a chord of length 10 cm divides a circle of radius 5 root 2 in two segments. find the area of minor segment
Answers
Answer:
the area of minor segment is 14.28 cm²
Step-by-step explanation:
Given radius = r = 5√2 cm = OA = OB
Length of chord AB = 10cm
n ΔOAB, OA = OB = 5√2
AB = 10cm
OA²+OB²=(5√2)²+(5√2)²=50+50=100=(AB)²
Pythagoras theorem is satisfied .
∴OAB is right triangle
= angle subtended by chord = ∠AOB = 90°
Area of segment (minor) = shaded region
= area of sector − area of ΔOAB
[θ/360 ] x πr² - 1/2 x OAxOB
=90/360 x 22/ 7 (5√ 2) 2 - 1/2 x 5 √2 x 5 √2
=275/7- 25= 100/7 cm²
=14.28 cm²
Answer:
Answer:
the area of minor segment is 14.28 cm²
Step-by-step explanation:
Given radius = r = 5√2 cm = OA = OB
Length of chord AB = 10cm
n ΔOAB, OA = OB = 5√2
AB = 10cm
OA²+OB²=(5√2)²+(5√2)²=50+50=100=(AB)²
Pythagoras theorem is satisfied .
∴OAB is right triangle
= angle subtended by chord = ∠AOB = 90°
Area of segment (minor) = shaded region
= area of sector − area of ΔOAB
[θ/360 ] x πr² - 1/2 x OAxOB
=90/360 x 22/ 7 (5√ 2) 2 - 1/2 x 5 √2 x 5 √2
=275/7- 25= 100/7 cm²
=14.28 cm²
pls mark brainliest
Step-by-step explanation: