Question

Anurag purchased a farmhouse which is in the form of a semicircle of diameter 70 m. He divides it into three parts by taking a point P on the semicircle in such a way that ∠PAB = 30° as shown in the following figure, where O is the centre of the semicircle. In part I, he planted saplings of Mango tree, in part II, he grew tomatoes and in part III, he grew oranges. Based on given information, answer the following questions.

1. What is the measure of ∠POA?   [1]
2. Find the length of wire needed to fence entire piece of land.   [1]
3. 1. Find the area of region in which saplings of Mango tree are planted.   [2]
                                       OR
   2. Find the length of wire needed to fence the region III.     [2]

Answer 1

i. Since O is the center of the semicircle, OA and OB are radii.

We observe that POA is an isosceles triangle, where OA = OP = 35 m.

Since ∠PAB = 30°, and AB is the diameter, ∠APB = 90° (as the angle subtended by a semicircle is always 90°).

Thus, in ΔPOA, using the property of a triangle:

∠POA = 90° − ∠PAB

= 90° − 30°

= 60

So, ∠POA = 60°.

ii. To fence the entire land, we need the perimeter of the semicircle + diameter.

Perimeter of semicircle = 12 × 2πr = πr

= 3.14 × 35

= 109.9 m

Diameter (AB) = 70 m

109.9 + 70 = 179.9 m

So, the length of wire needed is 179.9 m.

iii. (a) The area of Region I is the area of sector OAP.

Area = `θ/360°xxpir^2`

Here, θ = 60° and r = 35 m.

Area = `60/360xx3.14xx(35)^2`

= `1/6xx3.14xx1225`

= `(3.14xx1225)/6`

= `3843.5/6`

= 640.58 m²

So, the area of Region I (Mango trees) is 640.58 m².

OR

(b) To fence Region III, we need the arc length of sector OAP and the two line segments AP and PO.

Arc length of OAP:

L = `θ/360xx2pir`

= `60/360xx2xx3.14xx35`

= `1/6xx219.8`

= 36.63 m

AP and PO are both radii, so AP = PO = 35 m.

Total fencing required for Region III:

Arc length + AP + PO

= 36.63 + 35 + 35

= 106.63 m

So, the length of wire needed to fence Region III is 106.63 m.