Draw a regular pentagon and then a triangle of the same area. 

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 $AB$.

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 $PC$.

Steps of construction of triangle $\mathrm{A}\mathrm{B}\mathrm{C}$ in which $\mathrm{B}\mathrm{C}=8 \mathrm{c}\mathrm{m}, \mathrm{\angle }\mathrm{B}=45°$ and $\mathrm{A}\mathrm{B}-\mathrm{A}\mathrm{C}=1 \mathrm{c}\mathrm{m}$ are given below.

$\text{Step} 1\text{:}$ Draw $\mathrm{B}\mathrm{C}=8 \mathrm{c}\mathrm{m}$.

$\text{Step} 2\text{:}$ Construct $\mathrm{\angle }\mathrm{C}\mathrm{B}\mathrm{X}=45°$

$\text{Step} 3\text{:}$ Cut off $\mathrm{B}\mathrm{P}=9 \mathrm{c}\mathrm{m}$ on $\mathrm{BX}$. Join $\mathrm{C}\mathrm{P}$.

$\text{Step} 4\text{:}$ Draw perpendicular bisector for $\mathrm{P}\mathrm{C}$, meeting $\mathrm{BC}$ produced at $\mathrm{A}$.

$\text{Step} 5\text{:}$ Join $\mathrm{A}\mathrm{C}$. Then $△\mathrm{A}\mathrm{B}\mathrm{C}$ is the required triangle.

Which of the above step(s) is/are incorrect?

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

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 $QX$.

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 $AC=10 \mathrm{cm}$.

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 $△ABC$.

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

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 $ABC$.

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.