Chapter 10.6, Problem 3PSA

To determine

To Find: $\text{Radius of the circle}\odot P$

Answer to Problem 3PSA

  $10\text{mm}$

Explanation of Solution

Given information:

1) $\text{AC = 12 mm, Ba = 16 mm}$

2) $\overline{\text{BC}}\text{is a diameter}$

Formula Used:

1) $\text{Angle inscribed by the diameter is 90°}\text{.}$

2) ${\text{In a right angled triangle P}}^{\text{2}}{\text{+ B}}^{\text{2}}{\text{= H}}^{\text{2}}$

  $\begin{array}{l}\because \overline{BC}\text{ is a diameter hence }\angle \text{A = 90°}\\ \text{In right angled triangle }\Delta \text{ABC}\\ A{C}^2+B{A}^2=B{C}^2\\ {12}^2+{16}^2=B{C}^2\\ B{C}^2=144+256\\ B{C}^2=400\\ BC=20\text{mm}\\ \text{Now if Diameter = 20 mm}\\ \text{then Radius = 10 mm}\end{array}$