Question 15 Marks
Construct an isosceles right$-$angled triangle whose hypotenuse is of length $6 \ cm.$
AnswerLet, $\triangle UVW$ be the isosceles right$-$angled triangle,
right$-$angled at $U.$
Hypotenuse $VW = 6 \ cm$
$UV = UW$
$\Rightarrow \angle UWV = \angle UVW$
$\angle U = 90^\circ $
$\Rightarrow \angle UWV + \angle UVW = 90^\circ $
$\Rightarrow 2\angle UVW = 90^\circ $
$\Rightarrow \angle UWV = \angle UVW = 45^\circ $
Steps:
$1.$ Draw $VW = 6 \ cm.$
$2.$ Construct $\angle WVY = 45^\circ $ and $\angle VWX = 45^\circ $
$3.$ Ray $VY$ and ray $WX$ meet at $U.$
Thus, $\text{UVW}$ is the required triangle.
View full question & answer→Question 25 Marks
Construct an isosceles triangle using the given data: Altitude $XT = 6.8\ cm$ and vertex $\angle X = 30^\circ $
AnswerAltitude $XT = 6.8\ cm$ and vertex $\angle X = 30^\circ $
Steps of construction:
$1.$ Draw a line $SU$ of any length.
$2$. Take a point $T$ on $SU.$
$3$. Through the point $T$ on $SU$
draw $NT$ perpendicular to $SU.$
$4$. With $T$ as centre and radius $6.8 \ cm$, draw an arc to cut $NT$ at $X.$
$5$. Construct $\angle TXY =\angle TXZ =\frac{1}{2} \times 30^{\circ}=15^{\circ}$.
$(a)$ With $X$ as centre, draw an arc cutting $XT$ at $L..$
$(b)$ With $L$ as centre and same radius, cut the arc at $P$ and $Q$.
$(c)$ Join $PX$ and $QX.$
$(d)$ Bisect $\angle P X T$ and $\angle Q X T$. Let $XA$ and $XB$ be the bisectors.
$(e)$ Again bisect $\angle A X T$ and $\angle B X T$.
Let $X R$ and $X V$ be the bisectors. $X R$ and $X V$ make an angle of $15^{\circ}$ with $XT.$
$(f)$ Mark the points as $Y$ and $Z$ where $X R$ and $X V$ meet $SU.$
Thus, $\text{XYZ}$ is the required triangle. View full question & answer→Question 35 Marks
Construct an isosceles triangle using the given data: Altitude $AD = 4\ cm$ and vertex $\angle A = 90^\circ $
AnswerAltitude $AD = 4\ cm$ and vertex $∠A = 90°$
Steps of construction:
$1.$ Draw a line a $SU$ of any length.
$2.$ Take a point $D$ on $SU.$
$3$. Through the point $D$ on $SU$ draw $ND$ perpendicular to $SU.$
$4$. With $D$ as centre and radius $4\ cm$, draw an arc to cut $ND$ at $A$.
$5$. Construct $\angle DAB =\angle DAC =\frac{1}{2} \times 90^{\circ}=45^{\circ}$.
$(a)$ With $A$ as centre, draw an arc cutting $AD$ at $L$.
$(b)$ With $L$ as centre and same radius, cut the arc at $X$ and $Y$.
$(c)$ Using $X$ and $Y$, draw $P Q$ perpendicular to $A D$.
$(d)$ Bisect $\angle P A D$ and $\angle Q A D$. Let $A T$ and $A V$ are the bisectors. $A T$ and $A V$ make an angle of $45^{\circ}$ with $A D$.
$(e)$ Mark the points as $B$ and $C$
where $AR$ and $AV$ meet $SU.$
Thus, $\text{ABC}$ is the required triangle. View full question & answer→Question 45 Marks
Construct an isosceles triangle using the given data: Altitude $RM = 5\ cm$ and vertex $\angle R = 120^\circ $
AnswerAltitude $RM = 5\ cm$ and vertex $\angle R = 120^\circ $
Steps of construction:
$1.$ Draw a line $SU$ of any length.
