Question

Which figure is formed by joining the mid points of the adjacent sides of a square?

• A. Rhombus
• B. Square
• C. Reactangle
• D. Parallelogram

Answer 1

The correct option is B Square

Let $E,F,G,H$ are mid points of the sides $AB,BC,CD,DA$ repectively.

In $△AEH,$

$\angle A={90}^{\circ }$ [$ABCD$ is a square.]

So, by using Pythagoras theorem,

$\begin{array}{cc}H{E}^2& =A{E}^2+A{H}^2\\ b^2& ={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2\\ & =2\times \frac{a^2}{4}\\ & =\frac{a^2}{2}\\ b& =\frac{a}{\sqrt{2}}\end{array}$

Now, in $△EHF,$

$\begin{array}{cc}E{H}^2+E{F}^2& =2\cdot b^2\\ & =2\cdot {\left(\frac{a}{\sqrt{2}}\right)}^2\\ & =2\times \frac{a^2}{2}\\ & =a^2\\ & =H{F}^2\end{array}$

So, by converse of Pythagoras theorem $△EHF$ is a right-angled triangle with right angle at $E.$ Similarly we can prove that there is a right angle at each vertex of quadrilateral $EFGH$ and each side is equal to $\frac{a}{\sqrt{2}}.$ Hence, quadrilateral $EFGH$ is a square.