Question

AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find the area of the minor segment.

Answer 1

Answer:

The area of minor segment is (8π/3 - 4√3) cm².

Step-by-step explanation:

Given :  

Radius of circle, r (OA & OB) = 4 cm  

Length of chord , AB = 4cm  

∆OAB is an equilateral triangle, ∠AOB = 60°

Angle subtended at centre , θ = 60°  

Area of minor segment = (Area of sector OAB) – (Area of equilateral ΔAOB)

= θ/360 × πr²  - √3/4 a²

=( 60°/360°) × π × 4² - √3/4 × 4²

= ⅙ × 16π - √3/4 × 16

= 16π/6 - 4√3

= (8π/3 - 4√3) cm²

Area of minor segment = (8π/3 - 4√3) cm²  

Hence, the area of minor segment is (8π/3 - 4√3) cm².

HOPE THIS ANSWER WILL HELP YOU….

Here are more questions of the same chapter :

In the following figure, shows a sector of a circle, centre O, containing an angle  θ°. Prove that:

(i)Perimeter of the shaded region is $(tan\Theta+sec\Theta+\frac{\pi\Theta}{180}-1)$

(ii)Area of the shaded region is $\frac{r^{2}}{2} (tan\Theta-\frac{\pi\Theta}{180})$

https://brainly.in/question/9459553

In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110° at the centre of the circle. Find:

(i)the circumference of the circle

(ii)the area of the circle

(iii)the length of the arc AB,

(iv)the area of the sector OAB

https://brainly.in/question/9459071

Answer 2

Hey mate..

Answer :

Given: AB is a chord of a circle with centre O and radius 4cm. AB is of length 4cm and divides the circle into two segments.

To find: the area of the minor segment.

Solution:

Radius of circle = 4cm

Length of chord = 4cm

(Hence it makes an equilateral triangle at centre, in which all angle must be = 60°)

Area of sector =

$\frac{teta \: \: o }{360} \pi {r}^{2} \: \\ \\ = \frac{60}{360} \times \pi \times 4 \times 4 \\ \\ = \frac{1}{6} \times \pi \times 4 \times 4 = \frac{8\pi}{3} c {m}^{2}$

Area of equilateral ∆OAB =

$\frac{ \sqrt{3} }{4} {a}^{2} \\ \\ = \frac{ \sqrt{3} }{4} {a}^{2} = \frac{ \sqrt{3} }{4} \times 16 = 4 \sqrt{3} c {m}^{2}$

Area of minor segment = area of sector – area of ∆OAB

$( \frac{8\pi}{3} - 4 \sqrt{3)}c {m}^{2}$

Hope this helps u dude ✌

If helpful Mark it as Brainliest answer ⭐⭐

FOLLOW ME IF YOU CAN ❤️