$2.$ Take a point $M$ on $SU.$
$3.$ Through the point $M$ on $SU$ draw $NM$ perpendicular to $SU.$
$4.$ With $M$ as centre and radius $5 \ cm$, draw an arc to cut $N M$ at $R$.
$5.$ Construct $\angle MRP =\angle MRQ =\frac{1}{2} \times 120^{\circ}=60^{\circ}$.
$(a)$ With $R$ as centre, draw an arc cutting $RM$ at $L$.
$(b)$ With $L$ as centre and same radius, cut the arc at $X$ and $Y$.
$(c)$ Join $RX$ and $44$ and produce them to $T$ and $V$ respectively. $R T$ and $R V$ make an angle of $60^{\circ}$ with $Rm.$
$(d)$ Mark the points as $P$ and $Q$ where $RT$ and $RV$ meet $SU.$
Thus, $\text{PQR}$ is the required triangle. View full question & answer→Question 55 Marks
Construct an isosceles triangle in which: Base $DE = 6 \ cm$ and $\angle F = 45^\circ $
AnswerIn isosceles $\triangle DEF,$
Base $DE = 6\ cm$
$\angle F = 45^\circ ....($given$)$
$\Rightarrow \angle D = \angle E ....(\text{DEF}$ is isosceles triangle$)$
Now, $\angle D + \angle E + \angle F = 180^\circ $
$2\angle D + 45 = 180^\circ $
$2\angle D = 135^\circ $
$\Rightarrow \angle D = \angle E = 67.5^\circ $
Steps:
$1.$ Draw $DE = 6\ cm.$
$2.$ Construct $\angle DEP = 67.5^\circ $ and $\angle EDQ = 67.5^\circ $
$3$. Ray $EP$ and ray $DQ$ meets at $F.$
Thus, $\text{DEF}$ is the required triangle.
View full question & answer→Question 65 Marks
Construct an isosceles triangle in which: $XY = XZ, YZ = 5.5 \ cm$ and $\angle X = 60^\circ $
AnswerIn $\triangle XYZ,$
$XY = XZ ....($given$)$
$c \angle XZY = \angle XYZ ....(i)$
$\angle X = 60^\circ ....($given$)$
Now, $\angle X + \angle Y + \angle Z = 180^\circ $
$60^\circ + \angle Y + \angle Y = 180^\circ ....[$From $(i)]$
$2\angle Y = 120^\circ $
$\Rightarrow \angle Y = 60^\circ = \angle Z$
Steps:
$1.$ Draw $YZ = 5.5\ cm.$
$2.$ Construct $\angle YZP = 60^\circ $ and $\angle ZYQ = 60^\circ $
$3.$ Ray $ZP$ and $YQ$ meet at $x.$
Thus, $\text{XYZ}$ is the required triangle.
View full question & answer→Question 75 Marks
Construct an isosceles triangle in which: $AB = AC, BC = 6 \ cm$ and $\angle B = 75^\circ $
AnswerIn $\triangle ABC,$
$AB = AC ....($given$)$
$\Rightarrow \angle ACB = \angle ABC = 75^\circ $
Steps:
$1.$ Draw $BC = 6\ cm.$
$2$. Construct angle $\angle BCM = 75^\circ $ and $\angle CBN = 75^\circ $
$3.$ Ray $CM$ and ray $BN$ meets at $A.$
Thus, $\text{ABC}$ is required angle.
