Question

find the area of the segment of a circle of radius 14 cm the radius that encloses the sector region are perpendicular to each other pie = 22/7

Answer 1

Required Answer:-

We have:

• A circle of radius 14 cm.
• And two of the radius that encloses the sector region are perpendicular to each other$.$

To FinD:

• Area of the segment that is enclosed by the radius.

Step-by-Step Explanation:

The radius are perpendicular to each other which means angle subtended by the radius is 90°. Hence, the chord AB subtends an angle of 90°.

Then:

Area of the segment = Area of sector - Area of right angled triangle OAB.

Area of the sector:

= Θ / 360° × πr²

= 90° / 360° × 22/7 × (14)² cm²

= 1/4 × 22/7 × 14 × 14 cm²

= 22 × 7 cm²

= 154 cm²

Area of right-angled triangle:

= 1/2 × r²

= 1/2 × (14)² cm²

= 98 cm²

Now, Area of segment:

= Area of sector - Area of right angled ∆OAB.

= 154 cm² - 98 cm²

= 56 cm²

Hence:-

• The required area of the segment is 56 cm²

Answer 2

Answer:

$\huge \bf \: Given$

• Radius of circle = 14 cm
• Two of the radius that encloses the sector region are perpendicular to each other.

$\huge \bf \: To \: find$

Area of the segment that is enclosed by the radius.

$\huge \bf \: Solution$

It is given that the radius are perpendicular to each other therefore the radius will be subtend to 90⁰.

Area of the segment = Area of sector - Area of right angled triangle OAB.

$\sf \: \dfrac{90}{360} \times \dfrac{22}{7} \times 14 \times 14$

$\sf \: \dfrac{1}{4} \times 22 \times 2 \times 14$

$\sf \: \dfrac{1}{4} \times 616$

$\sf \: area \: of \: sector \: = 154 \: cm$

Now,

Finding area of right angled triangle

$\sf \: \dfrac{1}{2} \times {r}^{2}$

$\sf \: \dfrac{1}{2} \times 14 \times 14$

$\sf1 \times 7 \times 14$

$\sf \: 98 cm²$

Now, Area of segment:

Area of sector - Area of right angled ∆ OAB.

154 cm - 98 cm

56 cm²