Question
In Fig. triangle $AEC$ is right-angled at $E, B$ is a point on $EC, BD$ is the altitude of triangle $ABC, AC = 25\ cm, BC = 7\ cm$ and $AE = 15\ cm.$ Find the area of triangle $ABC$ and the length of $DB.$
✓
Answer
Given, $\mathrm{AC}=25 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}$, and $\mathrm{AE}=15 \mathrm{~cm}$
In $\triangle \mathrm{AEC}$, using Pythagores theorem,
$\mathrm{AC}^2=\mathrm{AE}^2+\mathrm{EC}^2$
$\Rightarrow \mathrm{EC}^2=\mathrm{AC}^2+\mathrm{AE}^2$
$\Rightarrow \mathrm{EC}^2=(25)^2-(15)^2=625-225=400$
$\mathrm{EC}=\sqrt{400}=20 \mathrm{~cm}$
And $\mathrm{EB}=\mathrm{EC}-\mathrm{BC}=20-7=13 \mathrm{~cm}$
$\text { Area of } \triangle \mathrm{AEC}=\frac{1}{2} \times \mathrm{AE} \times \mathrm{EC}$
$=\frac{1}{2} \times 15 \times 20=150 \mathrm{~cm}^2$
And Area of $\triangle \mathrm{AEB}=\frac{1}{2} \times \mathrm{AE} \times \mathrm{EB}=\frac{1}{2} \times 15 \times 13=97.5 \mathrm{~cm}^2$
$\therefore$ Area of $\triangle \mathrm{ABC}=$ Area of $\triangle \mathrm{AEC}-$ Area of $\triangle \mathrm{AEB}$
$=150-97.5$
$=52.5 \mathrm{~cm}^2$
Again, Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{BD} \times \mathrm{AC}$
$52.5=\frac{1}{2} \times \mathrm{BD} \times 25$
$\Rightarrow \mathrm{BD}=\frac{25.5 \times 2}{25}=4.2 \mathrm{~cm}$
Hence, the area of $\triangle \mathrm{ABC}$ is $52.5 \mathrm{~cm}^2$ and the length of $DB$ is $4.2 \ cm.$
In $\triangle \mathrm{AEC}$, using Pythagores theorem,
$\mathrm{AC}^2=\mathrm{AE}^2+\mathrm{EC}^2$
$\Rightarrow \mathrm{EC}^2=\mathrm{AC}^2+\mathrm{AE}^2$
$\Rightarrow \mathrm{EC}^2=(25)^2-(15)^2=625-225=400$
$\mathrm{EC}=\sqrt{400}=20 \mathrm{~cm}$
And $\mathrm{EB}=\mathrm{EC}-\mathrm{BC}=20-7=13 \mathrm{~cm}$
$\text { Area of } \triangle \mathrm{AEC}=\frac{1}{2} \times \mathrm{AE} \times \mathrm{EC}$
$=\frac{1}{2} \times 15 \times 20=150 \mathrm{~cm}^2$
And Area of $\triangle \mathrm{AEB}=\frac{1}{2} \times \mathrm{AE} \times \mathrm{EB}=\frac{1}{2} \times 15 \times 13=97.5 \mathrm{~cm}^2$
$\therefore$ Area of $\triangle \mathrm{ABC}=$ Area of $\triangle \mathrm{AEC}-$ Area of $\triangle \mathrm{AEB}$
$=150-97.5$
$=52.5 \mathrm{~cm}^2$
Again, Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{BD} \times \mathrm{AC}$
$52.5=\frac{1}{2} \times \mathrm{BD} \times 25$
$\Rightarrow \mathrm{BD}=\frac{25.5 \times 2}{25}=4.2 \mathrm{~cm}$
Hence, the area of $\triangle \mathrm{ABC}$ is $52.5 \mathrm{~cm}^2$ and the length of $DB$ is $4.2 \ cm.$
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