View full question & answer→Question 85 Marks
Construct a right$-$angled triangle in which: $QP = QR$ and hypotenuse $PR = 7 \ cm$
AnswerIn $\triangle PQR,$
$QP = QR ....($given$)$
$\Rightarrow \angle QPR = \angle QRP$
Since hypotenuse $PR = 7\ cm, \angle PQR = 90^\circ $
$\therefore \angle QPR + \angle QRP = 90^\circ $
$\Rightarrow \angle QPR = \angle QRP = 45^\circ $
Steps:
$1$. Draw $PR = 7\ cm.$
$2$. Draw a ray $PT$ such as $\angle RPT = 45^\circ $ and ray $RS$ such as $\angle PRS = 45^\circ $
$3$. Ray $RS$ and ray $PT$ meets at $Q.$
Thus, $\text{PQR}$ is the required triangle.
View full question & answer→Question 95 Marks
Construct a triangle using the given data: $DE = 5\ cm, \angle D = 75^\circ $ and $\angle E = 60^\circ $
Answer$DE = 5\ cm, \angle D = 75^\circ $ and $\angle E = 60^\circ $
Steps of Construction:
$1.$ Draw a line segment $DE = 5\ cm.$
$2.$ With $D $as centre, draw an arc cutting $DE$ at $P.$
$3.$ With $P$ as centre and same radius, cut the arc at $Q$ and then from $Q,$ with same radius, cut the arc at $R.$
$4$. With $Q$ and $R$ as centre bisect $\angle RDQ$ thus formed to draw a ray $XD.$
$5.$ Again bisect the $\angle XDQ$. Let $DY$ be the bisector. $DY$ makes an angle of $75^\circ $ with $DE.$
$6.$ With $E$ as centre, draw an arc meeting $DE$ at $S.$
$8.$ Produce $ET$ to $EZ. EZ$ makes an angle of $60^\circ $ with $DE.$
$9.$ Mark the point as $F$, where $DY$ and $EZ$ cut each other.
Thus, $\text{DEF}$ is the required triangle. View full question & answer→Question 105 Marks
Construct a triangle using the given data: $PQ = 6.2\ cm, \angle P = 105^\circ $ and $\angle = 45^\circ $
Answer$PQ = 6.2\ cm, \angle P = 105^\circ $ and $\angle = 45^\circ $
Steps of Construction:
$1$. Draw a line segment $PQ = 6.2\ cm.$
$2$. With $P$ as centre, draw an arc meeting $PQ$ at $A.$
$3$. With $A$ as centre and same radius, cut the arc at $B$ and with $BQ$ as centre and same radius, cut the arc at $C.$
$4$. With $B$ and $C$ as centre, cut arcs and draw $PM$ perpendicular to $PQ.$
$5$. Bisect $\angle MPC$. Let $PX$ be the bisector. $PX$ makes an angle of $105^\circ $ with $PQ.$
$6$. With $Q$ as centre, draw an arc meeting $PQ$ at $S.$
$7$. With $S$ as centre and same radius, cut the arc at $T$ and with $T$ as centre and same radius, cut the arc at $U.$
$8.$ With $T$ and $U$ as centre, cut arcs and draw $QN$ perpendicular to $PQ.$
$9$. Bisect $\angle NQP.$ Let $QY$ be the bisector. $QY$ makes an angle of $45^\circ $ with $PQ.$
$10.$ Mark the point as $R$, where $PX$ and $QY$ cut each other.
Thus,$\text{PQR}$ is the required triangle. View full question & answer→Question 115 Marks
Construct a triangle using the given data: $BC = 6.0\ cm, \angle B = 60^\circ $ and $\angle C = 45^\circ $
Answer$BC = 6.0\ cm, \angle B = 60^\circ $ and $\angle C = 45^\circ $
Steps of Construction:
$1$. Draw a line segment $BC = 6\ cm.$
$2$. With $B$ as centre, draw an arc meeting $BC$ at $M.$
$3$. With M as centre and same radius, cut the arc at $N.$
$4$. Produce $BN$ to $BX. BX$ makes an angle of $60^\circ $ with $BC.$
$5$. With $C$ as centre, draw an arc meeting $BC$ at $P.$
$6$. With $P$ as centre and same radius, cut the arc at $Q$ and with $Q$ as centre and same radius, cut the arc at $R.$
$7.$ With $Q$ and $R$ as centre, cut arcs and draw $CY$ perpendicular to $BC.$
$8$. Bisect $\angle YCB$. Let $CZ$ be the bisector. $CZ$ makes an angle of $45^\circ $ with $BC.$
$9$. Mark the point as $A$, where $CZ$ and $BX$ cut each other.
