Question

14. The radius of a circle with centre O is 7 cm. Two radii OA and OB are
drawn at right angles to each other. Find the areas of minor and major
segments.
​

Answer 1

Answer:

Minor Segment : 14

Major Segment : 231/4

Step-by-step explanation:

Minor Segment = Area of Sector-Area of Triangle

= Q/360 × pi × radius^2 - 1/2 × base × height

= 90/360 × 22/7 × 7 × 7 - 1/2 × 7 ×7

= 1/4 × 22 × 7 - 1/2 × 49

= 77/2 - 49/2

= 77-49/2. = 28/2

= 14

Major Segment = 360 - Q/ 360 × 22/7 × 7 ×7

= 360 - 90/ 360 × 22 ×7

= 270 / 360 × 154

= 3/4 ×154

= 231/4

Hope it would be correct.

Answer 2

Given:

• Radius of the Circle = r = 7cm
• Central Angle = $\theta$ = 90°

To Find:

• Area of Minor Segment
• Area of Major Segment

Solution:

Now,

we, know that

$\underline{ \boxed{\tt Area \: Of \: minor \: segment = \frac{ \pi {r}^{2} \theta }{360} - \frac{1}{2} {r}^{2} \sin \theta }}$

where,

• Radius r = 7cm
• $\theta$ = 90°
• π = $\frac{22}{7}$

So,

$</strong><strong> \tt \to Area \: Of \: minor \: segment = \frac{ \pi {r}^{2} \theta }{360} - \frac{1}{2} {r}^{2} \sin \theta$

$</strong><strong> \tt \implies \frac{22}{7} \times 7 \times 7 \times \frac{90}{360} - \frac{1}{2} \times 7 \times 7 \times \sin 90 \degree$

$\tt \implies \frac{77}{2} - \frac{49}{2}$

$\tt \implies\frac{77 - 49}{2}$

$\tt \implies \frac{28}{2}$

$\tt \implies \frac{ \cancel{28}}{ \cancel{2}} = 14 {cm}^{2}$

Area of Minor Segment = 14cm²

we, know that,

$\underline{ \boxed{ \tt Area \: Of \: Circle= \pi {r}^{2} }}$

where,

• where,Radius r = 7cm
• π = $\frac{22}{7}$

So,

$\tt \to Area \: Of \: Circle= \pi {r}^{2}$

$\tt \implies \frac{22}{7} \times 7 \times 7$

$</strong><strong> \tt \implies \frac{22}{ \cancel7} \times \cancel 7 \times 7$

$</strong><strong> \tt \implies 22 \times 7 = 154 {cm}^{2}$

Area of Circle = 154cm²

Now,

we, know that,

$\underline{ \boxed{\tt Area \: of \: Major \: Segment = Area \: of \: Circle - Area \: of \: Minor \: Segment }}$

where,

• Area of Minor Segment = 14cm²
• Area of Circle = 154cm²

So,

$\tt \to Area \: of \: Major \: Segment = Area \: of \: Circle - Area \: of \: Minor \: Segment$

$\tt \implies Area \: of \: Major \: Segment = 154 - 14$

$\tt \implies Area \: of \: Major \: Segment = 140 {cm}^{2}$

Area of Major Segment = 140cm²

Hence, The Area of Minor Segment will be 14cm²

and

Area of Major Segment will be 140cm²