Question

In the given figure, PQRS is a square inscribed in a circle of radius 4 cm. PQ is produced till point Y. From Y a tangent is drawn to the circle at point R. What is the length (in cm) of SY ?

A) 4√10 B) 2√10 C) 6√10 D) 3√5

Answer 1

Given:

PQRS is a square inscribed in a circle of radius 4 cm

PQ is produced to point Y

From Y a tangent is drawn to the circle at point R

To find:

The length (in cm) of SY

Solution:

From the figure attached below, we have

O is the centre of the circle

Radius = OS = OQ = 4 cm

RY is a tangent to the circle

PQ = QR = RS = SP = sides of the square

Diameter = SQ = Diagonal of a sqaure = 4 × 2 = 8 cm

Finding the side of the square:

We know,

$\boxed{\bold{Diagonal\: of\: a\: square\: = \sqrt{2} a}}$

where ⇒ "a" → side of a square

∴ PQ = QR = RS = SP = a = $\frac{8}{\sqrt{2}}$ = $4\sqrt{2} \:cm$

Finding the length of YP:

$\boxed{\bold{Tangent-Secant\:Formula }}}:$ If a tangent & a secant of a circle are drawn from an external point to the circle, then the product of the lengths of the secant and its external segment is equal to the square of the length of the tangent segment.

Here we have,

Tangent → RY = $\sqrt{QR^2 +YQ^2} = \sqrt{(4\sqrt{2} )^2 +YQ^2}$

Secant → YP = YQ + PQ = YQ + 4√2

So, according to the formula, we get

$RY^2 = YP^2 \times YQ^2$

by substituting the values of RY & YP

$\implies (4\sqrt{2} )^2 + YQ^2 = YQ(YQ+ 4\sqrt{2} )$

$\implies (4\sqrt{2} )^2 + YQ^2 = YQ^2+ (4\sqrt{2}\times YQ)$

$\implies (4\sqrt{2} )^2 = (4\sqrt{2}\times YQ)$

$\implies YQ = 4\sqrt{2}$

∴ YP = YQ + 4√2 = 4√2 + 4√2 = 8√2

Finding the length of SY:

Construction: Join S and Q

Now, in ΔSYP, using Pythagoras Theorem, we get

$SY^2 = SP^2 + YP^2$

$\implies SY = \sqrt{(4\sqrt{2} )^2 + (8\sqrt{2} )^2}$

$\implies SY = \sqrt{ 32 + 128}$

$\implies SY = \sqrt{ 160}$

$\implies SY = \sqrt{ 4\times 4\times 2\times 5}$

$\implies \bold{SY = 4\sqrt{ 10}\:cm}$ ← option (A)

Thus, the length (in cm) of SY is $\underline{4\sqrt{10}\:cm }$.

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