Question

find the area of the hexagon abcdef fig ac= 8cm,cd = 4cm bm = 3cm ne =10.2 do =5cm af =8.4 op=3.2cm​

Answer 1

Answer:

Step-by-step

ONCD is a square

\therefore NC=CD=OD=ON= 5 cm∴NC=CD=OD=ON=5cm

AC= 8 cm

AC= AN+NC

8= AN+5

\therefore AN=3 cm∴AN=3cm

A(\Delta ABC)=\frac{1}{2}\times AC\times BMA(ΔABC)=  

2

1

​  

×AC×BM

=\frac{1}{2}\times 8\times 3=12=  

2

1

​  

×8×3=12

A(\square ONCD)= (side)^{2}=5^{2}=25A(□ONCD)=(side)  

2

=5  

2

=25

NE=NO+OP+PE

10.2+5+3.2+PE

\therefore PE=2∴PE=2

A(\square ANPF)A(□ANPF) = length \times× breadth       (\because \square ANPF(∵□ANPF is a rectangle)

=AF\times AN=AF×AN                            \therefore (AN=PF)(AF=NP)∴(AN=PF)(AF=NP)

=8.4\times 3=8.4×3

=25.2=25.2

A(\Delta ODE)=\frac{1}{2}\times (OD\times DE)A(ΔODE)=  

2

1

​  

×(OD×DE)

=\frac{1}{2}\times 5\times (OP+PE)=  

2

1

​  

×5×(OP+PE)

=\frac{1}{2}\times 5\times (3.2+2)=\frac{1}{2}\times 5\times 5.2=  

2

1

​  

×5×(3.2+2)=  

2

1

​  

×5×5.2

=8

A(\Delta FPE)=\frac{1}{2}\times FP\times PEA(ΔFPE)=  

2

1

​  

×FP×PE

=\frac{1}{2}\times 3\times 2=3=  

2

1

​  

×3×2=3

\therefore∴ Area of hexagon ABCDEF=A(\Delta ABC)+A(\Delta FPE)+A(\Delta OPE)+A(\square ONCD)+A(\square ANPF)ABCDEF=A(ΔABC)+A(ΔFPE)+A(ΔOPE)+A(□ONCD)+A(□ANPF)

=12+3+8+25+25.2=12+3+8+25+25.2

=73.2 cm^{2}=73.2cm