Question

3. A circle is inscribed in a triangle with sides 8, 15 and 17 cm. The radius of
circle is
(a) 6 cm
(b) 5 cm
(c) 4 cm
30​

Answer 1

A circle is inscribed in a triangle with sides 8, 15 and 17 cm.

We have to find Radius of circle.

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☯ Let the triangle be ABC.

Where,

• AB = 8 cm

• BC = 15 cm

• AC = 17 cm

★ Applying Pythagoras theorem,

➯ H² = B² + P²

➯ AC² = AB² + AC²

➯ 17² = 8² + 15²

➯ 289 = 64 + 225

➯ 289 = 289

Therefore, LHS = RHS

So, We can say that ∆ABC is a right angled triangle.

We know that,

Area of ∆ = 1/2 × b × h

Therefore,

Area of ∆ABC,

Putting values in formula -

➯ 1/2 × 8 × 15

➯ 60 cm²

☯ Let r be the radius of incircle whose centre is O.

Now,

➯ ar( ∆AOB + ∆BOC + ∆COA)

➯ 1/2 × AB × r + 1/2 × BC × r + 1/2 × AC × r

➯ 1/2 × r × (AB + BC × AC)

Putting values,

➯ 1/2 × r × (8 + 15 + 17)

➯ 1/2 × r × 40

➯ 20r

We know that,

➯ ar( ∆AOB + ∆BOC + ∆COA) = ar( ∆ABC)

➯ 20r = 60

➯ r = 60/20

➯ r = 3 cm

∴ Radius of Circle is 3 cm.

Answer 2

Given ,

A circle is inscribed in a triangle with sides 8 , 15 and 17 cm

We know that , the pythagoras theorem is given by

$\boxed{ \tt{ {(h)}^{2} = {(b)}^{2} + {(p)}^{2} }}$

Thus ,

(17)² = (8)² + (15)²

289 = 64 + 225

289 = 289

$\therefore$

The given triangle is right angled triangle

We know that , the area of triangle is given by

$\boxed{ \tt{ Area = \frac{1}{2} \times base \times height}}$

Thus ,

Ar( ΔABC) = 1/2 × 8 × 15

Ar( ΔABC) = 4 × 15

Ar( ΔABC) = 60 cm²

Now ,

Ar( ΔABC) = Ar( ΔAOB + ΔBOC + ΔAOC)

Thus ,

$\tt \implies 60 = \{ \frac{1}{2} \times AB \times r \} + \frac{1}{2} \times BC \times r \} + \{ \frac{1}{2} \times AC \times r \}$

$\tt \implies 60 = \frac{r}{2} \{AB + BC + AC \}$

$\tt \implies 120 = r \{ 15 + 8 + 17\}$

$\tt \implies 120 = r \{40 \}$

$\tt \implies r = \frac{120}{40}$

$\tt \implies r = 3 \: \: cm$

Therefore , the radius of circle is 3 cm

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