Mahabir Singh Solutions for Chapter: Constructions, Exercise 2: ACHIEVERS SECTION (HOTS)

Following are the steps of construction of a $∆\mathrm{ABC}$ in which $\mathrm{AB}=5\mathrm{cm},\mathrm{\angle }\mathrm{A}=30°$ and $\mathrm{AC}-\mathrm{BC}=2.5\mathrm{cm}$. Arrange them and select the CORRECT option.
$\left(i\right)$ Draw $\mathrm{\angle }\mathrm{BAX}=30°$
$\left(ii\right)$ Draw the perpendicular bisector of $\mathrm{BD}$ which cuts $\mathrm{AX}$ at $\mathrm{C}.$
$\left(iii\right)$ Draw $\mathrm{AB}=5\mathrm{cm}$.
$\left(iv\right)$ Join $\mathrm{BD}$.
$\left(v\right)$ Join $\mathrm{BC}$ to obtain the required  $△\mathrm{ABC}$.
$\left(vi\right)$ From ray $\mathrm{AX}$, cut off line segment $\mathrm{AD}=\mathrm{AC}-\mathrm{BC}=2.5\mathrm{cm}$.

State $\mathrm{\text{'}}\mathrm{T}\mathrm{\text{'}}$ for true and $\mathrm{\text{'}}\mathrm{F}\mathrm{\text{'}}$ for false.

$\left(\mathrm{i}\right)$ A triangle whose sides measure $8\mathrm{cm},4\mathrm{cm}$ and $12\mathrm{cm}$ can be possible.
$\left(\mathrm{ii}\right)$ It is possible to construct an angle of $67.5°$ using ruler and compass only.
$\left(\mathrm{iii}\right)$ It is possible to construct a $∆XYZ$ in which $\angle X=60°,\angle Y=100°$ and $\angle Z=20°$

         $\left(\mathrm{i}\right)$          $\left(\mathrm{ii}\right)$          $\left(\mathrm{iii}\right)$

$\left(\mathrm{A}\right)$   T            F              T
$\left(\mathrm{B}\right)$   F            F              F
$\left(\mathrm{C}\right)$  F            T              T
$\left(\mathrm{D}\right)$  T            T              F

Let $\mathrm{ABC}$ be a triangle in which $\mathrm{BC}=5\mathrm{cm},\mathrm{\angle }\mathrm{B}=60°$ and $\mathrm{AC}+\mathrm{AB}=7.5\mathrm{cm}.$ Given below are the steps of constructing the $△\mathrm{ABC}.$ Which of the following steps is INCORRECT?
Step I: Draw a line segment $\mathrm{BC}$ of length $5\mathrm{cm}.$
Step II: Draw an $\mathrm{\angle }\mathrm{XBC}=60°$ at point $\mathrm{B}$ of line segment $\mathrm{BC}.$
Step III: Cut off  $\mathrm{PB}=3.5\mathrm{cm}$ on the ray $\mathrm{BX}.$
Step IV: Join $\mathrm{PC}.$.
Step V: Draw $\perp$bisector of $\mathrm{BC}$ which intersect ray $\mathrm{BX}$ at $\mathrm{A}.$ Join $\mathrm{AC}.$
Step VI: $△\mathrm{ABC}$ is the required triangle.

Following are the steps of construction of a rectangle $ABCD$ whose adjacent sides are of lengths $5cm$ and $3.5cm.$ Arrange them and select the CORRECT option.
$\left(\mathit{P}\right)$ Draw a line segment $BC$ of length $5cm$.
$\left(\mathit{Q}\right)$ With $A$ as centre, draw an arc of radius $5cm$.
$\left(\mathit{R}\right)$Draw an $\angle XBC=90°$ at point $B$ of line segment $BC.$
$\left(\mathit{S}\right)$ Cut a line segment $AB=3.5cm$ on $\stackrel{-}{BX}.$
$\left(\mathit{T}\right)$ With $C$ as centre, draw an arc of radius $3.5cm$ which intersects the arc at $D$.
$\left(\mathit{U}\right)$ Join $AD$ and $CD$.

Step $\mathrm{I}$ & Step $\mathrm{V}$ are in correct order while constructing an equilateral triangle one of whose altitudes measures $5\mathrm{cm}.$ Which of the following options is CORRECT while arranging the remaining steps in CORRECT order?
Step I: Draw a line $\mathrm{XY}.$
$\left(\mathit{i}\right)$ From $\mathrm{\angle }\mathrm{P},$ set off $\mathrm{PA}=5\mathrm{cm},$ cutting $\mathrm{PQ}$ at $\mathrm{A}$.
$\left(\mathit{i}\mathit{i}\right)$ From $\mathrm{P},$ draw $\mathrm{PQ}\perp \mathrm{XY}.$
$\left(\mathit{i}\mathit{i}\mathit{i}\right)$ Mark any point $\mathrm{P}$ on $\mathrm{XY}.$
Step V: Construct $\mathrm{\angle }\mathrm{PAB}=30°$ and $\mathrm{\angle }\mathrm{PAC}=30°$, meeting $\mathrm{XY}$ at $\mathrm{B}$ and $\mathrm{C}$ respectively.