find the area of quadrilateral abcd ,has ab=3cm,bc=4cm,cd=4cm,da=5cmand,ac=5cm
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ab = 3cm
bc = 4cm
cd = 4cm
ad = 5cm
ac = 5cm
Pythagoras theorem says c^2 = a^2 + b^2
c = ac = 5cm
a = ab = 3cm
b = bc = 4cm
c^2 = (5cm)^2
a^2 = (3cm)^2
b^2 = (4cm)^2
25 = 9 + 16
So it is a right angle.
A = (b×h)/2
A = (3×4)/2 = 12/2 = 6cm
abc = 6cm
A of triangle = √s(s-a)(s-b)(s-c)
s = (a+b+c)/2
s = (5+5+4)/2 = 14/2 = 7
√7(7-5)(7-5)(7-4)
√7×2^2×3
2√21cm
6+2√21cm
8√21cm
I hope it is correct ...
With regards
Achyuta Shrimate
bc = 4cm
cd = 4cm
ad = 5cm
ac = 5cm
Pythagoras theorem says c^2 = a^2 + b^2
c = ac = 5cm
a = ab = 3cm
b = bc = 4cm
c^2 = (5cm)^2
a^2 = (3cm)^2
b^2 = (4cm)^2
25 = 9 + 16
So it is a right angle.
A = (b×h)/2
A = (3×4)/2 = 12/2 = 6cm
abc = 6cm
A of triangle = √s(s-a)(s-b)(s-c)
s = (a+b+c)/2
s = (5+5+4)/2 = 14/2 = 7
√7(7-5)(7-5)(7-4)
√7×2^2×3
2√21cm
6+2√21cm
8√21cm
I hope it is correct ...
With regards
Achyuta Shrimate
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