The incircle of triangle ABC touches sides AB, BC and CA at P, Q and R respectively. If AP = BQ = CR = 2 cm, then which of the following statements are true?

Perimeter of $Δ A B C$ = 12 cm

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$∠ B = 60^{\circ}$

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BP + CQ + AR = 6 cm

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$△ A B C$ is equilateral

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Solution

The correct options are
A
Perimeter of $Δ A B C$ = 12 cm

B
$∠ B = 60^{\circ}$

C
BP + CQ + AR = 6 cm

D
$△ A B C$ is equilateral

We know that tangents from an external point are equal.

So, AP = AR = 2 cm
BP = BQ = 2 cm
CQ = CR = 2 cm

BP + CQ+ AR = 6cm

AB = BC = CA= 4 cm

$△ A B C$ is equilateral

$∠ B = 60^{\circ}$

Perimeter of $△ A B C$ =3 x 4 = 12 cm

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Similar questions

In the given figure, the incircle of $△ A B C$ touches the sides AB, BC and CA at the points P, Q, R respectively. Show that
$A P + B Q + C R = B P + C Q + A R$
$= \frac{1}{2} (P e r i m e t e r o f △ A B C)$
[3 MARKS]

Hello

Please help me with the following

The sides of a triangle touch a circle. Sides AB, BC, CA touch the circle at points P, Q, R respectively. Prove that:

1. AP+BQ+CR=BP+CQ+AR

2.AP +BQ+CR=1/2 *perimeter of 🔺 ABC

Q. If the incircle of

△ A B C

touches the sides

A B, B C

and

C A

P, Q

and

R

, find the value of :

A P + B Q + C R

Q. The incircle of

△ A B C

having integral side lengths and least perimeter, touches

A B, B C, C A

R, P, Q

respectively. If

\frac{2}{A R} + \frac{5}{B P} + \frac{5}{C Q} = \frac{6}{r}

, where

r

is the radius of the incircle, then

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