Find the area of the above pentagon ABCDE in which BG is perpendicular on AC, EF is perpendicular on AD and DH is perpendicular on AC. AD = 10 cm, AC = 12 cm, EF = 3 cm, DH = 7 cm, BG = 5 cm.
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Answer:
In the given figure,
ABCDE, AD=8cm,AH=6cm,AG=4cm,AF=3cm
BF=2cm,CH=3cm and EG=2.5cm
The given figure consists of 3 triangles and one trapezium.
Now
Area of △AED=
2
1
×AD×GE=
2
1
×8×2.5=10cm
2
Area of △ABF=
2
1
×AF×BF=
2
1
×3×2=3cm
2
Area of △CDH=
2
1
×HD×CH=
2
1
×(AD−AH)×3
=
2
1
×(8−6)×3=
2
1
×2×3=3cm
2
Area of trapezium BFHC=
2
1
×(BF+CH)×FH
=
2
1
×(2+3)×(AH−AF)
=
2
1
×5×(6−3)=
2
1
×5×3=7.5cm
2
Hence,
Total area of the figure = Area of △AED + Area of △ABF + Area of △CDH + Area of trapezium BFHC
=10+3+3+7.5=23.5cm
2
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