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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