Question

If the sides of a triangle are 4, 5 and 6 cms. Then the area of triangle is ______ sq.cms.

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{4}\sqrt{7} \)
• C. \( \frac{4}{15} \)
• D. \( \frac{4}{15}\sqrt{7} \)
• E. \( \frac{15}{4}\sqrt{7} \)

Hint:

Whenever all three sides of a triangle are given, directly apply Heron’s formula to avoid using trigonometry.

Answer 1

Answer: (E)

Concept: When the three sides of a triangle are known, the area can be found using Heron’s Formula: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] where \( s \) is the semi-perimeter: \[ s = \frac{a+b+c}{2} \]

Step 1: Find the semi-perimeter.
\[ s = \frac{4 + 5 + 6}{2} = \frac{15}{2} \]

Step 2: Compute the terms \( (s-a), (s-b), (s-c) \).
\[ s - 4 = \frac{15}{2} - 4 = \frac{7}{2} \] \[ s - 5 = \frac{15}{2} - 5 = \frac{5}{2} \] \[ s - 6 = \frac{15}{2} - 6 = \frac{3}{2} \]

Step 3: Apply Heron’s formula.
\[ \text{Area} = \sqrt{ \frac{15}{2} \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} } \]

Step 4: Simplify the expression.
\[ = \sqrt{ \frac{15 \cdot 7 \cdot 5 \cdot 3}{16} } = \frac{1}{4}\sqrt{15 \cdot 7 \cdot 5 \cdot 3} \] \[ 15 \cdot 3 = 45,\quad 45 \cdot 5 = 225 \] \[ = \frac{1}{4}\sqrt{225 \cdot 7} = \frac{1}{4} \cdot 15\sqrt{7} \]

Step 5: Final result.
\[ \text{Area} = \frac{15}{4}\sqrt{7} \]