Construct $∆PQR$ if $PQ = 5\ cm$, $m\angle\ PQR = 105^{\circ}$ and $m\angle QRP = 40^{\circ}$. [Hint: Recall angle-sum property of a triangle.]
Given: $PQ = 5\ cm$, $m\angle\ PQR = 105^{\circ}$ and $m\angle QRP = 40^{\circ}$.
To do: To construct $\triangle PQR$.
Solution:
Here given, $PQ = 5\ cm$, $m\angle PQR = 105^{\circ}$ and $m\angle QRP = 40^{\circ}$
Let us find the value of $\angle QPR=?$
$\angle PQR + \angle QRP + \angle QPR = 180^{\circ}$ [angle sum property of a triangle]
$\Rightarrow 105^{\circ} + 40^{\circ} + \angle QPR = 180^{\circ}$
$\Rightarrow \angle QPR = 35^{\circ}$
Steps of construction :
- Let us draw a line segment $PQ=5\ cm$.
- At $P$, let us draw a ray $PX$ making an angle of $35^{\circ}$ such that $\angle QPX=35^{\circ}$.
- At $Q$, let us draw a ray $QY$ making an angle of $105^{\circ}$ such that $\angle PQR=105^{\circ}$.
- Here, rays $PX$ and $QY$ intersect each other at the point $R$.
$\triangle PQR$ is the required triangle.
Related Articles
- In $∆ABC$, $\angle A=30^{\circ},\ \angle B=40^{\circ}$ and $\angle C=110^{\circ}$In $∆PQR$, $\angle P=30^{\circ},\ \angle Q=40^{\circ}$ and $\angle R=110^{\circ}$. A student says that $∆ABC ≅ ∆PQR$ by $AAA$ congruence criterion. Is he justified? Why or why not?
- Construct $∆ABC$, given $m\angle A = 60^{\circ}$, $m\angle B = 30^{\circ}$ and $AB = 5.8\ cm$.
- Construct the right angled $∆PQR$, where $m\angle Q = 90^{\circ},\ QR = 8cm$ and $PR = 10\ cm$.
- In triangles \( \mathrm{PQR} \) and \( \mathrm{MST}, \angle \mathrm{P}=55^{\circ}, \angle \mathrm{Q}=25^{\circ}, \angle \mathrm{M}=100^{\circ} \) and \( \angle \mathrm{S}=25^{\circ} \). Is \( \triangle \mathrm{QPR} \sim \triangle \mathrm{TSM} \) ? Why?
- Examine whether you can construct $∆DEF$ such that $EF = 7.2\ cm$, $m\angle E = 110^{\circ}$ and $m\angle F = 80^{\circ}$. Justify your answer.
- \( \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \). If \( 2 \angle \mathrm{P}=3 \angle \mathrm{Q} \) and \( \angle C=100^{\circ} \), find \( \angle B \).
- 10. Construct \( \triangle \mathrm{PQR} \) with \( \mathrm{PQ}=4.5 \mathrm{~cm}, \angle \mathrm{P}=60^{\circ} \) and \( \mathrm{PR}=4.5 \mathrm{~cm} . \) Measure \( \angle \mathrm{Q} \) and \( \angle \mathrm{R} \). What type of a triangle is it?
- Construct $∆ABC$ with $BC = 7.5\ cm$, $AC = 5\ cm$ and $\angle C = 60^{\circ}$.
- Construct a triangle \( \mathrm{PQR} \) in which \( \mathrm{QR}=6 \mathrm{~cm}, \angle \mathrm{Q}=60^{\circ} \) and \( \mathrm{PR}-\mathrm{PQ}=2 \mathrm{~cm} \).
- Construct $∆DEF$ such that $DE = 5\ cm$, $DF = 3\ cm$ and $\angle EDF = 90^{\circ}$.
- The exterior angle PRS of triangle PQR is $105^o$. If $\angle Q=70^o$, find $\angle P$. Is $\angle PRS > \angle P$.
- Name the types of following triangles:(a) Triangle with lengths of sides \( 7 \mathrm{~cm}, 8 \mathrm{~cm} \) and \( 9 \mathrm{~cm} \).(b) \( \triangle \mathrm{ABC} \) with \( \mathrm{AB}=8.7 \mathrm{~cm}, \mathrm{AC}=7 \mathrm{~cm} \) and \( \mathrm{BC}=6 \mathrm{~cm} \).(c) \( \triangle \mathrm{PQR} \) such that \( \mathrm{PQ}=\mathrm{QR}=\mathrm{PR}=5 \mathrm{~cm} \).(d) \( \triangle \mathrm{DEF} \) with \( \mathrm{m} \angle \mathrm{D}=90^{\circ} \)(e) \( \triangle \mathrm{XYZ} \) with \( \mathrm{m} \angle \mathrm{Y}=90^{\circ} \) and \( \mathrm{XY}=\mathrm{YZ} \).(f) \( \Delta \mathrm{LMN} \) with \( \mathrm{m} \angle \mathrm{L}=30^{\circ}, \mathrm{m} \angle \mathrm{M}=70^{\circ} \) and \( \mathrm{m} \angle \mathrm{N}=80^{\circ} \).
Kickstart Your Career
Get certified by completing the course
Get Started