find the area of a parallelogram ABCD in which a b is 14 cm BC is 10 cm and ac is 16 cm
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AB=14cm
BC=10cm
AC=16cm ( diagonal
area of parallelogram = area(∆ABD)+area(∆BDC)
∆ABD
AB=14cm
BD=10cm
AC=16cm
S=(14+10+16)/2=40/2=20
A=√S(S-a)(S-b)(S-c)
A=√20(20-14)(20-10)(20-16)
A=√20×6×10×4
A=√2×10×2×3×2×2
A=2×2×10√3
A=40√3
Similarly,
for second ∆
A=40√3
area of llgm=40√3+40√3
=80√3
=138.564cm²
BC=10cm
AC=16cm ( diagonal
area of parallelogram = area(∆ABD)+area(∆BDC)
∆ABD
AB=14cm
BD=10cm
AC=16cm
S=(14+10+16)/2=40/2=20
A=√S(S-a)(S-b)(S-c)
A=√20(20-14)(20-10)(20-16)
A=√20×6×10×4
A=√2×10×2×3×2×2
A=2×2×10√3
A=40√3
Similarly,
for second ∆
A=40√3
area of llgm=40√3+40√3
=80√3
=138.564cm²
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