Question

Draw a $△ABC$ with side $BC=7cm$, $\angle B={45}^{\circ },\angle A={105}^{\circ }$. Then, construct a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $△ABC$.

Answer 1

In $△ABC$, $\angle A+\angle B+\angle C={180}^{\circ }$

$\therefore {105}^{\circ }+{45}^{\circ }+\angle C={180}^{\circ }$

$\therefore \angle C={30}^{\circ }$

Constructing $△ABC$:

Step 1: Draw base $BC$ of length $7cm$.

Step 2: Draw a ray at an angle of ${45}^{\circ }$ with line $BC$ from point $B$.

Step 3: Draw a ray at an angle of ${30}^{\circ }$ from line $CB$ from point $C$ and mark the intersection point of the rays from $B$ and $C$ as $A$.

Step 4: Join $A-B$ and $A-C$.

$△ABC$ is the required triangle.

For constructing $△EBD\sim △ABC$

Step 1: Draw a ray $BX$ at an acute angle to line $BC$ on the opposite side of $A$.

Step 2: Mark $4$ equidistant points $B_1,B_2,B_3,B_4$ on ray $BX$.

Step 3: Join $B_3$ and $C$.

Step 4: Draw $B_{4}D$ parallel to $B_{3}C$ and label the intersection on the extension of $BC$ as $D$.

Step 5: Draw $DE$ parallel to $BA$ and label the intersection with the extension of $BA$ as $E$

$△EBD$ is the required triangle.