Question

In triangle ABC, AB=6,BC=8,AC=10, then what is the radius of the circle touching all the three sides ?

Answer 1

Hlw mate!!

Area ABC =√( 12*2*4*6) = 24 (cm)^2 using Heron's formula.

Area ABC is also = (1/2) 10*8 * sin B= 24, and so sin B = 0.6., B = 36.87 °

The line perpendicular to AB at X cuts BC at D. Let DX = h & BX = x.

Area BXD =(1/2) x h = 12 (cm)^2. ( given condition)

h = x tan B = 0.6 x.

Area BXD =(1/2) x ( 0.6x) = 12

x = BX = 6.33 cm

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Verify the result.

h = 6.33* sin B = 6.33* 0.6 = 3.798

Area (1/2) 6.33 *3.798 = 12 cm ^2

Answer 2

In triangle ABC, AB=6, BC=8, AC=10, then 1 is the radius of the circle touching all three sides.

Step-by-step explanation:

By Pythagoras theoram,

$AC^2 = AB^2 +BC^2$

Given in Δ ABC, AB= 6, BC= 8, AC=10

$10^2 =6^2 +8^2\\100=36+64\\100=100$

Δ ABC is a right-angled triangle and ∠B is a right angle.

We know that the radius of the circle through all the sides is a $(AB+BC-AC)/2$

The required radius of the circle $=(6+8-10)/2=12-10/2 =1$

Hence, the radius of the circle is 1.

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