AB=9CM,BC=5CM,CD=7CM,∆B=45°AND∆C=90°
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hello,
draw quadrilateral ABCD with ∠ABC=90°
now,join diagonal AC
in ΔABC,
area=1/2×base×height
here base is BC and height is AB
area=1/2×9cm×40cm
=180cm²...................................1
now we have to find AC,
by pythagoras theorem
AB²+BC²=AC²
9²+40²=AC²
1681=AC²
AC=41cm
now in ΔADC,
using herons formula,
s(semi perimeter)=(AC+AD+DC)/2
s=(15+28+41)/2
s=84/2
s=42cm
area=√s(s-a)(s-b)(s-c)
=√42(42-15)(42-28)(42-41)
=√42(27)(14)(1)
=√2×3×7×3×3×3×2×7
=2×3×3×7
=126cm².................................2
add 1 and 2
=306cm² which is area of quad.ABCD if u like it mark it as brainliest
draw quadrilateral ABCD with ∠ABC=90°
now,join diagonal AC
in ΔABC,
area=1/2×base×height
here base is BC and height is AB
area=1/2×9cm×40cm
=180cm²...................................1
now we have to find AC,
by pythagoras theorem
AB²+BC²=AC²
9²+40²=AC²
1681=AC²
AC=41cm
now in ΔADC,
using herons formula,
s(semi perimeter)=(AC+AD+DC)/2
s=(15+28+41)/2
s=84/2
s=42cm
area=√s(s-a)(s-b)(s-c)
=√42(42-15)(42-28)(42-41)
=√42(27)(14)(1)
=√2×3×7×3×3×3×2×7
=2×3×3×7
=126cm².................................2
add 1 and 2
=306cm² which is area of quad.ABCD if u like it mark it as brainliest
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