Question

PS is the diameter of a circle of radius 6cm. The points P and Q trisects the diameter PS. Semi circles are drawn on PQ and QS as diameter. Find the area of the shaded region.

Answer 1

PQ + QR + RS = PS

and PS = 12, diameter of the circle

Now, since Q & R Trisect the Diameter PS.

SO, PQ = QR= RS = 12/3 = 4cm.

Area of shaded region = area of semicircle with PQ as diameter + (Area of Semicircle with PS as diameter - Area of Semicircle with QS as diameter)

So area of Semicircle with PQ as diameter(A1) = π(6)²/2 = 36π = 18π

Area of circle with QR as diameter(A2) = π(4)²/2 = 16π = 8π

and, Area of semicircle with PQ as diameter(A3) = π(2)²/2 = 4π = 2π

Hence,

Area of shaded region = A3 + A1 -A2

= 2π + 18π - 8π = 12π cm²

[Question looks Complex but it is very easy, what we did is just Area of circle ( = πr²).]

Thankyou!!!

Answer 2

As PQ IS 12 cm
PQ TRISECTS THE DIAMETER PQ SO,
PS /3 = 2 CM
SO PQ=2 CM

WITH PQ AS DIAMETER
R= 1 CM
1/2.πR^2 (AREA OF SEMI CIRCLE)

IT IS 1.07 CM.................(1)

WITH PS AS DIAMETER
R =6 CM
ITS AREA IS 1/2.π.6^2
IT IS 56.57 cm..................(2)

WITH QS AS DIAMETER
R= 2CM
1/2.π2^2
IT IS 6.28 CM....................(3)

SUBTRACTING (3) FROM (2) AND ADDING (1).

IT GIVES THE AREA OF THE SHADED REGION. WHICH IS

51.36 cm square