Name: Tangents From an Exterior Point: Illustration
Uploaded: 2019-07-19
Duration: 4 min 47 s
Description: If you lift a circular object with a pair of tongs, then the tongs are two tangents to the circular object. Let us study the properties of tangents drawn fro...

Question

Draw a triangle of sides

4

centimetres,

5

centimetres,

6

centimetres and draw its incircle. Calculate its radius.

Accepted Answer

$1 .$ Draw a line $6$ cm

$2 .$ Taking B as centre, draw an arc of radius $5$ cm, and then taking $C$ as centre, draw an arc of $6$ cm which intersects the previous arc at $A$ .

$A B C$ is the required triangle.

$3 .$ Draw the angle bisector of $∠ A B C$ and $∠ A C B$ , which intersect each other at $O$ .

$4 .$ Draw $O P ⊥ B C$ , and taking $O$ as centre, draw a circle with a radius as $O P$ .

Calculation of radius:

We know, the radius of the incircle $= \frac{A}{S}$

Where, $A =$ area of the triangle

And $S =$ semi-perimeter of the triangle

Also, We know, area of the triangle $= \sqrt{(s (s - a) (s - b) (s - c))}$

Where, $a, b$ and $c$ are sides and $S$ is the semi-perimeter calculated as

$S = \frac{a + b + c}{2}$

$S = \frac{4 + 5 + 6}{2}$

$S = \frac{15}{2} = 7.5$

Area $= \sqrt{(7.5 (7.5 - 4) (7.5 - 5) (7.5 - 6)}$

$= \sqrt{(7.5 \times 3.5 \times 2.5 \times 1.5)}$

$= \sqrt{\frac{15}{2} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2}} = \frac{15}{4} \sqrt{7}$ cm

Therefore, the radius of the incircle $= \frac{\frac{15}{4} \sqrt{7}}{\frac{15}{2}} = \frac{\sqrt{7}}{2}$ cm.

Kerala Board Solutions for Chapter: Tangents, Exercise 6: Exercise 6

Kerala Board Mathematics Solutions for Exercise - Kerala Board Solutions for Chapter: Tangents, Exercise 6: Exercise 6

Questions from Kerala Board Solutions for Chapter: Tangents, Exercise 6: Exercise 6 with Hints & Solutions

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