Thus, $\text{ABC}$ is the required triangle. View full question & answer→Question 125 Marks
Construct a triangle using the given data: $PQ = 6.2\ cm, QR = 9.0\ cm$ and $\angle Q = 30^\circ $
Answer$PQ = 6.2\ cm, QR = 9.0\ cm$ and $\angle Q = 30^\circ $
Steps of Construction:
$1$. Draw a line segment $PQ = 6.2\ cm$
$2.$ With $Q$ as centre, draw an arc cutting $PQ$ at $M.$
$3$. With $M$ as centre and same radius, cut the arc at $N$. Join $QN.$
$4$. Bisect $\angle NQP$. Let $QY$ be the bisector. $QY$ makes an angle of $30^\circ $ with $PQ.$
$5$. With $Q$ as centre and radius $9\ cm$ cut an arc on $QY$. Mark the point as $R.$
$6$. Join $PR.$
Thus, $\text{PQR}$ is the required triangle. View full question & answer→Question 135 Marks
Construct a triangle using the given data: $XY = 5.2\ cm, XZ = 6.5\ cm$ and $\angle X = 75^\circ $
Answer$XY = 5.2\ cm, XZ = 6.5\ cm$ and $\angle X = 75^\circ $
Steps of Construction:
$1.$ Draw a line segment $XY = 5.2\ cm.$
$2.$ With $X$ as centre, draw an arc cutting $XY$ at $L.$
$3.$ With $L$ as centre and the same radius, cut the arc at $M$ and then from $M$, with same radius, cut the arc at $N$.
$4$. With $M$ and $N$ as centre bisect $\angle MXN$ thus formed to draw a ray $XP.$
$5$. Again bisect the $\angle MXP$. Let $XR$ be the bisector.$XR$ makes an angle of $75^\circ $ with $XY.$
$6$. With $X$ as centre and radius $6.5\ cm$ cut an arc on $XR$ and mark the point as $Z.$
$7$. Join $YZ.$
Thus, $\text{XYZ}$ is the required triangle. View full question & answer→Question 145 Marks
Construct a $\triangle ABC$, right$-$angled at $B$ with a perimeter of $10 \ cm$ and one acute angle of $60^\circ .$
Answer
Steps of construction:
$1.$ Draw $DE = 10 \ cm.$
$2.$ Draw $DP$ and $EQ$ such that $\angle PDE = 90^\circ $ and $\angle QED = 60^\circ .$
$3$. Draw $AD$ and $AE$, the bisectors of angles $\text{PDE}$ and $\text{QED}$ respectively, intersecting each other at $A.$
$4$. Draw perpendicular bisector of $AD$ and $AE$, interesting $DE$ at points $B$ and $C$ respectively.
$5$. Join $AB$ and $AC.$
Thus, $\text{ABC}$ is the required triangle. View full question & answer→Question 155 Marks
Construct a $\triangle PQR$ with $\angle Q = 60^\circ , \angle R = 45^\circ $ and the perpendicular from $P$ to $QR$ be $3.5 \ cm.$ Measure $PQ.$
Answer
Steps of construction:
$1.$ Draw a line segment $ST$ of any length.
