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Area of square = a^2
3.24 = a^2
a = under root 3.14
a = 1.8 m
Thus AB=BC=CD=DA=1.8
Now, M is the mid point of BC.
So, BM=MC= 0.9
Similarly AN=ND=0.9
In right triangle ABM we have;
AB=1.8 (base) , BM= 0.9 (perpendicular)
By Pythagorean’s theorem
h^2= p^2 +b^2
=(0.9)^2 + (1.8)^2
= 0.81 + 3.24
= 4.05
h = 2.01
Area of right triangle =
1/2 multiple base multiple height
3.24 = a^2
a = under root 3.14
a = 1.8 m
Thus AB=BC=CD=DA=1.8
Now, M is the mid point of BC.
So, BM=MC= 0.9
Similarly AN=ND=0.9
In right triangle ABM we have;
AB=1.8 (base) , BM= 0.9 (perpendicular)
By Pythagorean’s theorem
h^2= p^2 +b^2
=(0.9)^2 + (1.8)^2
= 0.81 + 3.24
= 4.05
h = 2.01
Area of right triangle =
1/2 multiple base multiple height
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