Construct a triangle $PQR$,in which $QR = 9\, cm$,$\angle Q = 60^{\circ}$ and $PR - PQ = 3\, cm$.
(N/A) Steps of construction:
$(1)$ Draw a ray $QX$ and cut off a line segment $QR = 9\, cm$ from it.
$(2)$ At point $Q$,construct a ray $QY$ such that $\angle YQR = 60^{\circ}$.
$(3)$ Produce the ray $QY$ backwards to form a ray $QZ$. Cut off a line segment $QS = 3\, cm$ from $QZ$.
$(4)$ Join $RS$. Draw the perpendicular bisector of the line segment $RS$.
$(5)$ Let the perpendicular bisector intersect the ray $QY$ at point $P$. Join $PR$.
$(6)$ Thus,$\Delta PQR$ is the required triangle.
$(1)$ Draw a ray $QX$ and cut off a line segment $QR = 9\, cm$ from it.
$(2)$ At point $Q$,construct a ray $QY$ such that $\angle YQR = 60^{\circ}$.
$(3)$ Produce the ray $QY$ backwards to form a ray $QZ$. Cut off a line segment $QS = 3\, cm$ from $QZ$.
$(4)$ Join $RS$. Draw the perpendicular bisector of the line segment $RS$.
$(5)$ Let the perpendicular bisector intersect the ray $QY$ at point $P$. Join $PR$.
$(6)$ Thus,$\Delta PQR$ is the required triangle.
Explore More
Similar Questions
The construction of a triangle $ABC$ in which $AB = 4 \, cm$ and $\angle A = 60^{\circ}$ is not possible when the difference of $BC$ and $AC$ is equal to: (in $cm$)
Medium
View SolutionDraw a line segment $PQ$ of length $7.4\,cm$. Then,construct the perpendicular bisector of the line segment $PQ$. Write the steps of construction.
Medium
View SolutionDraw an equilateral $\Delta XYZ$ with perimeter $21\,cm$. Write the steps of construction.
Medium
View SolutionWrite True or False and give reasons for your answer.
$A$ triangle $ABC$ can be constructed in which $BC = 6 \, cm$,$\angle C = 30^{\circ}$ and $AC - AB = 4 \, cm$.
$A$ triangle $ABC$ can be constructed in which $BC = 6 \, cm$,$\angle C = 30^{\circ}$ and $AC - AB = 4 \, cm$.
Easy
View SolutionWrite True or False and give reasons for your answer.
$A$ triangle $ABC$ can be constructed in which $\angle B = 60^{\circ}$,$\angle C = 45^{\circ}$ and $AB + BC + AC = 12 \text{ cm}$.
$A$ triangle $ABC$ can be constructed in which $\angle B = 60^{\circ}$,$\angle C = 45^{\circ}$ and $AB + BC + AC = 12 \text{ cm}$.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo