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

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Given:

Side $BC=7\ cm$,  $\angle B = 45^{o}$,  $\angle A = 105^{o}$.

To do:

We have to construct a triangle $ABC$ with side $BC=7\ cm,\ \angle B = 45^{o}$, $\angle A = 105^{o}$ and then construct another triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $ABC$.

Solution:

Here, $BC=7\ cm$, $\angle B =45^{o}$ , $\angle A=105^{o}$

In $\triangle ABC$,

$\angle A + \angle B + \angle C = 180^o$

$105^o + 45^o + \angle C = 180^o$

$150^o + \angle C = 180^o$

$\angle C = 30^o$

Steps of construction:

(i) Draw a line segment $BC = 7\ cm$.

At point $B$, draw an $\angle B = 45^o$ and at point $C$, draw an $\angle C = 30^o$ and get $\triangle ABC$.

(ii) Draw an acute angle $CBX$ on the base $BC$ at point $B$.

Mark the ray $BX$ with $B_1, B_2, B_3, B_4$ such that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$

(iii) Join $B_3$ to $C$.

(iv) Draw $B_4C’ \| B_3C$, where $C’$ is a point on extended line segment $BC$.

(v) At $C’$, draw $C’A’ \| AC$, where $A’$ is a point on extended line segment $BA$.

Therefore,

$\triangle A’BC’$ is the required triangle.

Updated on 10-Oct-2022 13:23:52