Question

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

Answer 1

BC = 7 cm $\angle B={45}^{\circ },\angle A={105}^{\circ }$
$\angle C={180}^{\circ }-(\angle B+\angle A)$
= ${180}^{\circ }-({45}^{\circ }+{105}^{\circ })$
= ${180}^{\circ }-{150}^{\circ }$
= ${30}^{\circ }$

Steps of construction
(i) Draw a line segment BC = 7 cm.
(ii) Draw an angle ${45}^{\circ }$ at B and ${30}^{\circ }$ at C. They intersect at A.
(iii) Draw an acute angle at B.
(iv) Divide ray BX in 4 equal parts as $B_1,B_2,B_3$ and $B_4$.
(v) Join $B_4$ to C.
(vi) From $B_3$ draw a line parallel to $B_{4}C$ intersecting BC at C'.
(vii) Draw another line parallel to CA from C' intersecting AB at A'.
(viii) $\Delta$ A'BC' is required triangle such that A'B'= $\frac{3}{4}$ AB.