Question

If the sides 50 m and 130 m of a triangular field meet at an angle of 72º, then find the area in which wheat is cultivated (shaded portion)

(a) 120 π m² 

(b) 150 π m² 

(c) 200 π m² 

(d) 180 π m²

Answer 1

Answer : (d) 180 π m² 

∠ABC = 90º (∵ OQ ⊥ BC and OD ⊥ AB Line through the center of a circle is perpendicular to the tangent of the circle at the point of contact.) 

∴ Given, ∠BAC = 72º ⇒ ∠ACB = 90º – 72º = 18º 

BC = \(\sqrt {{AC}^2 -{AB}^2}\) = \(\sqrt {{130}^2 -{50}^2}\) = \(\sqrt {14400}\) =120 m.

(Pythagoras' Theorem) 

Let BQ = x m ⇒ QC = (120 – x) m and AD = (50 – x) m. 

⇒ BD = x m, CP = (120 – x) m, AP = (50 – x) m (Tangents from the same point are equal) 

∴ BQ + QC + CP + AP + AD + DB = (130 + 120 + 50)m 

⇒ x + (120 – x) + (120 – x) + (50 – x) + (50 – x) + x = 300 

⇒ 340 – 2x = 300 

⇒ 2x = 40 

⇒ x = 20 m. 

⇒ Radius of circle = OQ = DB = 20 m (∵ ODBQ is a square)

∴ Area of circle = πr² = 400π m²

Now in cyclic quadrilateral POQC, ∠POQ = 180º – ∠ACB 

= 180º – 18º = 162º

∴ Area of sector POQ = \(\frac{162^o}{360^o} \times\) Area of circle

= \(\frac{162}{360} \times\) 400π 

= 180 π m².