The base of an isosceles is 14 cm and its perimeter is 36 cm. Find its area.

### Concept & Formula The perimeter of a triangle is the sum of all its sides. For an isosceles triangle with equal sides $a$ and base $b$, the perimeter is $2a + b$. Heron's formula can be used to find the area if all three sides are known: $$ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $$ where $s$ is the semi-perimeter. Alternatively, a specific formula for isosceles triangles is: $$ \text{Area} = \frac{b}{4}\sqrt{4a^2 - b^2} $$ ### Step-by-Step Solution 1. **Given:** Base ($b$) = 14 cm, Perimeter = 36 cm. Let the two equal sides be $a$. 2. **Find Equal Sides:** $2a + 14 = 36$ $2a = 22 \Rightarrow a = 11 \text{ cm}$. The sides are 11, 11, and 14. 3. **Calculate Semi-perimeter ($s$):** $s = \frac{36}{2} = 18 \text{ cm}$. 4. **Calculate Area using Heron's Formula:** Area = $\sqrt{18 \times (18-11) \times (18-11) \times (18-14)}$ Area = $\sqrt{18 \times 7 \times 7 \times 4}$ Area = $\sqrt{9 \times 2 \times 49 \times 4}$ Area = $3 \times 7 \times 2 \times \sqrt{2} = 42\sqrt{2} \text{ sq. cm.}$ ### Exam Strategy & Shortcut Instead of Heron's formula, using the altitude is faster if you know Pythagorean triplets, but here $\frac{b}{2} = 7$ and hypotenuse is $11$. Altitude $h = \sqrt{11^2 - 7^2} = \sqrt{121 - 49} = \sqrt{72} = 6\sqrt{2}$. Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 6\sqrt{2} = 42\sqrt{2}$. This requires less calculation. ### Common Pitfall Misidentifying which side is the base versus the equal sides when applying the perimeter formula leads to the wrong lengths for the unknown sides. ### Final Answer Therefore, the correct answer is **$42\sqrt{2}$ sq. cm.**