MCQ
$ABC$ 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 $\triangle ABC$. The radius of the circle is:
- A$1\ cm$
- ✓$2\ cm$
- C$3\ cm$
- D$4\ cm$
✓
Answer
Correct option: B.
$2\ cm$
In a right $\triangle\text{ABC},$ $\angle\text{B}=90^{\circ}$
$BC = 6\ cm, AB = 8\ cm$
$BC = 6\ cm, AB = 8\ cm$
$A C^2=A B^2+B C^2$ (Pythagoras Theorem)
$=(8)^2+(6)^2=64+36=100=(10)^2$
$AC = 10cm$
An incircle is drawn with centre 0 which touches the sides of the triangle $ABC$ at $P, Q$ and $R ,OP, OQ$ and $OR$ are radii and $AB, BC$ an $CA$ are the tangents to the circle.
$OP ⊥ AB, OQ ⊥ BC$ and $OR ⊥ CA$
$OPBQ$ is a square.
Let $r$ be the radius of the incircle.
$PB = BQ = r$
$AR = AP = 8 – r,$
$CQ = CR = 6 – r$
$AC = AR + CR$
$\Rightarrow 10 = 8 – r + 6 – r$
$\Rightarrow 10 = 14 – 2r$
$\Rightarrow 2r = 14 – 10 = 4$
$\Rightarrow r = 2$
Radius of the incircle $= 2cm.$
$=(8)^2+(6)^2=64+36=100=(10)^2$
$AC = 10cm$
An incircle is drawn with centre 0 which touches the sides of the triangle $ABC$ at $P, Q$ and $R ,OP, OQ$ and $OR$ are radii and $AB, BC$ an $CA$ are the tangents to the circle.
$OP ⊥ AB, OQ ⊥ BC$ and $OR ⊥ CA$
$OPBQ$ is a square.
Let $r$ be the radius of the incircle.
$PB = BQ = r$
$AR = AP = 8 – r,$
$CQ = CR = 6 – r$
$AC = AR + CR$
$\Rightarrow 10 = 8 – r + 6 – r$
$\Rightarrow 10 = 14 – 2r$
$\Rightarrow 2r = 14 – 10 = 4$
$\Rightarrow r = 2$
Radius of the incircle $= 2cm.$
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