Question

In the given figure if angle b is 90 degree BC is equal to 15 CM and AC is equal to 17 cm then find the radius of the circle with centre

Answer 1

Answer:

3cm

Step-by-step explanation:

BC = 15cm ,AC = 17cm

AB = √(17^2 - 15^2) = √64 = 8cm

radius of circle

= (8 + 15 - 17)/2 = 6/2 = 3 cm

or

radius of circle = 2 × area of ∆ABC/perimeter of ∆ABC

=2* 1/2*8*15/(8+15+17)= 3cm

Answer 2

The radius of circle=3 cm

Step-by-step explanation:

$\angle B=90^{\circ}$

$BC=15 cm$

AC=17 cm

Pythagoras theorem

$(hypotenuse)^2=(base)^2+(Perpendicular\;side)^2$

By using Pythagoras theorem

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

Substitute the values then we get

$(17)^2=AB^2+(15)^2$

$289=AB^2+225$

$AB^2=289-225=64$

$AB=\sqrt{64}=8$

AB=8 cm

Perimeter of triangle =Sum of sides of triangle=$AB+BC+AB=15+17+8=40 cm$

Area of triangle=$\frac{1}{2}\times base\times height$

Area of triangle ABC=$\frac{1}{2}\times 15\times 8=60 cm^2$

Radius of circle=$\frac{2\cdot area\;of\;triangle}{perimeter\;of\;triangle}$

Radius of circle=$\frac{2\times 60}{40}=3 cm$

Hence, the radius of circle=3 cm

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