Question

In ABC, circumradius (R), inradius (r) and exradius (r ) are 13/2

, 2, 3 respectively, then

the area of triangle is ..... sq.units.

1) 10

2) 20

( 3) 30

A 40

​

Answer 1

Answer:

20

Explanation:

Let △ABC be a triangle whose sides are of lengths a,b,c.

Then the area A of △ABC is given by:

A=abc4R

where R is the circumradius of △ABC.

Answer 2

The are of triangle is 30 sq. units

Given:

Circumradius(R), Inradius(r) , ex- radius (r1)

To Find:

The area of triangle in sq. units using circumradius, inradius, ex - radius

Solution:

We have,

$\frac{1}{r1} +\frac{1}{r2} + \frac{1}{r3}$ = $\frac{1}{r}$

$\frac{1}{r2} + \frac{1}{r3} = \frac{1}{r} - \frac{1}{r1}$

= $\frac{1}{2} - \frac{1}{3}$

$\frac{r2+r3}{r2r3} = \frac{1}{6}$ ..............(1)

and  r1+r2+r3 =  r + 4R

⇒ 3+r2+r3 = 2+26

r2+r3 = 25 ...........(2)

From (1) and (2) ,

We get,

r2r3 = 150

r2+r3 = 25

∴ r2-r3 = 5 ............(3)

From (2) and (3) ,

r2 = 15

r3 = 10

Again, $\frac{1}{r1} +\frac{1}{r2} + \frac{1}{r3}$ = $\frac{1}{r}$

⇒ r1r2+r2r3+r3r1 = $\frac{r1r2r3}{r}$

$s^{2}$ = $\frac{r1r2r3}{r}$

$s^{2}$ = 225

s = 15

Now, r =Δ/s, r = Δ/s-a

$\frac{s-a}{s} = \frac{r}{r1}$

= $\frac{2}{3}$

a = 5

Similarly,

b=13, c=12

Triangle is right angled

Area of ΔABC = $\frac{abc}{4R} = 30 sq.units$

The area of the triangle is 30 sq. units

#SPJ3