Question : Find the area of a triangle whose length of two sides are 4 cm and 5 cm and the angle between them is 45°.
Option 1: $4 \sqrt{2} \mathrm{~cm}^2$
Option 2: $7 \sqrt{2} \mathrm{~cm}^2$
Option 3: $5 \sqrt{2} \mathrm{~cm}^2$
Option 4: $6 \sqrt{2} \mathrm{~cm}^2$
Correct Answer: $5 \sqrt{2} \mathrm{~cm}^2$
Solution :
Given: 1st side = 4 cm
2nd side = 5 cm
Angle between them = $45^\circ$
We know that,
Area of triangle = $\frac{1}{2}\times\text{1st side × 2nd side}\times\sin\theta$
= $\frac{1}{2}\times4\times5\times\sin45^\circ$
= $10\times\frac{1}{\sqrt2}$
= $5\sqrt2$
Hence, the correct answer is $5\sqrt2\ cm^2$.
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