Q3. - Geometry

Question

 a. Draw a right triangle inscribed in a circle. 

 b. What do you know about the midpoint of the hypotenuse?

 c. Where is the center of the circle?

 d. If the legs of the right triangle are $6$ and $8$. Find the radius of the circle.

Step-by-Step Solution

Answer

a. 

b.  Midpoint of hypotenuse is centre of circle.

c.  $O$ is center of circle and  midpoint of hypotenuse $BC$.

d.  radius of circle is $5$. 

a.

Draw 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 . Step 1. Given:

A right triangle inscribed in a circle.

Step 2: solution:

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

In $\Delta ABC$, hypotenuse is $BC$.

The mid-point of hypotenuse is $O$ which the center of the circle is also. 

c. Step 1. Given:

A right triangle inscribed in a circle.

Step 2. Solution:

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

In $\Delta ABC$, hypotenuse is $BC$.

$O$ is the centre of circle and is the midpoint of the hypotenuse $BC$.

d. Step 1. Given:

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

Step 2. Concept used:

Pythagoras 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.

Step 3. Solution:

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

In triangle $\Delta ABC$, 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.

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

Therefore, radius of circle is $5$.