Question

A semi-circle is drawn with $AB$ as its diameter. From $C$, a point on $AB$, a line perpendicular to $AB$ is drawn meeting the circumference of the semi-circle at $D$. Given that $AC = 2 \ \text{cm}$ and $CD = 6 \ \text{cm}$, the area of the semi-circle (in sq. cm) will be:

• A. $32\pi$
• B. $50\pi$
• C. $40.5\pi$
• D. $81\pi$
• E. undeterminable

Hint:

When a perpendicular is drawn from a point on the diameter to the arc, use the radius property in the right-angled triangle to set up a Pythagoras relation.

Answer 1

The Correct Option is B

Let the center of the semicircle be $O$ and radius $R$. $AB$ is the diameter, so $AO = OB = R$. $C$ is a point on $AB$ such that $AC = 2 \ \text{cm}$. From $C$, $CD$ is perpendicular to $AB$ and $CD = 6 \ \text{cm}$ meets the semicircle at $D$. In right triangle $OCD$: $OC = R - AC = R - 2$ (since $O$ is between $A$ and $B$ and $C$ is between $A$ and $O$). Using Pythagoras: \[ OD^2 = OC^2 + CD^2 \] But $OD = R$ (radius). Therefore: \[ R^2 = (R - 2)^2 + 6^2 \] \[ R^2 = R^2 - 4R + 4 + 36 \] \[ 0 = -4R + 40 \quad \Rightarrow \quad R = 10 \ \text{cm} \] Area of semicircle: \[ \text{Area} = \frac{1}{2} \pi R^2 = \frac{1}{2} \pi (10^2) = 50\pi \ \text{sq. cm} \] \[ \boxed{50\pi} \]