Question
Triangle ABC is a right angled triangle at B. A point O is lying on the side BC. A semicircle with centre O is inscribed inside the triangle such that it touches AC at D. Find the area of quadrilateral ABOD if length AB = 30 units and ∠ BCA = 30∘.
Answer: Option A
:
A
∠ BCA = 30∘
∠ BAC = 60∘
∠ ODC = 90∘ (radius is perpendicular to the tangent)
∠ COD = 60∘
∠ BOD = 120∘
Now the sides OB = OD (Radii of same circle)
AB = AD (tangents from the same external point)
so, triangle ABO is congruent to triangle ADO
∠ BOA = ∠ DOA = 60∘
Now AB = 30, so OB = 30√3, so area ( AOB) = area (AOD) = 12* 30 * 30 √3 = 150 √3
So area (ABOD) = 300 √3 = 519.6.
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:
A
∠ BCA = 30∘
∠ BAC = 60∘
∠ ODC = 90∘ (radius is perpendicular to the tangent)
∠ COD = 60∘
∠ BOD = 120∘
Now the sides OB = OD (Radii of same circle)
AB = AD (tangents from the same external point)
so, triangle ABO is congruent to triangle ADO
∠ BOA = ∠ DOA = 60∘
Now AB = 30, so OB = 30√3, so area ( AOB) = area (AOD) = 12* 30 * 30 √3 = 150 √3
So area (ABOD) = 300 √3 = 519.6.
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