Question

Construct a triangle ABC with side BC = 8 cm, $\angle$ B = ${45}^{\circ }$, $\angle$ C = ${30}^{\circ }$. Then construct a triangle whose sides are $\left(\frac{4}{3}\right)$ times the corresponding sides of $\Delta$ ABC. [3 MARKS]

Answer 1

Construction : 2 Marks
Steps of construction : 1 Mark



Steps of construction :

1.Construct $\Delta$ ABC with given measurments.

2. Construct an acute angle $\angle$ CBZ at B on opposite side of vertex A of $\Delta$ ABC.

3. Mark-off four (greater 4 of and 3 in $\frac{4}{3}$) points $B_1,B_2,B_3,B_4$ on BZ such that
$B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$

4. Join $B_3$ (the third point) to C and draw a line through $B_4{C}^{{}^{\prime }}$ parallel to $B_{3}C$, intersecting the extended line segment BC at C'.

5. Draw a line through C' parallel to CA intersecting the extended line segment AB at A'.

6. Triangle A'BC' so obtained is the required triangle.