Question

In the figure, if the triangle$\vartriangle ABC$ is an isosceles triangle of perimeter 20, calculate the approximate area of the circle with center O.

A) 26.83
B) 33.65
C) 44.17
D) 57.33

Answer 1

Hint: The radius of the circle is needed to find the area of the circle. In an isosceles triangle, the length of the two sides is equal.
A circle is a collection of all the points from a fixed point, at a fixed distance. The fixed point is called the center. The fixed distance from the center is called the radius $r$.
The relation between diameter and radius of the circle: diameter is twice of radius
Pythagoras theorem can be applied to every right angle triangle. The angle between two adjacent sides of a rectangle is $90^\circ $.
In a right-angled triangle, the base and perpendicular are at the angle of $90^\circ $ each other, and hypotenuse is the longest side.

Complete step by step solution:
Step 1: Draw the given figure

Step 2: given that
$\vartriangle ABC$is an isosceles triangle
Hence, BC = AC …… (1)
AB = 4 cm …… (2)
The perimeter of the triangle$\vartriangle ABC$= 20
Step 3: use the perimeter of the triangle to find length AC:
The perimeter of a triangle is equal to the sum of the length of all three sides of the triangle.
In $\vartriangle ABC$, perimeter is equal to AB + BC + AC = 20
 4 + AC + AC = 20 (from (1) and (2))
 $
 2AC = 20 - 4 \\
 {\text{ }}AC = \dfrac{{16}}{2} \\
 $
 AC = 8 cm …… (3)
Step 4: use the below theorem to find $\angle ACD$
Theorem 1: The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

In the above circle, $\angle AOB = 2(\angle ACB)$

In the given circle with center O,
Chord AD is a diameter, as it is passing through the center.
$\angle AOD = 180^\circ $ ($\because $AOD is a straight line)
According to theorem 1:
$\angle AOD = 2(\angle ACD)$
$
 \Rightarrow \angle ACD = \dfrac{{\angle AOD}}{2} \\
 {\text{ = }}\dfrac{{180^\circ }}{2} \\
 {\text{ = 90}}^\circ \\
 $
 Step 5: find the length AD using Pythagoras theorem
Pythagoras theorem: square of the hypotenuse is equal to the sum of the square of base and square of perpendicular.

 $\mathop {{\text{Hypotenuse}}}\nolimits^{\text{2}} {\text{ = }}\mathop {{\text{ Base}}}\nolimits^{{\text{2 }}} {\text{ + }}\mathop {{\text{ Perpendicular}}}\nolimits^{\text{2}} $


$\vartriangle ACD$ is a right angle triangle, right-angled at C.
Using Pythagoras theorem on $\vartriangle ACD$
 $\mathop {{\text{Hypotenuse}}}\nolimits^{\text{2}} {\text{ = }}\mathop {{\text{ Base}}}\nolimits^{{\text{2 }}} {\text{ + }}\mathop {{\text{ Perpendicular}}}\nolimits^{\text{2}} $
 $\mathop {{\text{AC}}}\nolimits^2 + \mathop {{\text{CD}}}\nolimits^{\text{2}} {\text{ }} = {\text{ }}\mathop {{\text{AD}}}\nolimits^{\text{2}} $
 $\mathop { \Rightarrow {\text{ 8}}}\nolimits^{\text{2}} + \mathop 3\nolimits^{\text{2}} {\text{ }} = {\text{ }}\mathop {{\text{AD}}}\nolimits^2 $ (from (3) and given)
 $ \Rightarrow 64 + 9{\text{ }} = {\text{ }}\mathop {{\text{AD}}}\nolimits^2 $
 \[ \Rightarrow \mathop {{\text{ AD}}}\nolimits^{\text{2}} = 73\]
 \[ \Rightarrow {\text{AD = }}\sqrt {73} \]cm
${\text{AD }} \approx {\text{ 8}}{\text{.54 cm}}$ …… (4)
Step 7: find OD
We know, the diameter is twice of radius
$
 {\text{ }}d = 2r \\
 {\text{ AD = 2}} \cdot {\text{OD}} \\
 {\text{ 8}}{\text{.54 = 2}} \cdot {\text{OD}} \\
 {\text{ OD = }}\dfrac{{8.54}}{2} \\
 $
Radius, r, OD = 4.27 cm
Step 8: Find the required area of the circle.
Area of circle $ = \pi \mathop r\nolimits^2 $
 $
 = 3.14 \times \mathop {4.27}\nolimits^2 \\
 = 57.25 \\
 $
$ \approx 57.33$sq. units

The area of the given circle is 57.33 sq. units. The correct option is (D).

Note:
The perimeter of the circle is $2\pi r$units.
The angle in a semicircle is a right angle.

Consider point P at on the semicircle.
Line AB is the diameter of the semicircle with center O
Hence, $\angle AOB = 180^\circ $
Thus, using the theorem 1:
$
 \angle APB = \dfrac{1}{2}\angle AOB \\
 \Rightarrow {\text{ }} = \dfrac{1}{2}\left( {180^\circ } \right) \\
 $
Thus, $\angle APB = {90^ \circ }$
Do not expect the question will only be solved on one concept. Focus on all the given information, the approach of the solution lies within.