Q.194. Application of properties of triangles using trigonometry.

Question 194.
Find the minimum area of a square that can be drawn inside a triangle of sides 13, 14 and 15, with three of the vertices touching the sides of the triangle.

Answer 194.

Let ABC be the triangle with side AB = 13, BC = 14 and CA = 15.

Let the vertices of square (having side x), P touch AB, Q touch BC and R touch CA

Let PQ make angle θ with AB
=> PR makes angle 90° - θ with AB

AB = 13 = AP + PB = ARcosA + xsinθ + BQcosB + xcosθ .. (1)
 Using sine rule for triangles APR and BPQ,
AR/cosθ = x/sinA and BQ/sinθ = x/sinB
=> AR = xcosθ/sinA and BQ = xsinθ/sinB

Plugging in (1),
13 = xcosθ cotA + xsinθ + xsinθ cotB + xcosθ ... (2)

Using cosine rule to triangle ABC, it can be shown that
cosA = 99/195 and cosB = 5/13
=> cotA = 99/168 and cotB = 5/12

Plugging in (2),
13 = x (1 + 99/168) cosθ + x(1 + 5/12) sinθ
=> 13 = (1/168) (267cosθ + 238sinθ) x
=> x = (13 x 168) / (267cosθ + 238sinθ)

Area of sqaure is minimum when x is minimum and
267cosθ + 238sinθ is maximum
Maximum value of 267cosθ + 238sinθ
= √(267^2 + 238^2) = √(127933)
=> Minimum area of square, x^2
= (13 x 168)^2 / (127933)
= (168)^2 / 757
= 28224/757
≈ 37.284 sq. units.

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