Question

ABCD is a square field of side 80m. If $EF$ is perpendicular to $DC$, find the area of $\Delta EDC$ and the area of the shaded portion respectively.

A) $3200{m^2}, 3200{m^2}$
B) $3200{m^2}, 3500{m^2}$
C) $3600{m^2}, 3200{m^2}$
D) $3200{m^2}, 3600{m^2}$

Answer 1

Hint:
We are given a square field out of which a triangle is distinctively shown and are asked to find the area of the triangle and the area of the shaded region separately. Thus, we will first calculate the area of the triangle. Finally, we will subtract this value from the area of the square.

Formulae Used:
$ar(triangle) = \dfrac{1}{2} \times b \times h$
Where, $b$ is the base length of the triangle and $h$ is the height of the triangle.
$ar(square) = {s^2}$
Where, $s$ is the length of each side of the square.

Complete step by step solution:
Here,
Side length of the square$ABCD$, $s = 80m$
Thus,
$ar(ABCD) = {(80)^2} = 6400{m^2}$
Now,
As $EF \bot DC$
Thus,
$EF$ is the altitude or the height of $\Delta EDC$.
Now,
Clearly $EF = s = 80m$
Again,
$DC = s = 80m$
Thus,
$ar(\Delta EDC) = \dfrac{1}{2} \times 80 \times 80 = 3200{m^2}$
Now,
\[ar\left( {shaded{\text{ }}region} \right) = ar(ABCD) - ar(\Delta EDC)\]
Substituting the values, we get
\[ar\left( {shaded{\text{ }}region} \right) = 6400 - 3200 = 3200{m^2}\]

Hence, the answer to the question is $3200{m^2}, 3200{m^2}$ which is option (A).

Additional Information:
We have separate formulae for calculating the area of different triangles.
For equilateral triangle, we have $Area = \dfrac{{\sqrt 3 }}{4}{s^2}$
Where, $s$ is the side length of the triangle.
Similarly, for other types of triangle we have other formulae.
But, all the different formulae can be derived from the heron’s formula.
The formulation of the heron’s formula is:
$Area = \sqrt {s(s - a)(s - b)(s - c)} $
Where, $s$ is the semi perimeter of the triangle and $a,b,c$ are the side lengths of the triangle.
The semi perimeter as the name suggests is the half of the perimeter of the triangle.
Thus,
$s = \dfrac{{a + b + c}}{2}$

Note:
Students must note that $EF = s$ as it is a line segment between the two opposite edges of the square parallel to a side of a square. If the line segment $EF$ would have been in any other angle to the side $DC$, then the case would have been different and we could not have considered it as the altitude of the triangle.