Questions & Answers

Question

Answers

(a) 110.50 sq. m and 40.43 m

(b) 115.50 sq. m and 45.43 m

(c) 112.50 sq. m and 42.43 m

(d) 113.50 sq. m and 43.43 m

Answer
Verified

We are given that the length of the diagonal(d) is 15 m.

We will suppose the side of the given square as x,

Now, we will apply the Pythagoras theorem in the lower triangle of square as shown in figure because triangle formed is a right angled triangle as angle made at all four corners by sides of square is $ {{90}^{\circ }} $ , and

We know, Pythagoras theorem states that, for a right angled triangle

$ {{(hypotenuse)}^{2}}={{(base)}^{2}}+{{(height)}^{2}} $

So, we get

$ {{d}^{2}}={{x}^{2}}+{{x}^{2}} $

$ {{15}^{2}}=2{{x}^{2}} $

$ {{x}^{2}}=\dfrac{225}{2} $

\[\begin{align}

& {{x}^{2}}=112.5 \\

& x=\sqrt{112.5} \\

& x=\pm 10.61 \\

\end{align}\]

But we know that length cannot be negative, so

$ x=10.61 $

Hence length of the side of the square is 10.61 m

Now to find the Area(A) of the square we will use the formula,

$ A={{\left( side \right)}^{2}} $

Hence,

$ \begin{align}

& A={{\left( x \right)}^{2}} \\

& A={{\left( 10.61 \right)}^{2}} \\

& A=112.50 \\

\end{align} $

Hence area of the square is 112.50 sq. m

Now to find the Perimeter(P) of the square we will use the formula,

$ P=4\times side $

Hence,

$ \begin{align}

& P=4\times x \\

& P=4\times 10.61 \\

& P=42.43 \\

\end{align} $

Hence perimeter of the square is 42.43 m

This question can also be done by checking the values of option like we can divide perimeter of each option by 4 which will give us the length of side of the square and then we will find the length of diagonal by using Pythagoras theorem and if it matches with given diagonal length then it is right answer as all of the given perimeters in options are different we do not need to check length of diagonal through area given in options.

But we do not recommend you to do it by the method told above as it involves even more calculations just do it the way it is shown in detail.

×

Sorry!, This page is not available for now to bookmark.