Question

Find the area of a regular
hexagon MNOPQR of side 5 cm​

Answer 1

Answer:

Area of trapezium MNQR

=4× 11+5/2

=2×16=32cm²

The area of hexagon MNOPQR =2×32 = 64cm²

∆MNO and ∆RPQ are congruent triangles with altitude 3cm.

Area of MNO=8×3 =12cm²

Area of ∆RPQ

Area of rectangle MOPR =85=40cm²

Area of hexagon MNOPQR =40+12+12= 64cm²