Question

plz solve 2nd and 3rd question

Answer 1

your answer is here

Answer 2

1:

Shaded Area = Area of Bigger circle - area of smaller circle - Area of larger sector + area of Smaller sector.

Therefore,
Required Area:

$\pi \times {14}^{2} - \pi \times {7}^{2} - \frac{40°}{360° } \times \pi \times 14 ^{2} + \frac{40°}{360° } \times \pi \times 7 \\ = \pi(14^{2} - 7 ^{2} ) - \frac{\pi}{9} ( {14}^{2} - {7}^{2} ) \\ = 147\pi - \frac{147\pi}{9} \\ = 410.5 {cm}^{2}$

2:

Required Perimeter:

AB + PB +minor arc PA

Minor Arc PA =

$\frac{\theta}{360°} \times 2\pi {r} \\ = \frac{\theta\pi r}{180°}$


In triangle BOA, Since AB is tangent to the circle.

$\sec( \theta) = \frac{OB}{OA} \\ \\ \sec( \theta) = \frac{OB}{r} \\ r\sec( \theta) = OB \\ r\sec( \theta) =r + PB \\ r(\sec( \theta) - 1) = PB$

And

$\tan(\theta) = \frac{AB}{OB} \\ \tan(\theta) = \frac{AB}{r} \\ r\tan(\theta) = AB$
Adding the Above three,

$AB+PB+ \text{minor Arc} \: PA = r \tan(\theta) + r( \sec(\theta) - 1) + \frac{\theta\pi r}{180°} \\ \\ = r(\tan(\theta) + \sec(\theta) - 1 + \frac{\theta\pi }{180°}) \\ = r(\tan(\theta) + \sec(\theta) + \frac{\theta\pi }{180°} - 1) \\ = \text{required \: perimeter}$