Question

Find the area of polygon ABCDEF, if AD = 18cm, AQ = 14cm, AP = 12cm, AN = 8cm, AM =

4cm, FM = 5cm, EP=6 cm, QC=4 cm and BN=5 cm and FM, EP, QC and BN are

perpendiculars to diagonal AD.​

Answer 1

Answer:

I know you have got the answer so it is very easy e area means sum of all the side then 18 + 14 + 12 + 8 it is the answer of first and second 4 + 5 + 6 + 4 + 5 is answer of 2nd

Answer 2

The area of polygon ABCDEF is 118cm^2.

Step-by-step explanation:    

Given:

AD=18cm, AQ=14cm, AN=8cm, AM=4cm

QD=AD-AQ=18cm-14cm=4cm

NQ=AQ-AN=14cm-8cm=6cm

PD=AD-AP=18cm-12cm=6cm

MP=AP-PM=12cm-4cm=8cmM

$Area of \triangle AMF=\frac{1}{2} \times 5cm\times4cm=10cm^2$

$Area of \triangle EPD=\frac{1}{2} \times 6cm\times6cm=18cm^2$

$Area of \triangle QCD=\frac{1}{2} \times 4cm\times4cm=8cm^2$

$Area of \triangle BCQN=\frac{1}{2} \times (5+4)cm\times6cm=27cm^2$

$Area of \triangle EMFQ=\frac{1}{2} \times (5+6)cm\times6cm=55cm^2$

Area of polygon ABCDEF=$Area of \triangle AMF+Area of \triangle EPD+Area of \triangle QCD+Area of BCQN+Area of EMFQ$

=$(10+18+8+27+55)cm^2$

=$118cm^2$