Question

The area of square ABCD is \[16c{m^2}\]. Find the area of the square joining the midpoints of the sides.

A) \[8c{m^2}\]
B) \[3c{m^2}\]
C) \[5c{m^2}\]
D) \[9c{m^2}\]

Answer 1

Hint:
The problem is very easy to solve. Here in order to find the area of the square PQRS that is formed by joining the midpoints of the sides of that square ABCD we need the length of the side of the square. So first with the help of the area of ABCD we will find the length of the side of the main square and then by using the Pythagoras theorem we will find the side of inner square PQRS.

Complete step by step solution:
Given that area of square ABCD is \[16c{m^2}\].
Thus taking its square root we will get the length of the side of that square.
Length of side of square ABCD = \[\sqrt {16} = 4cm\]
Now PQRS are the midpoint of sides AB,BC,CD and AD respectively.
So AP=PB=BQ=QC=CR=RD=DS=SA=2cm. Is it?

Now look at the ∆ASP. Angle A is a right angle, also we have two sides of the triangle given. So we can use the Pythagoras theorem to find the length of the third side and that is the side of square PQRS.
Using the Pythagoras theorem for the ∆ASP.
\[S{P^2} = A{P^2} + A{S^2}\]
Taking the square root,
\[ \Rightarrow SP = \sqrt {A{P^2} + A{S^2}} \]
Putting the values
\[ \Rightarrow SP = \sqrt {{2^2} + {2^2}} \]
\[ \Rightarrow SP = \sqrt 8 \]
\[ \Rightarrow SP = 2\sqrt 2 \]

This is the side of inner square PQRS. So the area of the square will be,
\[area = side \times side\]
Putting the value of side,
\[ \Rightarrow 2\sqrt 2 \times 2\sqrt 2 \]
\[ \Rightarrow 2 \times 2 \times 2\]
\[ \Rightarrow 8c{m^2}\]

Thus the option A is correct.

Note:
Students may take the side of outer square as the side of inner square directly because both are the same shapes but this is not the correct way. We need to find the side of the square PQRS with the help of square ABCD.