Question

Question 4
Construct the following and give justification :
​​​​​​​
An equilateral triangle, if its altitude is 3.2cm.

Answer 1

We know that in an equilateral triangle all the sides are equal and all angles are equal i.e., each angle is ${60}^{\circ }$.
Given, the altitude of an equilateral triangle is 3.2cm.
Steps to construct the $\Delta ABC$.
(i) Draw a line PQ.
(ii) Take a point D on PQ and draw a ray $DE\perp PQ$.
(iii) Cut the line segment AD of length 3.2cm on the ray DE.
(iv) Make angles $\angle DAB={30}^{\circ }$ at A with line AB intersecting DP.
(v) Cut the line segment DC on DQ such that DC = BD.
(vi) Join AC, thus, $\Delta ABC$ is the required triangle.


Justification
$Here,\angle A=\angle BAD+\angle CAD$
$={30}^{\circ }+{30}^{\circ }={60}^{\circ }$
$Also,AD\perp BC$
$\therefore \angle ADB={90}^{\circ }$

$In\Delta ABD,\angle BAD+\angle ADB+\angle DBA={180}^{\circ }\left[anglesumproperty\right]$

${30}^{\circ }+{90}^{\circ }+\angle DBA={180}^{\circ }[\angle BAD={30}^{\circ },byconstruction]$

$\angle DBA={60}^{\circ }$

$Similarly,\angle DCA={60}^{\circ }$
$Thus,\angle A=\angle B=\angle C={60}^{\circ }$

Hence, $\Delta ABC$ is an equilateral triangle.