$ and sin ${90}^{\circ }=1$.
So, the ratio will be :
$\Rightarrow \frac{1}{\sqrt{2}}:\frac{1}{\sqrt{2}}:1$

On rationalising, we get
$\Rightarrow 1:1:\sqrt{2}$

So, the sides AB, BC & AC will be in the ratio $1:1:\sqrt{2}$.

Hence, $AB:BC:AC\equiv 1:1:\sqrt{2}$

Let, $AB=x,BC=x$ and $AC=\sqrt{2}x$

Since AC = 4, So $\sqrt{2}x=4$

$\Rightarrow x=2\sqrt{2}$

So, $AB=2\sqrt{2}cm,BC=2\sqrt{2}cm$ and $AC=4cm$

Area of the square : $Area=side\times side=AB\times BC=2\sqrt{2}\times 2\sqrt{2}=8c{m}^2$

Perimeter of the square: $Perimeter=4\left(AB\right)=4\times 2\sqrt{2}=8\sqrt{2}cm$