Question

In a abc , ab = 8cm , ïƒabc = 90 . then find the radius of the circle inscribed in the triangle.

Answer 1

In a ΔABC, the radius of the circle inscribed in the triangle is 2cm.

• Given,
• AB=8cm
• ⇒a=8cm
• BC=6cm
• ⇒b=6cm
• Since the triangle is right angled triangle, as ∠B=90°
• now,
• $c=\sqrt{a^2+b^2} \\\\c=\sqrt{8^2+6^2} \\\\c=10$
• we use the formula,
• radius of circle = area of triangle / semi perimeter of triangle
• $r=\dfrac{\frac{1}{2}*6*8}{\frac{6+8+10}{2}}\\\\r=\dfrac{24}{12}\\\\r=2 cm$

Answer 2

Given : In a right triangle abc, AB = 8 cm and AC = 6 cm. Find the radius of its in-circle.

To find : Radius of the circle inscribed in the triangle.

Solution :

AB = 8 cm  ;  AC = 6 cm.

By applying Pythagoras theorem,

BC² = AB² + AC²

BC² = 8² + 6²

       = 64 + 36    

BC² = 100

BC = 10 cm.

Formula :

Radius of triangle = $\frac{\text { Area of triangle }}{\text { Semi-perimeter of triangle }}$

Area of triangle = $\frac{1}{2}\times\{b\}\times\{h\}$

Semi-perimeter of triangle = $\frac{a+b+c}{2}$

Finding radius of the circle inscribed in the triangle :  

Area of triangle = $\frac{1}{2}\times\{b\}\times\{h\}$

                          = $\frac{1}{2}\times\{6\}\times\{8\}$

                          = $\frac{1}{2}\times\{48\}$

Area of triangle = 24                                      

Semi-perimeter of triangle = $\frac{a+b+c}{2}$

a = 8 ; b = 6 ; c = 10

                                            = $\frac{8+6+10}{2}$

                                            = $\frac{24}{2}$

Semi-perimeter of triangle = 12.

Radius of triangle = $\frac{\text { Area of triangle }}{\text { Semi-perimeter of triangle }}$

                              = $\frac{24}{12}$

Radius of triangle = 2 cm.

Therefore, the radius of the circle inscribed in the triangle is 2 cm.

To learn more...

brainly.in/question/2618477