Question
$\triangle\text{ABC}$ is right angled at $B$. Given that $AC = 15\ cm, AB = 9\ cm$ and $E$ and $D$ are the mid$-$points of sides $AC$ and $AB$ res. Calculate the area of $\triangle\text{ADE}.$
✓
Answer
As $D$ and $E$ are the midpoints of $AB$ and $AC.$
So, by mid$-$point theorem
$\text{DE} = \frac{\text{BC}}{2} = \frac{12}{2} = 6\text{cm}$
$\text{AD} = \frac{\text{AB}}{2} = \frac{9}{2} = 4.5\text{cm}$
Area of $\triangle\text{ADE} = 0.5 \times \text{DE} \times \text{AD}$
$= 0.5 \times 6 \times 4.5 = 13.5\text{cm}^2$
So, by mid$-$point theorem
$\text{DE} = \frac{\text{BC}}{2} = \frac{12}{2} = 6\text{cm}$
$\text{AD} = \frac{\text{AB}}{2} = \frac{9}{2} = 4.5\text{cm}$
Area of $\triangle\text{ADE} = 0.5 \times \text{DE} \times \text{AD}$
$= 0.5 \times 6 \times 4.5 = 13.5\text{cm}^2$
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