there is a pentagonal
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arc555:
area of th figure will be 337.5
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1st method.
Area of traperium ABCF = 1/2(Sum of parallel side) ×height
=>1/2(15+30) × 7.5
=>1/2(45)×7.5
=>168.75 cm^2
Area of Trap. CDEF =Area of trap. ABCF
Hence, area of the the pentalgon ABCDE =(168.75×2) cm^2
=>337.5 cm^2
2nd method.
Since, AB = CD and perpendicular
It is clers that ABCD is a square
Area if sq. ABCD = SIDE ×SIDE
=>15×15 = 225cm^2
Area of triangle DEA = 1/2(b× h)
= >1/2(15×15)
=>1/2(225)
=>112.5 cm^2
Area of the pentalgon = Area of sq. + Area of tri.
=>225 + 112.5
=>337.5 cm^2
We obseve that in both method area is equal.
Area of traperium ABCF = 1/2(Sum of parallel side) ×height
=>1/2(15+30) × 7.5
=>1/2(45)×7.5
=>168.75 cm^2
Area of Trap. CDEF =Area of trap. ABCF
Hence, area of the the pentalgon ABCDE =(168.75×2) cm^2
=>337.5 cm^2
2nd method.
Since, AB = CD and perpendicular
It is clers that ABCD is a square
Area if sq. ABCD = SIDE ×SIDE
=>15×15 = 225cm^2
Area of triangle DEA = 1/2(b× h)
= >1/2(15×15)
=>1/2(225)
=>112.5 cm^2
Area of the pentalgon = Area of sq. + Area of tri.
=>225 + 112.5
=>337.5 cm^2
We obseve that in both method area is equal.
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