Question

AB is chord of circle subtending an angle at centre of 90 . if radius of circle is 6 cms area of minor segment​

Answer 1

Answer:10.28

Step-by-step explanation:

Answer 2

Answer:

The area of the minor segment $&=10.28 \mathrm{~cm}^{2}$.

Step-by-step explanation:

Given:

AB is a chord of a circle subtending an angle at the center of 90.

If the radius of the circle is 6cm.

To find:

the area of the minor segment​.

Step 1

Area of sector $A O B D$

$=\frac{\theta}{360^{\circ}} \pi r^{2}$

$=\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6^{2}$

$\frac{1}{4} \times \frac{22}{7} \times 36$

Convert element to fraction:

$36=\frac{36}{1}$

$=\frac{1}{4} \times \frac{22}{7} \times \frac{36}{1}$

Step 2

Apply the fraction rule:

$\frac{a}{b} \times \frac{c}{d}=\frac{a \times c}{b \times d}$

$=\frac{1 \times 22 \times 36}{4 \times 7 \times 1}$

$\frac{1 \times 22 \times 36}{4 \times 7 \times 1}=\frac{792}{28}$

$=\frac{792}{28}$

Factor the number

$792=4 \times 198$

$=\frac{4 \times 198}{28}$

Step 3

Factor the number

$28=4 \times 7$

$=\frac{4 \times 198}{4 \times 7}$

Cancel the common factor

= 28.28571 ....

Area of minor segment = area of sector AOBD - area of $\triangle A O B$

$&=28.28-18$

$&=10.28 \mathrm{~cm}^{2}$

Therefore, the area of the minor segment $&=10.28 \mathrm{~cm}^{2}$.

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