# PQRS is a diameter of a circle of radius 6cm. The lengths PQ, QR and RS are equal. Semi circles are drawn on PQ and QS as diameters. Find the perimeter and area of the region so obtained. Solution : PS = 12 cm
As PQ = QR =RS
∴ PQ =QR =RS = 1/3 x PS = 1/3 x 12 = 4 cm.
QS = 2 PQ
QS = 2 x 4 = 8 cm
∴ Area of shaded region = Area of semicircle with PS as diameter + Area of semicircle with PQ as diameter – Area of semicircle with QS as diameter.
= ½ [ 3.14 x 62 + 3.14 x 22 - 3.14 x 42]
= ½ [ 3.14 x 36 + 3.14 x 4 – 3.14 x 16 ]
= ½ [ 3.14 ( 36 + 4 – 16)]
= ½ ( 3.14 x 24 ) = ½ x 75.36
∴ Area of shaded region = 37.68 cm2

• 98

Diameter of the circle i.e., PQRS = 2 * 6 = 12 cm

PORS is divided into 3 parts,therefore PQ = RS = QR = 1/3 *12  = 4 cm

area of semi circler with PQ as diameter = 1/2/pie r2

• 1/2 * 22/7 * 22
• 22/7 * 2
• 44/7                      ... ( i )

Length of QS = 2/3 * 12 =8 cm

area of semi circle with QS as diameter = 1/2 pie r

• 1/2 * 22/7 * 42
•  22/7 * 2*4
• 22/7 * 8
• 176/7                  ... ( ii )

area of semicircle with PQRS as diameters = 1/2 pie r2

• 1/2 * 22/7 * 12 *12
• 22/7 * 6 *12
• 22/7 * 72
• 1584 / 7                                          ... ( iii )

area of semicircle with PQRS as diameter - area of semicircle with QS as diameter  + area of semi circle with PQ as diameter

1584/7 - 176/7 + 44/7

= 1408 / 7 + 44/7

= 1452/ 7

Hope its correct

• -26

Diameter of circle(PQRS)=6*2=12cm

PQ=QR=RS=12/3=4cm

QS=QR+RS

=4+4

=8cm

=Arc of semicircle of radius 6cm+

Arc of semicircle of radius 4cm+

Arc of semicircle of radius 2cm

=2πr1/2+2πr2/2+2πr3/2

=π(6+4+2)

=π(12)

=12 πcm2

=Area of semicircle of radius 6cm+

Area of semicircle of radius 2cm-

Area of semicircle of radius 4cm

=πr12/2+πr22/2+πr32/2

=π/2(r12 +r3-r22 )

=π/2[(6)2+(2)2-(4)2]

=π/2(36+4-16)

= π/2*24

=12 πcm2

Therefore,

=264/7

=37.71cm2