AB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, If the area of the quadrilateral ACBD is $Δ$ , then find the value of $\frac{Δ}{27}$ .

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Solution

$T h e d i a m e t e r o f t h e g i v e n c i r c l e i s A B w h i c h s u b t e n d s ∠ A C B & ∠ A D B a t t h e c i r c u m f e r e n c e . ∴ ∠ A C B = 90^{2} = ∠ A D B . S o Δ A C B & Δ A D B a r e r i g h t t r i a n g l e s . N o w i t i s g i v e n t h a t A B = 15 c m, A C = 12 c m a n d B D = 9 c m . S o i n Δ A C B B C = \sqrt{{A B}^{2} - {A C}^{2}} = \sqrt{15^{2} - 12^{2}} c m = 9 c m = B D . ∴ B e t w e e n Δ A C B & Δ A D B w e h a v e B C = B D & t h e h y p o t e n u s e A B i s c o m m o n t o b o t h . ⟹ Δ A C B & Δ A D B a r e c o n g r u e n t (b y R H S t e s t) i . e t h e i r a r e a s a r e e q u a l . ∴ a r . A C B D = 2 \times (a r Δ A C B o r a r . Δ A D B) . = 2 \times \frac{1}{2} \times A C \times B C = A C \times B C = 12 \times 9 {c m}^{2} = 108 {c m}^{2} . A n s - a r . A C B D = 108 {c m}^{2} .$
$∴ Δ = 108$
$∴ \frac{Δ}{27} = 4$

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AB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, If the area of the quadrilateral ACBD is Δ, then find the value of Δ27.

AB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, If the area of the quadrilateral ACBD is $Δ$ , then find the value of $\frac{Δ}{27}$ .