A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centr of the circle. find the area of the minor as well as the major segment.(π=3.14,√=1.73
Answers
Answer:
HEY MATE !!!!!!!!
Step-by-step explanation:
Given :
Radius ( r ) = 15 cm
angle at the centre ( x ) = 60°
i ) Area of the sector ( OPAQ )
= ( x/360 ) × πr²
= ( 60/360 ) × πr²
= (1/6 ) × 3.14 × 15²
= 117.75 cm² --( 1 )
ii ) Area of the equilateral triangle OPQ
= ( √3/4 ) × side²
= (1.73/4) × 15²
= 0.4325 × 225
= 97.3125 cm²----( 2 )
iii ) Area of the circle ( A ) = πr²
= 3.14 × 15²
= 706.5 cm² ---( 3 )
iii ) Area of the minor segment ( PAQ )
= ( 1 ) - ( 2 )
= 117.75 - 97.3125
= 20.4375 cm² ---( 4 )
iv ) Area of the major segment
= ( 3 ) - ( 4 )
= 706.5 cm² -20.4375 cm²
= 686.0625cm²
HOPE IT'S HELPFUL^_^
Answer:
In the mentioned figure,
O is the centre of circle,
AB is a chord
AXB is a major arc,
OA=OB= radius = 15 cm
Arc AXB subtends an angle 60 °
at O.
Area of sector AOB=
=60/360 ×π×r ²
= 60/360 ×3.14×(15) ²
=117.75cm ²
Area of minor segment (Area of Shaded region) = Area of sector AOB− Area of △ AOB
By trigonometry,
AC=15sin30
OC=15cos30
And, AB=2AC
∴ AB=2×15sin30=15 cm
∴ OC=15cos30=15'2
=15× 2
1.73
=12.975 cm
∴ Area of △AOB=0.5×15×12.975=97.3125cm ²
∴ Area of minor segment (Area of Shaded region) =117.75−97.3125=20.4375 cm ²
Area of major segment = Area of circle − Area of minor segment
=(3.14×15×15)−20.4375
=686.0625cm²