In the figure given below, ABC is a right angle triangle where AB = 7 cm and BC = 10 cm.
Given that AEB and BCD are isoscales right angle triangle. Area of the shaded region is
[NSTSE 2011]
C
B
72
Answers
Step-by-step explanation:
ABC is a right angled triangle where ∠A=90
∘
BC=10cm and AB=6cm
Let O be the centre and r be the radius of the in-circle.
AB,BC and CA are the tangents to the circle at P,M and N
∴IP=IM=IN=r(radius of the circle)
In △BAC,
BC
2
=AB
2
+AC
2
(by pythagoras theorem)
⇒10
2
=6
2
+AC
2
⇒AC
2
=100−36=64
∴AC=8cm
Area of △ABC=
2
1
bh=
2
1
×AC×AB=
2
1
×8×6=24sq.cm
Area of △ABC=Area of △IAB+Area of △IBC+ Area of △ICA
⇒24=
2
1
r(AB)+
2
1
r(BC)+
2
1
r(CA)
⇒24=
2
1
r(AB+BC+CA)
⇒24=
2
1
r(6+8+10)
⇒24=12r
∴r=
12
24
=2cm
Area of the circle=πr
2
=
7
22
×2
2
=12.56sq.cm
Area of shaded region=Area of △ABC−Area of the circle.
=24−12.56=11.44sq.cm
Answer:
85 1/7 cm2
Step-by-step explanation:
85 1/7 cm2