Let be a quadrilateral with area 18, with side parallel to the side and . Let be perpendicular to and . If a circle is drawn inside the quadrilateral touching all the sides, then its radius is
- A.3
- B.2
- C.
- D.1
Show answer and solution
Correct option: B
Solution
Let CD = a, so AB = 2a. AD is perpendicular to both AB and CD, making ABCD a right trapezoid with height AD = h.
Area = (1/2)(AB + CD)·h = (1/2)(2a + a)·h = (3ah)/2 = 18, so ah = 12.
For an inscribed circle of radius r: the circle touches AB (bottom), CD (top), and AD (left side). Since AD ⊥ AB and AD ⊥ CD, the height h = AD = 2r (diameter fits between the two parallel sides), so r = h/2.
Also, the circle touches AD, so its center is at distance r from AD, meaning r = r ✓. The circle touches the slant side BC. For a tangential quadrilateral, AB + CD = BC + AD, so 2a + a = BC + h, giving BC = 3a - h.
From geometry, BC² = (AB - CD)² + h² = a² + h². So (3a - h)² = a² + h², giving 9a² - 6ah + h² = a² + h², so 8a² = 6ah, meaning h = (4a/3).
From ah = 12: a·(4a/3) = 12, so a² = 9, a = 3, h = 4.
Therefore r = h/2 = 2.
Area = (1/2)(AB + CD)·h = (1/2)(2a + a)·h = (3ah)/2 = 18, so ah = 12.
For an inscribed circle of radius r: the circle touches AB (bottom), CD (top), and AD (left side). Since AD ⊥ AB and AD ⊥ CD, the height h = AD = 2r (diameter fits between the two parallel sides), so r = h/2.
Also, the circle touches AD, so its center is at distance r from AD, meaning r = r ✓. The circle touches the slant side BC. For a tangential quadrilateral, AB + CD = BC + AD, so 2a + a = BC + h, giving BC = 3a - h.
From geometry, BC² = (AB - CD)² + h² = a² + h². So (3a - h)² = a² + h², giving 9a² - 6ah + h² = a² + h², so 8a² = 6ah, meaning h = (4a/3).
From ah = 12: a·(4a/3) = 12, so a² = 9, a = 3, h = 4.
Therefore r = h/2 = 2.
[AIEEE 2007]