Question

Let ABCD be a quadrilateral in which AB || CD, AB ⊥ AD, AB = 3CD and area of quadrilateral is 4 sq. units. If a circle can be drawn touching all the sides of the quadrilateral, whose radius is r then find 4r².

Answer 1

Answer:

$\huge\underbrace\mathbb\blue{AnSwEr}$

$<b><marquee>$

Let P,Q,R,S be the points of contact of in-circle with the sides AB, BC, CD, DA respectively. Since AD is perpendicular to AB and AB is parallel to DC, we see that AP = AS = SD = DR = r, the radius of the inscribed circle. Let BP = BQ = y and CQ = CR = x. Using AB = 3CD, we get r + y = 3(r + x) Since the area of ABCD is 4. we also get 4 = 1/2AD(AB CD) = 1/2(2r)(4(r + x)). Thus we obtain r (r + x) = 1. Using Pythagoras theorem, we obtain BC2 = BK2 + CK2 . However BC = y + x ,BK = y - x and CK = 2r. Substituting these and simplifying, we get xy = r2 . But r + y = 3 (r + x) gives y = 2r + 3x. Thus r2 = x(2r + 3x) and this simplifies to get (r – 3x)(r + x) = 0. We conclude that r = 3x. Now the relation r(r + x) = 1 implies that 4r2 = 3, gives r = √3/2