Chapter 9.1, Problem 3WE

(a)

To determine

To Draw: The right-angled triangle inscribed in a circle.

Explanation of Solution

Given information:

A right triangle inscribed in a circle.

Since, angle inscribed by a semicircle is always right angle therefore right triangle is inscribed in a circle is drawn below.

  

(b)

To determine

To Define: midpoint of the hypotenuse of the right triangle inscribed in a circle.

Answer to Problem 3WE

Explanation of Solution

Given:

A right triangle inscribed in a circle.

A right triangle is inscribed in a circle is drawn below.

  

In triangle ABC, hypotenuse is BC.

The mid-point of hypotenuse is O which is also the canter of the circle

(c)

To determine

To find: The location of centre of circle if a right triangle inscribed in a circle.

Explanation of Solution

Given information:

A right triangle inscribed in a circle.

A right triangle is inscribed in a circle is drawn below.

  

In triangle ABC, hypotenuse is BC.

O is the centre of circle and is the midpoint of the hypotenuseBC.

(d)

To determine

To find: The radius of circle having right triangle inscribed.

Answer to Problem 3WE

radius of circle is $5$ .

Explanation of Solution

Given:

A right triangle in a circle, the legs of the triangle is 6 and 8.

Formula Used

:Pythagorus theorem is used which is.

  $B{C}^2=A{B}^2+A{C}^2$

Radius is equal to the half of diameter in a circle.

Calculations

A right triangle is inscribed in a circle is drawn below.

  

In triangle ABC, hypotenuse is BC.

Diameter is hypotenuse BC.

  $\begin{array}{l}BC=\sqrt{A{B}^2+A{C}^2}\\ BC=\sqrt{6^2+8^2}\\ BC=\sqrt{100}\\ BC=10\end{array}$

Radius is.

  $\begin{array}{l}r=OB=OC=\frac{BC}{2}\\ =\frac{10}{2}\\ =5\end{array}$

Therefore, radius of circle is $5$ .