Question

Construct a $△ABC$ in which $AB=5cm$. $\angle B={60}^{\circ }$ altitude $CD=3cm$. Construct a $△AQR$ similar to $△ABC$ such that side of $△AQR$ is $1.5$ times that of the corresponding sides of $△ACB$.

Answer 1

We have to construct a $△ABC$

$\to$ Draw a line $\overline{AB}=5cm$

$\to$ Take $B$ as point by using compass measure an angle of ${60}^o$

$\to$ Marks that angle as $ABZ$

$\to$ Draw a line $EF$ which passes through (intersects) $BX$, which us at a height of $3cm$ from $AB$

$\to$ The point where $EF$ intersect $BX$ mark as $C$

$\to$ Now join $AC$

$\to$ Now Draw $CD\perp$ to $AB$

$CD$ is attitude of $ABC$ having $3cm$

Now we have to construct $△AQR$ similar to $△ABC$

As given side of $△AQR$ is $1.5$ times of $△ACB$

$\to$ So extend the line $AB$ and from $B$ take $1.5cm$ and make a point $R$

$\to$ From $R$ take ${60}^o$ angle By extending the line $AC$

$\to$ The point where the angle intersect the line mark it as $q$

$\to$ Hence we construct $△ABC$ & $△AQR$