Question

$△ABC\sim △LBN.$ In $△$
$ABC,AB=5.1cm,\angle B=$
${40}^{\circ },BC=4.8cm,\frac{AC}{LN}=\frac{4}{7}$.
Construct $△ABC$ and $△LBN$.

Answer 1

Analysis:
As shown in the figure,
Let B – C – N and B – A – L.

$△ABC\sim △LBN\dots [$ Given $]$
$\therefore \angle ABC\cong \angle {LBN}_{\dots }$ [Corresponding angles of similar
triangles]
$\frac{AB}{LB}=\frac{BC}{BN}=\frac{AC}{LN}\dots$ (i) [Corresponding sides of similar
triangles]
But. $\frac{AC}{LN}=\frac{4}{7}\dots$ (ii) [Given]
$\therefore \frac{AB}{LB}=\frac{BC}{BN}=\frac{AC}{LN}=\frac{4}{7}\dots [$ From(i)and(ii)]
$\therefore$ sides of $△LBN$ are longer than corresponding sides of $△$
$ABC$.
$\therefore$ If seg $BC$ is divided into 4 equal parts, then seg $BN$ will be
7 times each part of seg $BC$.
So, if we construct $△ABC$, point $N$ will be on side $BC$, at a
distance equal to 7 parts from $B$.
Now, point $L$ is the point of intersection of ray $BA$ and a line
through $N$, parallel to $AC$.
$△LBN$ is the required triangle similar to $△ABC$.
Steps of construction:
i. Draw $△ABC$ of given measure. Draw ray $BD$ making an acute angle with side $BC$.
ii. Taking convenient distance on compass, mark 7 points $B_1,B_2,B_3,B_4,B_5,B_6$ and $B_7$ such that
${BB}_1=B_1{B}_2=B_2{B}_3{B}_3=B_{4}4=B_4{B}_5=B_5{B}_6=B_6{B}_7$.
iii. Join $B_{4}C$. Draw line parallel to $B_{4}C$ through $B_7$ to intersects ray $BC$ at $N_s$
iv. Draw a line parallel to side $AC$ through $N$. Name the point of intersection of this line and ray $BA$ as $L_s$
$△LBN$ is the required triangle similar to $△ABC$.