Question
Construct the following and give justification: An equilateral triangle if its altitude is $3.2\ cm.$
✓
Answer
Steps of Construction,
$1.$ Draw a line $l.$
$2.$ Mark any point $D$ on the line $l.$
$3.$ At point $D$, draw $\overline{\text{DX}}\perp\text{l}$ with the help of ruler and compass and cut $DA = 3.2\ cm$ on $\overline{\text{DX}}.$
$4.$ At the point $A$, construct $AB$ and $AC$ which meets the $l$ as points $B$ and $C$ respectively such that $\angle\text{DAB}=30^\circ$ and $\angle\text{DAC}=30^\circ$
Then $\triangle\text{ABC}=30^\circ$ is the required equilateral triangle because,
$\angle\text{ABC}=180^\circ-(90^\circ+30^\circ)=60^\circ$
$\angle\text{ACB}=180^\circ-(90^\circ+30^\circ)=60^\circ$
And $\angle\text{BAC}=30^\circ+30^\circ=60^\circ$
$1.$ Draw a line $l.$
$2.$ Mark any point $D$ on the line $l.$
$3.$ At point $D$, draw $\overline{\text{DX}}\perp\text{l}$ with the help of ruler and compass and cut $DA = 3.2\ cm$ on $\overline{\text{DX}}.$
$4.$ At the point $A$, construct $AB$ and $AC$ which meets the $l$ as points $B$ and $C$ respectively such that $\angle\text{DAB}=30^\circ$ and $\angle\text{DAC}=30^\circ$
Then $\triangle\text{ABC}=30^\circ$ is the required equilateral triangle because,
$\angle\text{ABC}=180^\circ-(90^\circ+30^\circ)=60^\circ$
$\angle\text{ACB}=180^\circ-(90^\circ+30^\circ)=60^\circ$
And $\angle\text{BAC}=30^\circ+30^\circ=60^\circ$
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