Chapter 10.2, Problem 15PSC

(a)

To determine

To find: the radius of the circle.

Answer to Problem 15PSC

  $r=8$ .

Explanation of Solution

Given:

  

Concept used:

A perpendicular from the centre to the chord bisects the chord.

In right angle triangle Pythagoras theorem can be used as:

  $H^2=P^2+B^2$ .

Distance or radius cannot be negative.

Calculation:

According to the given:

  $\begin{array}{l}AB=AC,AF=6,ED=1.\\ \text{Also, }AD\perp BC\text{ and EF}\perp AC.\end{array}$

A perpendicular from the centre to the chord bisects the chord.

  $\begin{array}{l}\Rightarrow AF=FC=6\text{ and }BD=DC.\\ AB=AC=AF+FC=12.\end{array}$

Join $EC$ and consider:

  $\begin{array}{l}\Delta \text{'}sADC\text{ and }EDC.\\ \Rightarrow A{C}^2=A{D}^2+D{C}^2.\\ \Rightarrow D{C}^2=E{C}^2-E{D}^2.\end{array}$

By putting the main equation:

  $\begin{array}{l}A{C}^2={\left(AE+ED\right)}^2+E{C}^2-E{D}^2.\\ AE=EC=radii=r.\end{array}$

By putting the value:

  $\begin{array}{l}{12}^2=r^2+2r+1+r^2-1.\\ r^2+r-72=0.\\ \left(r+9\right)\left(r-8\right)=0.\end{array}$

Since, radius or distance cannot be negative therefore, positive value is taken.

Hence, $r=8$ .

(b)

To determine

To find: the perimeter of triangle $\Delta ABC$ .

Answer to Problem 15PSC

Perimeter is $24+6\sqrt{7}\text{unit}$ .

Explanation of Solution

Given:

  

Concept used:

Perimeter of the shape having sides are the sum of all distance of the respective sides.

In right angle triangle Pythagoras theorem can be used as:

  $H^2=P^2+B^2$ .

Distance or radius cannot be negative.

Calculation:

  $\begin{array}{l}r=8.\\ \Rightarrow D{C}^2=8^2-1^2.\\ \Rightarrow \sqrt{63}=3\sqrt{7}.\\ BC=6\sqrt{7}.\end{array}$

Perimeter:

  $\begin{array}{l}AB+BC+AC=12+6\sqrt{7}+12.\\ \Rightarrow 24+6\sqrt{7}.\end{array}$

Hence, perimeter is $24+6\sqrt{7}\text{unit}$ .