Areas Related to Circles
Length of Arc and Area of Sector of Circle
There is a design on a square window glass as shown below.
The shaded portion is coloured yellow. They are the segments of a circle whose radius is 7 cm and the angle of the sector is 90°. Rest all the portion is transparent. Can we calculate the area of the shaded portion?
To calculate the area of the shaded portion, we should know how to calculate the area of segments of the circle.
Firstly, let us know the formula to find the area of the segment of a circle.
The above figure shows a circle with radius r. OPXQ is a sector of the circle. The angle of sector is θ. PXQP is a segment of the circle.
From the figure, we can see that
Area of the segment PXQP = area of the sector OPXQ − area of the triangle OPQ
− area of ΔOPQ
Using this formula, we can calculate the area of the shaded portion in the figure discussed in the beginning.
In that figure, the radius of the circle is given as 7 cm and angle of the sector 90°. To calculate the area of the shaded portion, we will calculate the area of each segment. Let us draw the circle whose segment has been shown in the given figure. On drawing the circle, we obtain the following figure.
The sector has the angle 90° at the centre, i.e. AOB = 90°.
Let us draw a perpendicular from O to AB, i.e. OFAB.
Now in Δ OAF and Δ OBF,
OA = OB (radius of the circle)
∠OFA = ∠OFB (both 90°)
OF = OF (common side)
By RHS Congruence rule,
ΔOAF ≅ ΔOBF
Therefore, AF = BF = AB (by CPCT)
And ∠AOF = ∠BOF = ∠AOB (by CPCT)
= × 90° = 45°
But we know that
⇒ AB = 2 AF
Base = cm
And corresponding height = cm
Thus, area of ΔOAB = × base × height
Given, r = 7 cm
θ = 90°
Thus, area of sector OADB
Therefore, area of the minor segment ADB
= Area of sector OADB − area of ΔOAB
We have calculated the area of one such segment. But we have 4 such segments.
Thus, the area of shaded portion is 56 cm2.
It can be seen that it was a very lengthy process to find the area of triangle in the above discussed case.
The area of triangle can be calculated in the other way as well. Let us see how.
Let us consider ΔABC such that AB = c and AC = b.
BD is perpendicular to AC. Thus, BD is the height of ΔABC corresponding to base AC.
In right-angled triangle ΔADB, we have
⇒ BD = AB sin A
⇒ BD = c sin A
Similarly, we can find the area of the triangle inside the circle…
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