Question

A square whose diagonal is 10 cm long has area equal to
(a) \[{\rm{28 c}}{{\rm{m}}^{\rm{2}}}\]
(b) \[{\rm{34 c}}{{\rm{m}}^{\rm{2}}}\]
(c) \[{\rm{5}}0{\rm{ c}}{{\rm{m}}^{\rm{2}}}\]
(d) None of these

Answer 1

Hint: To solve this question, we will first find out the side of the square whose area is to be calculated using pythagoras theorem and then after side is calculated, we will find out the area by the formula for the area of the square which is given as:
\[{\rm{Area of square}} = {\left( {{\rm{side}}} \right)^2}\]

Complete step-by-step answer:
In this question, we must first know what a square is. A square is a regular quadrilateral, which means that it has four equal sides and four equal angles. It can also be divided as a rectangle in which two adjacent sides have equal length. Also the angle between the adjacent sides are \[{\rm{9}}0^\circ \].It is given in the question that the diagonal has length = 10 cm.

In the above figure, we can see that the diagonal of length 10 cm is made by joining the corners B and D. To calculate the area of the square, we must know the length of the sides of the square. To calculate the length of the side, we are going to apply the pythagorean theorem in the right-angled triangle BCD. The pythagoras theorem says that:

\[{{\rm{H}}^2}\, = \,{P^2} + \,\,{B^2}\]

where H is the hypotenuse, in our case it is diagonal BD. P is the perpendicular, in our case it is side BC. B is the base, in our case it is the side CD. Thus applying the formula we get:
\[ \Rightarrow {\left( {BD} \right)^2}\, = \,{\left( {BC} \right)^2}\, + \,{\left( {CD} \right)^2}\]
\[ \Rightarrow {\left( {10} \right)^2}\, = \,{\left( {side} \right)^2}\, + \,{\left( {side} \right)^2}\]
\[ \Rightarrow 50\, = \,{\left( {side} \right)^2}\]
\[ \Rightarrow side\, = \,\sqrt {50} \,\]
\[ \Rightarrow \,side\, = \,5\sqrt 2 \,cm\]
Now, we have got the length of the side, so now we can calculate the area of the square. The area of the square is given by the formula:
 \[{\rm{Area = }}{\left( {{\rm{side}}} \right)^2}\]
\[ \Rightarrow \,{\rm{Area = }}{\left( {5\sqrt 2 cm} \right)^2}\]
\[ \Rightarrow {\rm{Area of square}}\,{\rm{ = 50 c}}{{\rm{m}}^2}\]


Note: The area of square can also be calculated by first calculating the area of the triangle OAB and then multiplying it by 4.The area of the triangle will be
\[{\rm{Area(}}\Delta {\rm{OAB) = }}\dfrac{1}{2} \times OA \times OB \times \sin \theta \]
In our case, \[\theta \, = {90^ \circ }\]and OA = OB = Half the length of diagonal.