In the following figure, AB = 36 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region.
Answers
Answer:
The area of shaded region is 45π cm².
Step-by-step explanation:
Given :
AB = 36 cm and M is mid-point of AB
Therefore, AM = BM = ½ × AB = ½ × 36 = 18 cm
AM = BM = 18 cm
AP = PM = MQ = QB = 9 cm
Let the Radius of circle with centre C be ‘r’ i.e CR = r
Join P to C and M to C , MC _|_ AB
MR = AM = 18 cm
CM = MR - CR
CM = (18 - r ).........(1)
PC = PE + CE
PC = (9 + r).......(2)
In ∆ PCM , by using Pythagoras theorem
PC² = PM² + MC²
(9 + r)² = 9² + (18 - r)²
[From eq 1 & 2]
81 + r² + 18r = 81 + 324 + r² - 36r
18r + 36r = 324
54r = 324
r = 324/54
r = 6
Radius of circle with C as a centre = 6 cm
Area of shaded region , A = Area of semicircle with diameter AB - Area to semicircles with diameter AM and MB - Area of circle with C as a centre
A = ½ π(36/2)² - 2× ½ π(18/2)² - π(6)²
A = ½ π(18)² - π(9)² - π(6)²
A = ½ π × 324 - 81π - 36π
A = 162π - 81π - 36π
A = 162π - 117π
A = 45π cm²
Area of shaded region = 45π cm²
Hence, the area of shaded region is 45π cm².
HOPE THIS ANSWER WILL HELP YOU….
Step-by-step explanation:
Let the radius of the circle be r.
and △PCM is right △
PC2=PM2+MC2
(9+r)2=(9)2+(16−r)2
81 + r.218.r = 81 + 324 + r.2 − 36.r
54.r = 324
r = 32454=6cm
Areaofshadedregion=324π2−81π+36π
= 162π−117π
Therefore area of the shaded region =45πcm2