Question

find the area of polygon PQRSTU shown in adjoining figure, if PS=11cm, PY=9cm, PX=8cm, PW=5cm, PV=3cm, QV=5cm, UW=4cm, RX=6cm, TY=2cm.

Answer 1

According to question
It is given that
PS = 11 Cm
PY  = 9 Cm
PX = 8 Cm
PW = 5 Cm
PV = 3 Cm
QV = 5 Cm
QW  =4 Cm
RX  = 6 Cm
TY = 2 Cm

Thus Total re of the polygon  = Ar( PVQ ) +  Ar (PUW) +  Ar ( SYT) + Ar(RSX) +  Ar (QVXR) + Ar( WXTU)
         Area = 1/2 * 5 * 3 + 1/2 * 5 * 4 +  1/2 * 2 * 2 +  1/2 * 3 * 6 +  12 * 5 *1 +  5*5 +  1/2 * 4 * 2 +  4 * 2

      Area = 15/2 + 10 + 2 + 9 + 60 + 25 + 4 + 8
             = 125.5 Cm^2

Answer 2

Solution :-

1) Area of Δ PQV = 1/2*Base*Height

PV = Base = 3 cm and QV = Height = 5 cm

⇒ Area = 1/2*3*5

= 15/2

= 7.5 sq cm

2) Area of Trapezium VQRX = 1/2*(Sum of two parallel sides)*height

Px - PV = VX

⇒ 8 cm - 3 cm = VX (Height) 

= 5 cm

Two parallel sides are QV = 5 cm and RX = 6 cm

⇒ 1/2*(6 + 5)5

⇒ 1/2*11*5

⇒ 55/2

= 27.5 sq cm

3) Area of Δ RSX = 1/2*Base*Height

PS - PX = XS (Base)

⇒ 11 cm - 8 cm = 3 cm

So, XS = Base = 3 cm and RX = Height = 6 cm

⇒ 1/2*3*6

⇒ 18/2

= 9 sq cm

4) Area of Δ STY = 1/2*Base*Height

PS - PY = SY (Base)

⇒ 11 - 9 = SY (Base) = 2 cm

Height = TY = 2 cm

1/2*2*2

= 2 sq cm

5) Area of trapezium TUWY = 1/2(Sum of two parallel side)*Height

Two parallel sides = UW = 4 cm and TY = 2 cm

 PY - PW = WX (Height) 

⇒ 9 - 5 = WX = 4 cm (Height)

Area of trapezium = 1/2*(4 + 2)*4

⇒ 1/2*6*4

⇒ 24/2

= 12 sq cm

6) Area of Δ PUW = 1/2*Base*Height

⇒ UW = 4 cm (Base) and PW (Height) = 5 cm

Area = 1/2*4*5

⇒ 20/2

= 10 sq cm

Area of the figure PQRSTU = Area of Δ PQV + Area Δ of Trapezium VQRX +

Area of Δ RSX + Area of Δ STY + Area of Trapezium TUWY + Area of Δ

PUW

⇒ 7.5 + 27.5 + 9 + 2 + 12 + 10

⇒ 68 sq cm

Answer.