Question

If the length of a diagonal of a square is a + b, then the area of the square is:
(a) \[{{\left( a+b \right)}^{2}}\]
(b) \[\dfrac{1}{2}{{\left( a+b \right)}^{2}}\]
(c) \[{{a}^{2}}+{{b}^{2}}\]
(d) \[\dfrac{1}{2}\left( {{a}^{2}}+{{b}^{2}} \right)\]

Answer 1

Hint: Assume all the sides of the square equal to ‘x’. Use Pythagoras theorem, in the right angle triangle formed by a diagonal and two sides, given by: - \[{{h}^{2}}={{b}^{2}}+{{p}^{2}}\], where h = hypotenuse, b = base and p = perpendicular. Find the value of ‘\[{{x}^{2}}\]’ in terms of ‘a’ and ‘b’ which will be our answer. Use the information: - Area of square = \[{{\left( side \right)}^{2}}\].

Complete step by step answer:
We have been provided with a square whose length of diagonal is a + b and we have to find the area of the square. So, let us consider the diagram constructed below,

We have constructed a square ABCD with the length of each side to ‘x’ and length of diagonal AC equal to a + b. Now, we know that each angle of a square is \[{{90}^{\circ }}\]. Therefore, \[\Delta ABC\] is a right angle triangle at B.
In \[\Delta ABC\], applying Pythagoras theorem given by: - \[{{h}^{2}}={{b}^{2}}+{{p}^{2}}\], where h = hypotenuse, b = base and p = perpendicular, we have,
AC = Hypotenuse
AB = Base
BC = perpendicular
\[\begin{align}
  & \Rightarrow A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} \\
 & \Rightarrow {{\left( a+b \right)}^{2}}={{x}^{2}}+{{x}^{2}} \\
 & \Rightarrow 2{{x}^{2}}={{\left( a+b \right)}^{2}} \\
\end{align}\]
Dividing both sides with 2, we get,
\[\Rightarrow {{x}^{2}}=\dfrac{1}{2}{{\left( a+b \right)}^{2}}\] - (i)
Now, we know that,
Area of a square = \[{{\left( side \right)}^{2}}\], here side = x.
\[\Rightarrow \] Area = \[{{x}^{2}}\]
Using relation (i), we get,
\[\Rightarrow \] Area = \[{{x}^{2}}\]\[=\dfrac{1}{2}{{\left( a+b \right)}^{2}}\]

So, the correct answer is “Option B”.

Note: One may note that we have considered AB as base and BC as perpendicular in the right angle triangle ABC. You may consider BC as base and AB as perpendicular. It will not bother the answer. There can be another method to solve this question. We can assign some values to ‘a’ and ‘b’ and then find the area and check from the options by substituting the assumed values of ‘a’ and ‘b’ in the options. But this process can only be applied if options are present as above.