Construct a $∆ABC$ in which $AB + AC = 5.6\ cm, BC = 4.5\ cm$ and $\angle B = 45^o$.
Given:
A $∆ABC$ in which $AB + AC = 5.6\ cm, BC = 4.5\ cm$ and $\angle B = 45^o$.
To do:
We have to construct the given triangle.
Solution:
Steps of construction:
(i) Draw a line segment $BC = 4.5\ cm$.
(ii) At $B$, draw a ray $BX$ making an angle of $45^o$ and cut off $BE = 5.6\ cm$ and join $CE$.
(iii) Draw the perpendicular bisector of $CE$ which intersects $BE$ at $A$.
(iv) Join $AC$.
Therefore,
$\triangle ABC$ is the required triangle.
Related Articles
- Construct a $∆ABC$ in which $BC = 3.6\ cm, AB + AC = 4.8\ cm$ and $\angle B = 60^o$.
- Construct a $\triangle ABC$ in which $BC = 3.4\ cm, AB - AC = 1.5\ cm$ and $\angle B = 45^o$.
- Construct $\vartriangle ABC$ in which $BC=7\ cm,\ \angle B=75^{o}$ and $AB+AC=12\ cm$.
- Construct $∆ABC$ such that $AB=2.5\ cm$, $BC=6\ cm$ and $AC=6.5\ cm$. Measure $\angle B$.
- Using ruler and compasses only, construct a $∆ABC$, given base $BC = 7\ cm, \angle ABC = 60^o$ and $AB + AC = 12\ cm$.
- ∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and ∆LMN such that $\frac{BC}{MN} \ =\ \frac{5}{4}$.
- Construct $∆ABC$ with $BC = 7.5\ cm$, $AC = 5\ cm$ and $\angle C = 60^{\circ}$.
- Tick the correct answer and justify : In $∆ABC, AB = 6\sqrt3\ cm, AC = 12\ cm$ and $BC = 6\ cm$. The angle B is:(a) $120^o$(b) $60^o$(c) $90^o$(d) $45^o$
- Construct a triangle \( \mathrm{ABC} \) in which \( \mathrm{BC}=8 \mathrm{~cm}, \angle \mathrm{B}=45^{\circ} \) and \( \mathrm{AB}-\mathrm{AC}=3.5 \mathrm{~cm} \).
- Construct a $\vartriangle ABC$ in which $AB\ =\ 6\ cm$, $\angle A\ =\ 30^{o}$ and $\angle B\ =\ 60^{o}$, Construct another $\vartriangle AB’C’$ similar to $\vartriangle ABC$ with base $ AB’\ =\ 8\ cm$.
- Using ruler and compasses only, construct a $\triangle ABC$, from the following data:$AB + BC + CA = 12\ cm, \angle B = 45^o$ and $\angle C = 60^o$.
- Construct a triangle \( \mathrm{ABC} \) in which \( \mathrm{BC}=7 \mathrm{~cm}, \angle \mathrm{B}=75^{\circ} \) and \( \mathrm{AB}+\mathrm{AC}=13 \mathrm{~cm} \).
- Construct $∆ABC$, given $m\angle A = 60^{\circ}$, $m\angle B = 30^{\circ}$ and $AB = 5.8\ cm$.
- Construct $∆XYZ$ in which $XY=4.5\ cm,\ YZ=5\ cm$ and $ZX=6\ cm$.
- Construct a $\vartriangle ABC$ in which $CA= 6\ cm$, $AB= 5\ cm$ and $\angle BAC= 45^{o}$, Then construct a triangle whose sides are $\frac {3}{5}$ of the corresponding sides of ABC.
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies