9. Fig. 1 consists of a circle C with centre O. OA and OB are perpendicular to each other. The area of the triangle AOB is 32 square units. If AD = BD, the area of the triangle ABD is Fig.1 D A B O
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32√2 + 32 or 32√2 - 32 cm² is area of ΔABD
Step-by-step explanation:
OA = OB = Radius
Area of ΔAOB = (1/2) * OA * OB = 32
=> Radius * Radius = 64
=> Radius² = 8²
=> Radius = 8 cm
OA = OB = 8 cm
AB² = OA² + OB²
=> AB = 8√2 cm
Let say M is mid point of AB
then AM = BM = 4√2
OM = 4√2 ( as OM² = OA² - AM²)
AD = BD => DM will pass through center of circle O
DM ⊥ AB
DM = DO + OM or DO - OM ( depending on position of D)
DO = Radius = 8 cm
DM = (8 + 4√2) or (8 - 4√2)
Area of ΔABD = (1/2) AB * DM
= (1/2) (8√2) (8 + 4√2) or (1/2) (8√2) (8 - 4√2)
= 32√2 + 32 or 32√2 - 32
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