Question

A chord AB of a circle of radius 10 cm subtends a right angle at the centre. Find the area of major and minor segments.
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Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

A chord AB of a circle of radius 10 cm subtends a right at the center .

$\underline{\bf{Explanation\::}}$

Firstly, attachment a figure according to the given question.

As we know that formula of minor segment;

$\boxed{\bf{Area = r^{2} \bigg[\dfrac{\pi \theta}{360 \degree} - \frac{1}{2} sin \theta \bigg]}}$

We have :

• Radius, (r) = 10 cm
• Ф = 90°

A/q

$\mapsto\tt{Area\:_{(minor\:segment)} = r^{2}\bigg[\dfrac{\pi \theta}{360 \degree} - \dfrac{1}{2} sin \theta \bigg]}$

$\mapsto\tt{Area\:_{(minor\:segment)} = (10)^{2}\bigg[\dfrac{3.14 \times 90}{360 \degree} - \dfrac{1}{2} sin 90 \degree \bigg]}$

$\mapsto\tt{Area\:_{(minor\:segment)} = 100 \bigg[\dfrac{3.14 \times \cancel{90}}{\cancel{360 }\degree} - \dfrac{1}{2} sin 90 \degree \bigg]}$

$\mapsto\tt{Area\:_{(minor\:segment)} = 100 \bigg[\dfrac{3.14 }{4} - \dfrac{1}{2} \times 1 \bigg] \:\: [\therefore sin 90 \degree = 1]}$

$\mapsto\tt{Area\:_{(minor\:segment)} = 100\bigg[\dfrac{3.14 }{4} - \dfrac{1}{2} \bigg]}$

$\mapsto\tt{Area\:_{(minor\:segment)} = 100\bigg[\dfrac{3.14 - 2}{4} \bigg]}$

$\mapsto\tt{Area\:_{(minor\:segment)} = 100\bigg[\dfrac{1.14}{4} \bigg]}$

$\mapsto\tt{Area\:_{(minor\:segment)} = \cancel{100} \times \dfrac{1.14}{\cancel{4}} }$

$\mapsto\tt{Area\:_{(minor\:segment)} = 25 \times 1.14}$

$\mapsto\bf{Area\:_{(minor\:segment)} = 28.5\:cm^{2}}$

Now.

→ Area of major segment = Area of circle - area of minor segment (ACB)

→ Area of major segment = πr² - 28.5 cm²

→ Area of major segment = 3.14 × 10 ×10 - 28.5 cm²

→ Area of major segment = 3.14 × 100 - 28.5 cm²

→ Area of major segment = 314 cm² - 28.5 cm²

→Area of major segment = 285.5 cm² .

Answer 2

Answer:

radius r=10cm

angle =90

area of sector A=90/360pir^2

A=3.14x10x10/4

A=25x3.14

A=78.5sq.cm

Let the angle subtended and radius form an arc AOB,then

area of AOB=rxrsin90/2

AOB=10x10x1/2

ar. of AOB =50sq.cm

area of minor segment =78.5-50

=28.5aq.cm

Then area of circle =pir^2

=3.14x10x10

=314 Sq.cm

area of major segment =area of circle -area of minor segment

=314-28.5

=285.5sq.cm