Question

P is a point at a distance of 10 cm from the centre O of a circle and PQ and PR are tangents to the circle. If radius of the circle is 6 cm, then the area of quadrilateral PQOR is equal to 1. 24 cm2 2. 36 cm2 3. 48 cm2 4. 96 cm2​

Answer 1

Given:

Distance of point P from the centre O = 10cm

Radius of the circle = 6cm

To Find:

Area of quadrilateral PQOR.

Solution:

On drawing the quadrilateral, we get the required shape as shown in the figure. Here, PQ and PR are tangents to the circle. These tangents are perpendicular to the radii of the circle drawn towards them. The length of P from O which is the centre of the circle is given as 10cm. Let M be the point where the line joining O and P is cutting the circle. The distance OM is given by the radius of the circle, which is 6cm. The remaining distance, MP can be determined by subtracting the total length from the radius.

$OP=OM+MP$

$10=6+MP$

$MP=10-6$

∴ $MP=4cm$

Now, we have two triangles, ΔPRO and ΔPQO which are right triangles. According to the Pythagoras theorem, the square of the hypotenuse is the sum of squares of the other two sides. The side opposite to the 90° is the hypotenuse side.

In ΔPRO,

$OP^{2} =OR^{2}+PR^{2}$

$10^{2} =6^{2} +RP^{2}$

$RP^{2} =10^{2} -6^{2}$

$RP^{2} =100-36$

$RP^{2} =64$

$RP=8 cm$

In ΔPQO,

$OP^{2} =OQ^{2}+PQ^{2}$

$10^{2} =6^{2} +PQ^{2}$

$PQ^{2}=10^{2} -6^{2}$

$PQ=8cm$

To determine the area of quadrilateral, determine the areas of the triangles and add them. Since the values of both the triangles are the same, find the area of one triangle, multiply it by 2 and thus, we can determine the area of quadrilateral PQOR.

Area of ΔPQO = $\frac{1}{2}$ × base × height

$ar(PQO)=\frac{1}{2}* PQ*OQ$

$ar(PQO)=\frac{1}{2}*6*8$

$ar(PQO)=3*8=24cm^{2}$

Area of quadrilateral PQOR = 2 × Area of ΔPQO

$ar(PQOR)=2*ar(PQO)$

$ar(PQOR)=2*24=48cm^{2}$

Thus, the area of quadrilateral PQOR is 48cm².

The area of quadrilateral PQOR where O is the centre of a circle having a radius of 6 cm, and PQ and PR are tangents to the circle is 48cm².