Question

In figure arc AXB is a semicircle.
m∆ PAB = 40°

Hence, find the values of
1) m ∆APВ
2) m (arc PYB)
3) m (arc AZP)
​

Answer 1

Answer:

Answer

In right angle triangle ACB,

AB

2

=AC

2

+BC

2

=(12)

2

+(16)

2

=144+256=400

AB=

400

=20m=Diameter

Radius =

2

20

=10m

Arc length ACB =

2

1

×2πr=

2

1

×2×3.142×10=31.42m

Perimeter of shaded region =31.42+12+16=59.42cm

Area of shaded region =

2

1

πr

2

−

2

1

×AC×CB=

2

1

(3.142×10

2

−12×16)

=

2

1

(314.2−192)=

2

1

×122.2=61.1cm

2

Step-by-step explanation:

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