Question

In the given figure, \[\vartriangle ABC\] is an equilateral triangle the length of whose side is equal to \[10\] cm, and \[\vartriangle DBC\] is right-angled at D and BD \[ = 8\] cm. Find the area of the shaded region. [Take \[\sqrt 3 = 1.732\] ]

Answer 1

Hint:: To find the area of the shaded region we have to subtract the area of the unshaded region from the total area. First we have to find the area of triangle ABC and then find the area of triangle BDC. After this, subtract the area of triangle BDC from the area of triangle ABC and you will get the area of the unshaded region.

Complete step by step solution:
The triangle ABC given to us is equilateral triangle whose all three sides are equal that is
AB \[ = \] BC \[ = \] AC \[ = 10cm\]
And the triangle BDC is a right-angled triangle because one angle of this triangle is \[90^\circ \] that is \[\angle D = 90^\circ \] also in triangle BDC, BD \[ = 8cm\] and BC \[ = 10cm\]
According to the question, we have to find the area of the shaded region as shown in the above figure. And to find the area of the shaded region we have to subtract the area of the unshaded region from the total area. That is
Area of shaded region \[ = \] Area of \[\vartriangle ABC\] \[ - \] Area of \[\vartriangle BDC\] ------------ (i)
As we know that the area of an equilateral triangle \[ = \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}\]
Therefore, the area of \[\vartriangle ABC\] \[ = \dfrac{{\sqrt 3 }}{4} \times {\left( {10} \right)^2}\]
\[ \Rightarrow \dfrac{{\sqrt 3 }}{4} \times 100\]
On solving it further we have
\[ \Rightarrow \sqrt 3 \times 25\]
As it is given to us that \[\sqrt 3 = 1.732\] . Therefore,
\[ \Rightarrow 25 \times 1.732\]
That is
\[ \Rightarrow 43.3\]
Thus the required area of the triangle ABC is \[43.3{\text{ }}c{m^2}\]
Also the formula to find the area of a right angled triangle \[ = \dfrac{1}{2}{\text{ }} \times {\text{ }}base{\text{ }} \times {\text{ }}height\] ------- (ii)
The side opposite to the right angle is called the hypotenuse and the other two sides adjacent to the right angle are called base and perpendicular. According to the Pythagoras theorem, in a right triangle,
Hypotenuse \[^2\] \[ = \] Base \[^2\] \[ + \] Height \[^2\]
\[ \Rightarrow \] BC \[^2\] \[ = \] DB \[^2\] \[ + \] Height \[^2\]
\[ \Rightarrow {\left( {10} \right)^2} = {\left( 8 \right)^2} + {\left( {height} \right)^2}\]
On opening the squaring we get
\[ \Rightarrow 100 = 64 + {\left( {height} \right)^2}\]
Now shift \[64\] to the left hand side
\[ \Rightarrow 100 - 64 = {\left( {height} \right)^2}\]
By doing subtraction on the left hand side we get
\[ \Rightarrow 36 = {\left( {height} \right)^2}\]
That is
Height \[ = 6{\text{ }}cm\]
By putting this value and the given values in the equation (ii) we have
area of a right angled triangle BDC \[ = \dfrac{1}{2}{\text{ }} \times {\text{ 8 }} \times {\text{ 6}}\]
\[ \Rightarrow \dfrac{1}{2}{\text{ }} \times {\text{ 48}}\]
By dividing the numerator with denominator we get
\[ \Rightarrow 24\]
Therefore the required areas of triangles ABC and BDC are \[43.3{\text{ }}c{m^2}\] and \[24{\text{ }}c{m^2}\] . By substituting these values of areas in equation (i) we get
Hence, area of shaded region \[ = {\text{ }}43.3 - 24\]
\[ \Rightarrow 19.3{\text{ }}c{m^2}\]

Note:
Remember all the formulas of different types of triangles. Keep in mind that Pythagoras Theorem is used in right-angled triangles only. In this question we used this theorem because the height of the triangle is not given to us but the other two sides are given. Remember that the hypotenuse is greater than any one of the other sides, but less than their sum.