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Sagot :
The area of the triangle will increase at a rate of 3.81 m² per second
Given,
The length of the sides of the triangle are base, 12 m and hypotenuse 15 m
Area of the triangle = 1/2 bh
We have to find h;
That is,
Sin ∅ = opposite side /hypotenuse = h/15
h = 15Sin ∅
d∅/dt = 0.06rad/s
Now,
A = bh/2 = 12 × 15sin∅/2
A = 90sin∅,
By differentiating this equation with respect to A, we have
d(A)/dt = 90cos∅d∅/dt
d(A)/dt = 90cos∅(0.06)
∅ = π/4, cosπ/4 = 0.7071
90×0.7071×0.06
= 3.81m²/s
That is,
The area of the triangle will increase at a rate of 3.81 m² per second
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