Constructions of Triangles

Is it possible to construct triangles with the given measurement $\left(4 \mathrm{cm}, 5 \mathrm{cm}, 6 \mathrm{cm}\right)$

What is the measure of the side of an equilateral triangle constructed rounded off to one decimal place in $\mathrm{cm}$, if each of whose altitudes measures $5.4 \mathrm{cm}$?

 Can you construct $△\mathrm{ABC}$, in which $\mathrm{BC}=5.2 \mathrm{cm}$, $\angle \mathrm{C}=45°$ and perimeter of $△\mathrm{ABC}$ is $10 \mathrm{cm}$?

Constructing a triangle $\mathrm{X}\mathrm{Y}\mathrm{Z}$ in which $\mathrm{\angle }\mathrm{Y}=30°, \mathrm{\angle }\mathrm{Z}=90°$ and $\mathrm{X}\mathrm{Y}+\mathrm{Y}\mathrm{Z}+\mathrm{Z}\mathrm{X}=11 \mathrm{c}\mathrm{m}$ is possible.

Construction of $△\mathrm{PQR}$, in which $\mathrm{PQ}-\mathrm{PR}=2.4 \mathrm{cm}$, $\mathrm{QR}=6.4 \mathrm{cm}$ and $\angle \mathrm{PQR}=55°$ is possible.

A triangle, $△ABC$, with $\angle B=70°, \angle C=60°$ and $AB+BC+AC=11.2 \mathrm{cm}$ can be constructed.

We can construct a $△ABC$ in which $\angle B=45°, \angle C=60°$ and the perpendicular from the vertex $A$ to base $BC$ is $4.5 \mathrm{cm}$.

We can construct a right triangle whose one side is $3.5 \mathrm{cm}$ and the sum of the other side and the hypotenuse is $5.5 \mathrm{cm}$.

The incorrect step(s) of construction of a $△ABC$ whose perimeter is $11.6 \mathrm{cm}$ and the base angles are $45°$ and $60°$ is/are

What is the measure of the side of an equilateral triangle constructed rounded off to one decimal place in $\mathrm{cm}$, if each of whose altitudes measures $5.4 \mathrm{cm}$?

We can construct a triangle whose perimeter is $10.4 \mathrm{cm}$ and the base angles are $45°$ and $120°$.

We can construct a right triangle whose base is $12 \mathrm{c}\mathrm{m}$ and sum of its hypotenuse and other side is $18 \mathrm{c}\mathrm{m}$.

We can construct an equilateral triangle, given its side.

Arrange the following steps of construction of $\Delta ABC$ in which $AB=5.8 \mathrm{cm},BC+CA=8.4 \mathrm{cm}$ and $\angle B=60°$ in correct sequence.

Step I : Join$AD$.

Step II : From ray $BX$, cut off line segment $BD=BC+CA=8.4 \mathrm{cm}$

Step III : Draw a line segment $AB$ of length $5.8 \mathrm{cm}$.

Step IV : Draw a perpendicular bisector of $AD$ meeting $BD$ at point $C$. Join $AC$ to obtain $△ABC$.

Step V: Draw $\angle ABX=60°$ at point $B$ of line segment $AB$.

Which of the following steps of construction is incorrect while constructing a $\Delta ABC$ of perimeter $50 \mathrm{cm}$ and $\angle B=60°,\angle C=70°$.

Step I : Draw a line segment $XY$ equal to the perimeter $(=50 \mathrm{cm})$ of $\Delta \mathrm{ABC}$.

Step II : Construct $\angle YXD=\angle B$ and $\angle XYE=\angle C$.

Step III : Draw bisectors of angles $\angle YDX$ and $\angle XEY$ and mark their intersection point as $A$.

Step IV : Draw the perpendicular bisectors of $XA$ and $YA$ meeting $XY$ in $B$ and $C$ respectively.

Step V: Join $AB$ and $AC$ to obtain the required triangle $ABC$.

Arrange the following steps of construction of a $△PQR$ in which $QR=5 \mathrm{cm},\angle PQR=60°$ and $QP-PR=2 \mathrm{cm}$ in correct sequence.

Step I : Make an angle$XQR=60°$ at point $Q$ of base $QR$.

Step II : Join $SR$ and draw the perpendicular bisector of $SR$ that intersect $QX$ at $P$.

Step III : Draw the base $QR$ of length $5 \mathrm{cm}$.

Step IV : Join$PR$ to obtain $△PQR$.

Step V: Cut the line segment $QS=QP-PR=2 \mathrm{cm}$ from the ray $QX$.

Arrange the following steps of construction of a $△ABC$ in which $BC=8 \mathrm{cm}, \angle B=60°$ and the difference between the other two sides is $3 \mathrm{cm}$ in correct sequence.

Step I : Set off $BP=3 \mathrm{cm}$.

Step II : Draw $BC=8 \mathrm{cm}$.

Step III: Construct $\angle CBX=60°$

Step IV: Join $AC$. Then, $△ABC$ is the required triangle.

Step V: Draw the right bisector of  $PC$, meeting $PB$ produced at $A$.

Step VI: Join $PC$.

Arrange the following steps of construction of a $△ABC$, in which $BC=3.8 \mathrm{cm},\angle B=45°$ and $AB+AC=6.8 \mathrm{cm}$ in correct sequence.

Step $\mathrm{I}$ : Draw the perpendicular bisector of $CD$ meeting $BD$ at $A$.

Step $\mathrm{II}$ : Draw $BC=3.8 \mathrm{cm}$.

Step $\mathrm{III}$ : Join $CD$.

Step $\mathrm{IV}$ : From ray $BX$, cut off line segment $BD$ equal to $AB+AC$ i.e, $6.8 \mathrm{cm}$.

Step $\mathrm{V}$: Draw $\angle CBX=45°$

Step $\mathrm{VI}$: Join $CA$ to obtain the required $△ABC$.

Study the statements carefully and select the correct option.

Statement-I: It is possible to construct a triangle whose sides measure $7 \mathrm{cm}, 5 \mathrm{cm}$ and $12\mathrm{cm}$.

Statement-II : It is possible to construct an angle of $22.5°$ using ruler and compass only.

Study the statements carefully and select the correct option.

Statement-I : The sum of any two sides of a triangle is always greater than the third side.

Statement-II : It is possible to construct a $△ABC$ in which $AB=5 \mathrm{cm}, BC=5 \mathrm{cm}$ and $AC=10 \mathrm{cm}$.