Question:
What is the area of an isosceles triangle whose perimeter is 11 cm and base is 5 cm?
What is the area of an isosceles triangle whose perimeter is 11 cm and base is 5 cm?
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Altitude bisects base in isosceles triangle
Updated On: Apr 21, 2026
- \(10~\text{cm}^2\)
- \(\frac{5}{4}\sqrt{11}~\text{cm}^2\)
- \(20\sqrt{11}~\text{cm}^2\)
- \(\frac{4}{5\sqrt{11}}~\text{cm}^2\)
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The Correct Option is B
Solution and Explanation
Step 1: Let the equal sides of the isosceles triangle be \(a\).
Given perimeter \(= 11\) cm and base \(= 5\) cm:
\(2a + 5 = 11 \Rightarrow 2a = 6 \Rightarrow a = 3\) cm
Step 2: In an isosceles triangle, the altitude from the vertex bisects the base. So, half of base \(= \frac{5}{2} = 2.5\) cm
Step 3: Apply Pythagoras theorem to find height: \(h = \sqrt{3^2 - (2.5)^2} = \sqrt{9 - 6.25} = \sqrt{2.75}\)
Step 4: Simplify: \(\sqrt{2.75} = \sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2}\)
Step 5: Area of triangle: \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times \frac{\sqrt{11}}{2}\)
Step 6: Final area: \(\frac{5\sqrt{11}}{4}\)
Step 2: In an isosceles triangle, the altitude from the vertex bisects the base. So, half of base \(= \frac{5}{2} = 2.5\) cm
Step 3: Apply Pythagoras theorem to find height: \(h = \sqrt{3^2 - (2.5)^2} = \sqrt{9 - 6.25} = \sqrt{2.75}\)
Step 4: Simplify: \(\sqrt{2.75} = \sqrt{\frac{11}{4}} = \frac{\sqrt{11}}{2}\)
Step 5: Area of triangle: \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times \frac{\sqrt{11}}{2}\)
Step 6: Final area: \(\frac{5\sqrt{11}}{4}\)
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