Question

Question 6
Draw a triangle ABC with side $BC=7cm,\angle B={45}^{\circ },\angle A={105}^{\circ }$. Then, construct a triangle whose sides are 4/3 times the corresponding sides of $\Delta ABC$.

Answer 1

Sum of all side of triangle $={180}^{\circ }$
$\therefore \angle A+\angle B+\angle C={180}^{\circ }$
$\angle C={180}^{\circ }-{150}^{\circ }={30}^{\circ }$
Steps of Construction:
Step I: BC = 7 cm is drawn.

Step II: At B, a ray is drawn making an angle of ${45}^{\circ }$ with BC.
Step III: At C, a ray making an angle of ${30}^{\circ }$ with BC is drawn intersecting the previous ray at A. Thus, $\angle A={105}^{\circ }$.
Step IV: A ray BX is drawn making an acute angle with BC opposite to vertex A.
Step V 4 points $B_1,B_2,B_{3}and{B}_4$ are marked at equal distances on BX.
Step VI: $B_{3}C$ is joined and $B_4{C}^{\prime }$ is made parallel to $B_{3}C$.
Step VII: C'A' is made parallel CA.
Thus, A'BC' is the required triangle.
Justification:
$\angle B={45}^{\circ }$ (Common)
$\angle C=\angle C^{\prime }$
$\Delta A{B}^{\prime }{C}^{\prime }$ ~ $\Delta ABC$ by AA similarity condition. $\therefore \frac{BC}{B{C}^{\prime }}=\frac{AB}{A^{\prime }B}=\frac{AC}{A^{\prime }{C}^{\prime }}$ also,
$\frac{BC}{B{C}^{\prime }}=\frac{B{B}_3}{B{B}_4}=\frac{3}{4}$
$\Rightarrow AB=\frac{4}{3}A{B}^{\prime },BC=\frac{4}{3}B{C}^{\prime }andAC=\frac{4}{3}{A}^{\prime }{C}^{\prime }$