Question

A semicircle is drawn with AB as its diameter. From C, a ppint on AB, a line perpendicular to Ab is drawn meeting the circumference of the semicircle at D. Given that AC = 2cm and CD =6cm, what is the area of the semicircle ???
A) 32 π
B) 50 π
C) 40 π
D) 36 π
Please explain it

Answer 1

Okay. I could not find a good diagram for this. So, you will have to visualise the solution.

I will re-phrase the problem statement a bit.

AB is the diameter of the semicircle. D is a point on the semicircle. A perpendicular line is drawn from point D to AB which meets AB at C.

AC = 2 units; [problem statement does not talk about the unit, so I will just keep it as unit only]

CD = 6 units;

<ACD=90degrees<ACD=90degrees

Now, connect A with D and B with D.

In ∆ACD,AD=√(AC)2+(CD)2;ACD,AD=√(AC)2+(CD)2;[Pythagorus theorem]

=>AD=√(22+62)=>AD=√(22+62)

=>AD=√40units=>AD=√40units

Now look at ∆ ABD

<ADB=90degrees<ADB=90degrees; [lines connecting diametrically opposite points of a circle to a point on circle makes right angle. Property of a circle]

AD = √40√40

AC = 2 units

Let’s assume CB = x

=> AB = (2+x) units

From Pythagorus theorem,

(AD)2+(DB)2=(AB)2(AD)2+(DB)2=(AB)2

=>40+y2=(2+x)2=>40+y2=(2+x)2 —————— eq(1)

From ∆DCB

(CD)2+(BC)2=(BD)2(CD)2+(BC)2=(BD)2

=>36+x2=y2=>36+x2=y2 —————— eq(2)

Now replace value of y2y2 from eq(2) in eq(1)

40+36+x2=(2+x)240+36+x2=(2+x)2

=>76+x2=x2+4x+4=>76+x2=x2+4x+4

=>4x=72=>4x=72

=>x=18=>x=18

We knownAB=AC+CBAB=AC+CB

=>AB=2+x=>AB=2+x

=>AB=2+18=>AB=2+18

=>AB=20units=>AB=20units

So radius of the semi circle = 10 units.

Now it is simple, right?

Area of the semi circle = ((π∗r2)/2π∗r2)/2

=> Area of the semi circle = (π∗102)/2=50π(π∗102)/2=50π

Hence your answer is 50π