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A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is

Answer (Detailed Solution Below)

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The Rhombus and the inscribed circle can be shown according to the following figure:

ABCD is the rhombus, with diagonals 16 cm and 12 cm in length.

The circle has the center at O, and the radius is r.

So, we get:

AO = OC = 16/2 = 8 cm

And DO = OB = 12/2 = 6 cm

Applying Pythagoras theorem to the ΔADO, we get:

(AD)2 = (AO)2 + (OD)2

⇒ (AD)2 = 82 + 62

⇒ AD = 10 cm

Now, on considering the ΔADO, we can calculate its area in two ways:

Using hypotenuse AD as the base, or Using OD as the base. The area will remain the same. So,

(1/2) × AO × OD = (1/2) × r × AD

⇒ r = (8 × 6)/10 = 4.8 cm

So, the area of the circle = π × 4.82 = 23.04π cm2 —-(i)

And the area of the rhombus = (1/2) × 12 × 16 = 96 cm2 —-(ii)

∴ The ratio of the area of the circle to the area of rhombus = 23.04π/96 = 6π/25 cm2

Additional Information

Area of a Rhombus = (1/2) × (product of the diagonal lengths)

Diagonals of a rhombus intersect each other at the right angle

Area of a right-angled triangle = (1/2) × (product of the sides other than the hypotenuse)

Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Last updated on Jul 31, 2023

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