Question

Find the area of polygon ABCDEF, if AD = 18 cm, AQ = 14 cm, AP = 12 cm, AN = 8 cm, AM = 4 cm, and FM, EP, QC and BN are perpendiculars to diagonal AD.


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Answer 1

Answer:

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=8cm

Area of △AMF=

2

1

×5cm×4cm=10cm

2

Area of △EPD=

2

1

×6cm×6cm=18cm

2

Area of △QCD=

2

1

×4cm×4cm=8cm

2

Area of BCQN=

2

1

×(5+4)cm×6cm=27cm

2

Area of EMFQ=

2

1

×(5+6)cm×10cm=55cm

2

Area of polygon ABCDEF=Area of △AMF+Area of △EPD+Area of △QCD+Area of BCQN+Area of EMFQ=(10+18+8+27+55)cm

2

=118cm

2

Step-by-step explanation:

here is your answer hope it will help you...

Answer 2

$\huge \cal \pink{answer}$

Step-by-step explanation:

AD =18cm,AQ=14cm,An=8cm,AM=4 CM

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

NQ=AQ−AN=14cm−8cm=6cm

PD=AD−AP=18cm−12cm=6cm

MP=AP−PM=12cm−4cm=8cm

$area \: of \: △ \: amf \: = \frac{1}{2} \times 5cm \times 4cm = 10 {cm}^{2} \\ area \: of \: △epd = \frac{1}{2} \times 6cm \times 6cm = 18 {cm}^{2} \\ area \: of \: qcd \: = \frac{1}{2} \times 4cm \times 4cm \: = 8 {cm}^{2} \\ area \: of \: bcqn = \frac{1}{2} \times (5 + 4)cm \times 6cm \times 10cm = 55 55 {cm}^{2}$

Area of polygon ABCDEF=Area of △AMF+Area of △EPD+Area of △QCD+Area of BCQN+Area of

$EMFQ = (10 +18 + 8 + 27 + 55 {cm}^{2} = 18 {cm}^{2}$