Question

Draw a triangle $ABC$ with side $BC=6cm,AB=5cm$ and $\angle ABC={60}^o$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $ABC$.

Answer 1

Constructing $△ABC$:

Step 1: Draw base $BC$ of length $6cm$ using a ruler.

Step 2: Draw a ray from point $B$ at an angle of ${60}^{\circ }$.

Step 3: Mark an arc of radius $5cm$ on the ray is drawn from $B$ and label the intersection point as $A$. Join $A-C$

$△ABC$ is the given triangle.

Constructing $△EBD$ with sides $\frac{3}{4}$ times the sides of $△ABC$

Step 1: Draw a ray $BX$ 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 $B_4$ to point $C$.

Step 4: Draw $B_{3}D$ parallel to $B_{4}C$ and mark the intersection point with line $BC$ as $D$.

Step 5: Draw $DE$ parallel to $CA$ and mark the intersection point with $AB$ as $E$.

$△EBD$ is the required triangle.