Question

Question 1
Construct the following and give justification :

A triangle if its perimeter is 10.4cm and two angles are ${45}^{\circ }and{120}^{\circ }$.

Answer 1

Let ABC be a triangle.
Given perimeter = 10.4cm i.e., AB + BC + CA = 10.4cm and two angles are ${45}^{\circ }and{120}^{\circ }$.
$Let\angle B={45}^{\circ }and\angle C={120}^{\circ }$

Steps to construct the $\Delta ABC$
(i) Draw a line segment XY which is equal to the perimeter i.e., AB + BC + CA = 10.4cm

(ii) Make $\angle LXY=\angle B={45}^{\circ }and\angle MYX=\angle C={120}^{\circ }$
(iii) Bisect $\angle LXYand\angle MYX$ and let these bisectors intersect at a point A.
(iv) Draw perpendicular bisectors PQ and RS of AX and AY, respectively.
(v) Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC. Thus, $\Delta ABC$is the required triangle.

Justification
Since, B lies on the perpendicular bisector PQ of AX.
$\therefore$ XB = AB
Since, C lies on the perpendicular bisector RS of AY.
$\therefore$ CY = AC
Thus, AB + BC + CA = XB + BC + CY = XY
$Again,\angle BAX=\angle AXB[\because \Delta AXB,AB=XB]\cdots \left(i\right)$
$Also,\angle ABC=\angle BAX+\angle AXB[\because \angle ABCisanexteriorangleof\Delta AXB]$
$=\angle AXB+\angle AXB[fromEq.(i\left)\right]$
$=2\angle AXB=\angle LXY[\because AXisabisectorof\angle LXB]$
$Also,\angle CAY=\angle AYC[\because in\Delta AYC,AC]$
$\therefore \angle ACB=\angle CAY+\angle AYC[\because \angle ACBisanexteriorangleof\Delta AYC]$
$=\angle CAY=\angle CAY$
$=2\angle CAY=\angle MYX[\because AYisabisectorof\angle MYX]$
Thus, our construction is justified.