Draw a $ \triangle A B C $ in which base $ B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm} $ and $ \angle A B C=60^{\circ} $. Then construct another triangle whose sides are $ \frac{3}{4} $ of the corresponding sides of $ \triangle A B C $.

Given:

A \( \triangle A B C \) in which base \( B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm} \) and \( \angle A B C=60^{\circ} \).

To do:

We have to draw a \( \triangle A B C \) in which base \( B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm} \) and \( \angle A B C=60^{\circ} \). Then construct another triangle whose sides are \( \frac{3}{4} \) of the corresponding sides of \( \triangle A B C \).

Solution:

Steps of construction:

(i) Draw a triangle $ABC$ with sides $BC = 6\ cm, AB = 5\ cm$ and $\angle ABC = 60^o$.

(ii) Draw a ray $BX$, which makes an acute angle $\angle CBX$ below the line $BC$.

(iii) Locate four points $B_1, B_2, B_3$ and $B_4$ on $BX$ such that $BB_1 = B_1B_2=B_2B_3 = B_3B_4$.

(iv) Join $B_4C$ and draw a line through $B_3$ parallel to $B_4C$ intersecting $BC$ to $C’$.

(v) Draw a line through $C’$ parallel to the line $CA$ to intersect $BA$ at $A’$.

Justification of the construction:

$\mathrm{B}_{4} \mathrm{C} \| \mathrm{B}_{3} \mathrm{C}^{\prime}$     (By construction)

This implies,

$\frac{\mathrm{BB}_{3}}{\mathrm{BB}_{4}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}$       (Basic proportionality theorem)

$\frac{\mathrm{BB}_{3}}{\mathrm{BB}_{4}}=\frac{3}{4}$      (By construction)

This implies, $\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{3}{4}$.........(i)

Therefore,

$\mathrm{CA} \| \mathrm{C}^{\prime} \mathrm{A}^{\prime}$       (By construction)

$\Delta \mathrm{BC}^{\prime} \mathrm{A}^{\prime} \sim \Delta \mathrm{BCA}$

From equation (i),

$\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C}=\frac{3}{4}$     (By Basic Proportionality Theorem)

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

41 Views

Kickstart Your Career

Get certified by completing the course

Get Started