A triangle can be constructed by taking two of its angles as

Which of the following steps is incorrect while constructing $△XYZ$ if it is given that $XY=6 \mathrm{cm}$, $\angle ZXY=30°$ and $\angle XYZ=100°$

Step $1:$ Draw line $XY$ of length $6 \mathrm{cm}$.

Step $2:$ At $X$, draw a ray $XP$ making an angle of $30°$ with $XY$.

Step $3:$ At $Y$, draw a ray $YO$ making an angle of $100°$ with $YX$.

Step $4:$ The point of intersection of the two rays $XY$ and $YQ$ is $Z$.

Arrange the given steps in correct order, while constructing $△PQR$ where $PM\perp QS$ and it is given that $QR=4.2 \mathrm{cm}$, $\angle Q=120°$ and $PQ=3.5 \mathrm{cm}$.

Step $1:$ Now extend $RQ$ to $S$ and with $P$ as centre and with a sufficient radius, draw an arc, cutting $SQ$ at $A$ and $B$.

Step $2:$ Along $QX$, set off $QP=3.5 \mathrm{cm}$.

Step $3:$ Draw a line segment $QR=4.2 \mathrm{cm}$ and construct $\angle RQX=120°$.

Step $4:$ Joint $PR$.

Step $5:$ Joint $PC$, meeting $RQ$ produced at $M$. Then, $PM\perp QS$.

Step $6:$ With $A$ as centre and radius more than half $AB$, draw an arc. Now with $B$ as centre and with the same radius draw another arc, cutting the previous arc at $C$.

State ‘T’ for true and ‘F’ for false.

$\left(1\right)$ In a triangle, the measure of exterior angle is equal to the sum of the measure of interior opposite angles.

$\left(2\right)$ The sum of the measures of the three angles of a triangle is $90°$.

$\left(3\right)$ A perpendicular is always at $90°$ to a given line or surface.

Which of the following steps is incorrect while constructing $△\mathrm{LMN}$, right angled at $\mathrm{M}$, given that $\mathrm{LN}=5 \mathrm{cm}$ and $\mathrm{MN}=3 \mathrm{cm}$?

Step $1.$ Draw $\mathrm{MN}$ of length $3 \mathrm{cm}$.

Step $2.$ At $\mathrm{M}$, draw $\mathrm{MX}\perp \mathrm{MN}$. ($\mathrm{L}$ should be somewhere on this perpendicular).

Step $3.$ With $\mathrm{N}$ as centre,draw an arc of radius $5\mathrm{cm}$. ( $\mathrm{L}$ must be on this arc, since it is at a distance of $5 \mathrm{cm}$ from $\mathrm{N}$).

Step $4.$ $\mathrm{L}$ has to be on the perpendicular line $\mathrm{MX}$ as well as on the arc drawn with centre $\mathrm{N}$. Therefore, $\mathrm{L}$ is the meeting point of these two and $△\mathrm{LMN}$ is obtained.

Arrange the steps marked $\left(\mathrm{i}\right)$ to $\left(\mathrm{v}\right)$ in correct order while constructing a line parallel to a given line, through a point not on the line using ruler and compasses only.

Step $1.$ Take a line $l$ and a point $A$ outside $l$.

Step $2.$ Take any point $B$ on $l$ and join $B$ to $A$.

$\left(\mathrm{i}\right)$ Now with $A$ as centre and the same radius as in previous step, draw an arc $EF$ cutting $AB$ at $G$.

$\left(\mathrm{ii}\right)$ With the same opening as in previous step draw an arc cutting the arc $EF$ at $H$.

$\left(\mathrm{iii}\right)$ With $B$ as centre and a convenient radius, draw an arc cutting $l$ at $C$ and $BA$ at $D$.

$\left(\mathrm{iv}\right)$ Now, join $AH$ to draw a line $‘m’$.

$\left(\mathrm{v}\right)$ Now with $C$ as centre, open the compass for the radius equal to the arc $CD$.