Question

Find the area of the hexagon MNOPQR given below. Given that: MP = 8 cm, MJ = 6 cm, MI = 5 cm, MH = 3 cm, MG = 2.5 cm and RG, NH, QI and OJ are perpendiculars on diagonal MP from the vertices R, N, Q and O respectively. with steps

Answer 1

AB=MB−MA=4−2=2cm

AC=MC−MA=6−2=4cm

BD=MD−MB=7−4=3cm

CP=MP−MC=9−6=3cm

DP=MP−MD=9−7=2cm

Area of trapezium=

2

1

h(a+b), where a and b are parallel sides and h is the height of the trapezium.

Area of triangle=

2

1

bh where b is the base of the triangle and h is the height of the triangle.

Area(MNOPQR)=Area(MNA)+Area(ANOC)+Area(COP)+Area(DQP)+Area(BRQD)+Area(MBR)

⇒Area(MNOPQR)=

2

1

(2×2.5+4×(3+2.5)+3×3+2×2+3×(2+2.5)+4×2.5)

⇒Area(MNOPQR)=

2

1

(5+22+9+4+13.5+10)=

2

63.5

cm

2

=31.75cm

2