In an isosceles triangle, the measure of each of the equal sides is 10 cm and the angle between them is 45°. The area of the triangle is
Aptitude
Area
Difficulty: Easy
Choose an option
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A$25 \text{ cm}^2$
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B$\frac{25}{2}\sqrt{2} \text{ cm}^2$
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C$25\sqrt{2} \text{ cm}^2$
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D$25\sqrt{3} \text{ cm}^2$
Answer
Correct Answer: $25\sqrt{2} \text{ cm}^2$
Explanation
### Concept & Formula
The area of a triangle when two sides and the included angle are given is:
$$ \text{Area} = \frac{1}{2}ab\sin(\theta) $$
### Step-by-Step Solution
1. Here, $a = 10 \text{ cm}$, $b = 10 \text{ cm}$, and $\theta = 45^\circ$.
2. Area $= \frac{1}{2} \times 10 \times 10 \times \sin(45^\circ)$.
3. Area $= 50 \times \frac{1}{\sqrt{2}}$.
4. Rationalizing the denominator: $50 \times \frac{\sqrt{2}}{2} = 25\sqrt{2} \text{ cm}^2$.
### Exam Strategy & Shortcut
Recognize that $\frac{1}{\sqrt{2}}$ is equivalent to $\frac{\sqrt{2}}{2}$. Mental calculation: $100/2 = 50$, $50/\sqrt{2} = 25\sqrt{2}$.
### Common Pitfall
Forgetting the $\frac{1}{2}$ in the area formula or substituting the wrong value for $\sin(45^\circ)$.
### Final Answer
Therefore, the correct answer is **$25\sqrt{2} \text{ cm}^2$**.