Question

If the base of an isosceles triangle is \(3\sqrt{2}\) feet and the two equal sides are increasing at the rate of \(1\) ft/s, then the rate of increase of its area (in sq.ft/sec) when the angle between the equal sides is a right angle is

• A. \(3\sqrt{3}\)
• B. \(\sqrt{3}\)
• C. \(9\)
• D. \(3\)

Hint:

For related rates involving triangles, first use the geometric condition to determine the instantaneous dimensions of the figure. Only then differentiate the area formula with respect to time.

Answer 1

The Correct Option is D

Concept: This is a related rates problem involving an isosceles triangle. Let each equal side be \(a\). Then the area of a triangle with two sides \(a,a\) and included angle \(\theta\) is \[ A=\frac12 a^2\sin\theta. \] The base is fixed at \(3\sqrt2\). When the included angle becomes \(90^\circ\), the geometry of the triangle determines the corresponding value of \(a\).

Step 1: Determine the side length when the included angle is \(90^\circ\). Using the cosine rule, \[ b^2=a^2+a^2-2a^2\cos\theta. \] When \[ \theta=90^\circ, \] \[ \cos90^\circ=0. \] Therefore, \[ (3\sqrt2)^2=2a^2. \] \[ 18=2a^2. \] \[ a^2=9. \] \[ a=3. \]

Step 2: Write the area formula. Since the included angle is \(90^\circ\), \[ A = \frac12 a^2. \] Differentiating with respect to time, \[ \frac{dA}{dt} = a\frac{da}{dt}. \]

Step 3: Substitute the given rate. Given \[ \frac{da}{dt}=1\ \text{ft/s}. \] Also, \[ a=3. \] Hence, \[ \frac{dA}{dt} = 3(1) = 3. \] Therefore, \[ \boxed{3\ \text{sq.ft/sec}}. \]