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)
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✠ Question Given :
- 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)
✠ Required Solution :
✩ Values Given to us :
- ➼ PS = 5 cm
- ➼ PQ = 12 cm
- ➼ (Use π =22/7)
✩ Construction Needed :
- ➼ Draw a Diagonal in given figure in rectangle PQRS. As shown in Figure Attached Above !
✠ In ∆ PQS
✩ By Pythagoras Theorum :
- ➼ H² = P² + B²
- ➼ SQ² = PQ² + PS²
- ➼ SQ² = (12)² + (5)² cm
- ➼ SQ² = 144 + 25 cm
- ➼ SQ² = 169 cm
- ➼ SQ² = √ 169 cm
- ➼ SQ² = 13 cm [ Hypotenuse ]
✩ Diagonal = Diameter
- ➼ Radius of Circle = 13 / 2
✩ Area of Circle = π r²
- ➼ Area = 22/ 7 × (13/ 2)²
- ➼ Area = 132.66 cm²
✩ Area of Rectangle PQRS :
- ➼ Area = l × B
- ➼ Area = 12 × 5 cm
- ➼ Area = 60 cm²
✩ Area of Unshaded Region :
- ➼ Area of Circle - Area of Rectangle
- ➼ Area of Unshaded region = 132.66 - 60 cm²
- ➼ Area of Unshaded region = 72.66 cm²
Area of Unshaded region = 72.66 cm²
_____________________________
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Answer:
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
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