Question

AB and PQ are parallel chords of a circle centred at O. If distance between AB and PQ is 7 cm, radius of circle is 25 cm then area (ABQP) is equal to:

Answer 1

Answer:

175cm.

$thanks$

Answer 2

Answer:

The area of the trapezium (ABQP) = 343 cm^2

Step-by-step explanation:

Let the midpoint of AB is C

i.e, AB = 2AC

Given that: Distance between AB and PQ = 7 cm

So, OC = 7 cm

And radius of circle (r) = 25 cm

From the figure we can say;

OA = OP = radius (r) = 25 cm

PQ = 2OP = 50 cm

And from Pythagoras theorem we know that;

$oa = \sqrt{ {ac}^{2} + {co}^{2} }$

So we get;

$ac = \sqrt{ {oa}^{2} - {oc}^{2} } \\ ac = \sqrt{ {25}^{2} - {7}^{2} } \: cm \\ ac = 24 \: cm$

So we get, AB = 2AC = 2 × 24 cm = 48 cm

So, the area of trapezium (ABQP) ;

$(ABQP) = \frac{a + b}{2} \times h \\ (ABQP) = (\frac{ab + pq}{2}) \times h \\ (ABQP) = (\frac{50 + 48}{2}) \times 7 \: {cm}^{2} \\ (ABQP) = 343 \: {cm}^{2}$

So the area of the trapezium (ABQP) = 343 cm^2