Isosceles Triangle Area Bounds

Question

Let ABC be an isosceles triangle. Suppose that the sides AB and AC are equal and let the length of AB be $x$ cm. Let $b$ denote the angle $\angle ABC$ and $\sin b = 3/5$. If the area of the triangle ABC is M square cm, then which of the following is true about M?

Accepted Answer

E. In isosceles $\Delta ABC$, $AB = AC = x$, angle at base $\angle B = \angle C = b$. Given $\sin b = 3/5 \Rightarrow \cos b = 4/5$. Vertical angle $A = 180^\circ - 2b$. $\sin A = \sin(180^\circ - 2b) = \sin 2b = 2 \sin b \cos b = 2 \left(\frac{3}{5}\right)\left(\frac{4}{5}\right) = \frac{24}{25}$. Area $M = \frac{1}{2} x^2 \sin A = \frac{1}{2} x^2 \left(\frac{24}{25}\right) = \frac{12}{25} x^2$. Since $\frac{1}{4} = 0.25$ and $\frac{1}{2} = 0.50$, $0.48 x^2$ lies in $\frac{x^2}{4} \le M < \frac{x^2}{2}$.

Answer

$M < \frac{x^2}{4}$

Answer

$\frac{3x^2}{4} \le M < x^2$

Answer

$M \ge x^2$

Answer

$\frac{x^2}{2} \le M < \frac{3x^2}{4}$

Answer

$\frac{x^2}{4} \le M < \frac{x^2}{2}$

Isosceles Triangle Area Bounds

Explanation

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