Construction of Triangle

Construct a triangle $\mathrm{A}\mathrm{B}\mathrm{C}$ in which $\mathrm{B}\mathrm{C}=7.5 \mathrm{c}\mathrm{m},$ $\mathrm{\angle }\mathrm{B}=46°$ and $\mathrm{A}\mathrm{B}+\mathrm{A}\mathrm{C}=13 \mathrm{c}\mathrm{m}$.

Paritosh wants to construct a triangle $\mathrm{RST}$, in which angle $\mathrm{S} = 45°, \mathrm{ST}= 10 \mathrm{cm}$ long and $\mathrm{RS}-\mathrm{RT} = 2 \mathrm{cm}$. He has completed construction of some steps.

Step $1:$ Draw the base $\mathrm{ST}$ of the triangle.

Step $2:$ At point $\mathrm{S},$ make an angle $\mathrm{RST}$ of measure $45°$.

What should be Paritosh’s next step?

Construct a triangle whose one of the base angle is $60°$ with base length $5 \mathrm{cm}$ and the sum of other two sides is $7 \mathrm{cm}$.

Construct a triangle with base $5.5 \mathrm{c}\mathrm{m}$, one of the base angle of value $60°$ and the sum of other two sides $8 \mathrm{c}\mathrm{m}$.
 

Write the steps for the construction of $△\mathrm{PQR}$, in which $\mathrm{PQ}-\mathrm{PR}=2.4 \mathrm{cm}$, $\mathrm{QR}=6.4 \mathrm{cm}$ and $\angle \mathrm{PQR}=55°$.

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.

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 $5.5 \mathrm{cm}$.

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

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 $△PQR$ 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 $△ABC$.

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?

Construct a $△$$ABC$, in which $AC = 6.2$cm, $\angle A= 60°$and $AB -BC$$=1.5$cm.

Construct a $△ABC$ in which $\mathrm{BC}=5 \mathrm{cm}$, $\angle B=45°$ and $AB-\mathrm{AC}=2 \mathrm{cm}$.

Construct a $△$ ABC, in which  AB$= 6.2$cm, $\angle$A = 60° and BC + AC $= 9.5$ cm

. Construct a $△$ ABC, in which  BC = 5.5 cm, $\angle$ B = 75° and AB + AC = 9.2 cm

Draw an isosceles triangle ABC, in which AB=AC, draw altitude AI from vertex A. Measure and state whether the following statements are true or not. $\mathrm{BL}=\mathrm{LC}$

In a triangle, how many vertices (vertices can be possible)?

The point at the plane of a triangle, which is at an equal distance from the sides of the triangle is: