In the given figure PQRS is a rectangle inscribed in the circle. If PS = 5 cm and PQ = 12 cm, find the area of un-shaded region. (Use π =22/7)
Answers
Answer:
△ABC and△PQR in which AB=PQ,BC=QR and AM=PN.
Since AM and PN are median of triangles ABC and PQR respectively.
Now, BC=QR ∣ Given
⇒
2
1
BC=
2
1
QR ∣ Median divides opposite sides in two equal parts
BM=QN... (1)
Now, in △ABM and△PQN we have
AB=PQ ∣ Given
BM=QN ∣ From (i)
and AM=PN ∣ Given
∴ By SSS criterion of congruence, we have
△ABM≅△PQN, which proves (i)
∠B=∠Q ... (2) ∣ Since, corresponding parts of the congruent triangle are equal
Now, in △ABC and△PQR we have
AB=PQ ∣ Given
∠B=∠Q ∣ From (2)
BC=QR ∣ Given
∴ by SAS criterion of congruence, we have
△ABC≅△PQR, which proves (ii)
Answer:
Medium
Solution
verified
Verified by Toppr
Correct option is A)
Take △PQS
Applying pythaooras theorem, we get
PQ
2
+PS
2
=QS
2
⇒12
2
+5
2
=QS
2
⇒QS
2
=169
⇒QS=13
QS is a diameter to the circle.
Hence, radius of a circle will be
2
13
cm
Area of rectangle =l×b
=5×12=60 sq. cm
Area of circle =πr
2
=
7
22
×(
2
13
)
2
=
14
1859
sq. cm
Therefore, area of the shaded region =Area of circle−Area of rectangle=
14
1859
−60=
14
1019
sq. cm