Question

Find the area of the hexagon shown here Given that MP = 12 cm, MD = 8 cm MC = 6 cm MB = 4cm, MA= 2 cm, AN = 5 cm oc=5 cm DQ = 6 cm and BR = 6 cm​

Answer 1

Answer:

Step-by-step explanation:

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