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If isosceles triangle PQR, $PQ = PR = 5 cm$ is inscribed in a circle of radius 15 cm. Find the area of the triangle.

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69.84

{cm}^{2}

84.69

{cm}^{2}

72.69

{cm}^{2}

69.72

{cm}^{2}

answer is A.

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Detailed Solution

Given,
A circle radius (r) = 15 cm
An isosceles triangle in the circle = PQR

PQ = PR = 5 cm

Here, radius

AP = AQ = AR = 15 cm

Then, a quadrilateral = ARPQ
Where,

AP ⊥ QR

AX ⊥ QR

Suppose that

PX = x cm

Using the Pythagoras theorem in triangle of PXQ,

(side 1)^{2} + (side 2)^{2} = (hypotenuse)^{2}

(PX)^{2} + (QX)^{2} = (PQ)^{2}

(x)^{2} + (QX)^{2} = (5)^{2}

Rearrange the term

Q X^{2} = 5^{2} - x^{2}

Q X^{2}

= 25 - x^{2}

…………... (1)
Now the radius

AP = AX + PX

We put the value then we obtain,

\Rightarrow AX = AP - PX

\Rightarrow AX = 15 - x cm

………… (2)
In the triangle ∆ AXQ,
Again, using Pythagoras theorem,

A X^{2} + Q X^{2} = A Q^{2}

Form the equations (1) and (2) we have

\Rightarrow (15 - x)^{2} + 25 - x^{2} = (15)^{2}

[

(a - b)^{2} = a^{2} + b^{2} - 2 ab]

\Rightarrow 225 + x^{2} - 30 x + 25 - x^{2} = 225

\Rightarrow - 30 x = - 25

\Rightarrow x = \frac{- 25}{- 30}

\Rightarrow x = \frac{5}{6} cm

…………(3)
Put value of

x = \frac{5}{6} cm

in the equation (1)

(Q {X)}^{2} = 25 - {(\frac{5}{6})}^{2}

(Q {X)}^{2} = 25 - \frac{25}{36}

(Q {X)}^{2} = \frac{900 - 25}{36}

(Q {X)}^{2} = \frac{875}{36}

QX = \sqrt{\frac{875}{36}}

QX = \frac{1}{6} \sqrt{875} cm

Put the value equation (3) into equation (2) we have,

AX = AP - PX

AX = 15 - \frac{5}{6}

AX = \frac{90 - 5}{6}

AX = \frac{85}{6} cm

Now

△ PQR

isosceles triangle

AX ⊥ QR

then,

QX = RX

QR = 2 QX

QR = 2 \times \frac{\sqrt{875}}{6}

[

QX = \frac{1}{6} \sqrt{875}]

QR = \frac{2 \sqrt{875}}{6} cm

In the triangle

△ PQR

△ PQR = \frac{1}{2} \times QR \times AX

△ PQR = \frac{1}{2} \times \frac{2 \sqrt{875}}{6} \times \frac{85}{6}

△ PQR = \frac{85 \sqrt{875}}{36} {cm}^{2}

PQR =

69.842 {cm}^{2}

Hence area of triangle

= 69.842 {cm}^{2}

Correct option is 1.

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