ans this q in a proper manner so that i can understand properly
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lisharaj043p3xl5p:
hey in hurry i asked this q in sc sub so plzz dont mind:(
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at first the width can be found by 2 methods
1)BC-FG÷2
2)DC-HG÷2
IN both cases the width is 4cm
Now trapizium HGDC AREA FORMULAE is 1/2×h×(sum of parallel sides)
thus
1/2×4×(20+28)
=1/2×4×48=96
now area of HGDC=ABFE
Now trapizium BFGC has area
1/2×4×(24+32)
=1/2×4×56=212
now area of BFGC=AEHD
hope it helps you
plz mark it as most brainliest answer
1)BC-FG÷2
2)DC-HG÷2
IN both cases the width is 4cm
Now trapizium HGDC AREA FORMULAE is 1/2×h×(sum of parallel sides)
thus
1/2×4×(20+28)
=1/2×4×48=96
now area of HGDC=ABFE
Now trapizium BFGC has area
1/2×4×(24+32)
=1/2×4×56=212
now area of BFGC=AEHD
hope it helps you
plz mark it as most brainliest answer
thank u so much
but why u divided it by 2
and ur ans is wrong
correct ans is 112
oh sorry 56×2=112
i divided it by 2 because there are 2 distances left when subtracting FG from BC but we need only one for our sum
and that was just a multiplication mistake
but still sorry
its ok:)
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