ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in A ABC. The radius of the circle is
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Answers
Solution :
In a right ΔABC, ∠B = 90°
BC = 6 cm, AB = 8 cm
AC² = AB² + BC² (Pythagoras Theorem)
= (8)² + (6)²
= 64 + 36
= 100² (10)
∴ AC = 10 cm
An incircle is drawn with centre O which touches the sides of the triangle ABC at P, Q and R
OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle
OP ⊥ AB, 0Q ⊥ BC and OR ⊥ CA
OPBQ is a square
Let r be the radius of the incircle
PB = BQ = r
AR = AP = 8 - r,
CQ = CR = 6 - r
AC = AR + CR
10 = 8 - r + 6 - r
10 = 14 - 2r
2r = 14 - 10
2r = 4
r = 4/2
r = 2
∴ Radius of the incircle = 2 cm (b)
Answer:
Using Pythagoras theorem in △ABC
(AB)
2
+(BC)
2
=(AC)
2
(8)
2
+(6)
2
=(AC)
2
⇒(AC)
2
=64+36=100
⇒AC=10cm
Now inradius of triangle =r=
s
Δ
where Δ is the area of triangle and s is semi perimeter
Δ=
2
1
×8×6=24
s=
2
8+6+10
=
2
24
=12
⇒x=r=
s
Δ
=
12
24
=2cm
Step-by-step explanation: