Question

In the given figure, triangle ABC is circumscribed touching the circle at P, Q, R if AP = 4 cm, BP = 6 cm, ACC= 12 cm, then find radius of circle. ​

Answer 1

Step-by-step explanation:

please show the figure of the question

Answer 2

Final answer: The radius of the circle $=\frac{4\sqrt{6}}{3}$  $cm$

Given that: We are given,

I) Δ ABC is circumscribed touching the circle at P, Q, R.

II) In Δ ABC, AP = 4 $cm$, BP = 6 $cm$, AC= 12 $cm$

To find: We have to find the radius of circle.

Explanation:

• First draw a diagram with the given details. That diagram shown bellow.

• We have AP = 4 $cm$, BP = 6 $cm$ and AC = 12 $cm$.

• From figure, OP = OR = OQ = r.

• Where, r = radius of the circle.

• We know that, length of the tangents that drawn from an external point of a circle, are equal.

Tangent from A: AP = AR = 4 $cm$

Tangent from B: BP = BQ = 6 $cm$

Tangent from C: CR = CQ = ?

• CR =  AC - AR

CR = 12 - 4 = 8 $cm$

• CR = CQ = 8 $cm$

• BC = BQ + QC = 6 + 8 = 14 $cm$

• AB = AP + PQ = 4 + 6 = 10 $cm$

• AC = 12 $cm$

• We have to calculate the semi perimeter of Δ ABC.

Semi perimeter, $s= \frac{a+ab+c}{2}$

Where,

a, b, c = Length of three sides of the triangle.

Semi perimeter of Δ ABC:

$s = \frac{AB+BC+AC}{2}$

$s=\frac{10+14+12}{2}=\frac{36}{2} =18$ $cm$

• The area of the Δ ABC can be calculated by using Heron's formula:

Area of the triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

Where,

s = Semi perimeter of triangle

a, b, c = Length of three sides of the triangle.

Area of Δ ABC $=\sqrt{s(s-AB)(s-BC)(s-AC)}$

                         $=\sqrt{18(18-10)(18-14)(18-12)} \\\\ =\sqrt{18(8*6*4)}\\\\=\sqrt{ 18*192}\\\\=\sqrt{3456}\\=\sqrt{6*576}\\\\=24\sqrt{6}$

Area of Δ ABC $=24\sqrt{6}$ $cm^{2}$

• Area of Δ ABC = Area of Δ AOB+ Area of Δ BOC+ Area of Δ AOC    

                                   $=24\sqrt{6}$ $cm^{2}$

Area of a triangle $= \frac{1}{2}* base length*height length$

Area of Δ AOB $= \frac{1}{2}*AB*OP = \frac{1}{2} *10*r = 5r$ $cm$

Area of Δ BOC $= \frac{1}{2}*BC*OQ = \frac{1}{2} *14*r = 7r$ $cm$

Area of Δ AOC  $= \frac{1}{2}*AC*OR = \frac{1}{2} *16*r = 8r$ $cm$

Hence, area of Δ ABC  $= 5r+7r+6r =24\sqrt{6}$ $cm^{2}$

     $18r=24\sqrt{6}$

        $r=\frac{24\sqrt{6}}{18}\\\\r=\frac{(6*4)\sqrt{6}}{(6*3)} \\\\r= \frac{4\sqrt{6}}{3}$

• Hence radius of the circle $=\frac{4\sqrt{6}}{3}$  $cm$

To know more about the concept please go through the links

https://brainly.in/question/15230584

https://brainly.in/question/6190680

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