Question

Hello there Brainliacs, try to solve this question....

ABCD is a rectangle enclosed inside a half circle.

Prove the ratio of the perimeter of the rectangle to the diameter of the rectangle is 3:[√2]

I want a detailed explanation.

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Answer 1

refer to the attachment....

therefore the perimeter of the rectangle CDGH

or the rectangle inscribed in a semicircle of radius r is

$= 2( \sqrt{2} r + \frac{ \sqrt{2} }{2} r) \\ = 3 \sqrt{2} r$

now diameter of the semicircle is=2r

therefore the ratio of the perimeter of the rectangle to the diameter of the circle is

$= \frac{3 \sqrt{2} r}{2r} \\ = \frac{3}{ \sqrt{2} }$

=3:√2 (proved)

.

Answer 2

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Question:-

ABCD is a rectangle enclosed inside a half circle.

Prove the ratio of the perimeter of the rectangle to the diameter of the rectangle is 3:[√2]

Step-by-step explanation:

Let the length of the rectangle is 2x and the breadth is x.

And the radius of the semicircle is assumed to be r.

Now, see the diagram attached.

From, Δ OAB, OA = x and AB = x and OB = r and ∠ OAB = 90°.

So, applying Pythagoras Theorem r² = x² + x²

⇒ r = √2x.

Now, perimeter of the rectangle is 2(2x + x) = 6x

And the diameter of the semicircle is 2r = 2√2x

Hence, the ratio of the perimeter of the rectangle and the diameter of the semicircle will be 6x : 2√2x = 3 : √2 (Answer

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