$2.$ From any point $Y$ on $ST$, draw $XY$ perpendicular to $ST.$
$3.$ With $Y$ as centre and radius $3.5 \ cm$ mark a point $P$ on $XY.$
$4.$ With $P$ as centre, draw an arc cutting $XY$ at $L.$
$5.$ With $L$ as centre and same radius, cut the arc at $O$ and $M$. With $M$ as centre and same radius cut the arc at $N.$
$6$. Draw $PZ$ perpendicular to $XY$ using $M$ and $N.$
$7.$ Bisect angles $OPY$ and $ZPY$ making $30^\circ $ and $45^\circ $ angles with $PY$ respectively. $($In $\triangle PQY, \angle PQY = 60^\circ , \angle QYP = 90^\circ ; \therefore \angle QPY = 30^\circ $ and in $\triangle PYR, \angle YRP = 45^\circ , \angle RYP = 90^\circ ; \therefore \angle YPR = 45^\circ ).$
$8$. Join $PQ$ and $PR. \text{PQR}$ is the required triangle.
$9$. On measuring, $PQ = 4.1 \ cm.$ View full question & answer→Question 165 Marks
Construct a $\triangle PQR$ with $PQ = 5.4 \ cm, QR = 4.6 \ cm$ and $\angle Q = 60^\circ $. Draw the perpendicular $PS$ at $QR,$ measure the lengths of $SP$ and $SQ,$
AnswerSteps:
$1$. Draw $PQ = 5.4 \ cm.$
$2$. At $P$, construct $\angle PQT = 60^\circ .$
$3.$ With $Q$ as centre and radius $4.6 \ cm$, draw an arc intersecting ray $QT$ at $R.$
$4$. Join $PR.$
Thus, $\text{PQR}$ is the required triangle.
Draw $PU$, the perpendicular bisector of $QR$ intersecting $QR$ at $S.$
Then, we have
$SQ = 2.3 \ cm$ and $SP = 4.8 \ cm$
View full question & answer→Question 175 Marks
Construct a $\triangle XYZ$ with $YZ = 7.5 \ cm, \angle Y = 60^\circ $ and $\angle Z = 45^\circ $. Draw the bisectors of $\angle Y$ and $\angle Z$. If these bisectors meet at $O$, measure $\angle YOZ.$
Answer
Steps of construction:
$1.$ Draw a line segment $YZ = 7.5 \ cm.$
$2.$ With $Y$ as centre, draw an arc cutting $YZ$ at $L.$
$3.$ With $L$ as centre and same radius, cut the arc at $M.$
$4.$ Join $Y$ and $M$. Produce $YM$ to $S. YS$ makes an angle of $60^\circ $ with $YZ.$
$5$. With $Z$ as centre, draw an arc cutting $YZ$ at $P.$
$6.$ With $P$ as centre and same radius, cut the arc at $Q,$ and with $Q$ as centre and same radius cut the arc at $R$. Using $Q$ and $R$, Draw $UZ$ perpendicular to $YZ.$
$7.$ Bisect $\angle UZY.$ Let $TZ$ be the bisector. $TZ$ makes an angle of $45^\circ $ with $YZ.$
$8.$ Bisect $\angle SYZ$ and $\angle TZY.$
$9.$ Mark the point as $O$ where the bisectors of $\angle SYZ$ and $\angle TZY$ meet.
$10.$ On measuring $\angle YOZ = 127.5^\circ .$ View full question & answer→Question 185 Marks
Construct a triangle using the given data: Perimeter of triangle is $10.6 \ cm$, and the base angles are $60^\circ $ and $45^\circ $
AnswerPerimeter of triangle is $10.6 \ cm$, and the base angles are $60^\circ $ and $45^\circ $
Steps of construction:
$1.$ Draw $DE = 10.6 \ cm.$
$2.$ Draw $DP$ and $EQ$ such that $\angle PDE = 90^\circ $ and $\angle QED = 60^\circ .$
$3$. Draw $AD$ and $AE$, the bisectors of angles $PDE$ and $QED$ respectively, intersecting each other at $A.$
$4$. Draw perpendicular bisectors of $AD$ and $AE$, intersecting $DE$ at points $B$ and $C$ respectively.
$5$. Join $AB$ and $AC.$
Thus, $\text{ABC}$ is the required triangle. View full question & answer→Question 195 Marks
Construct a triangle using the given data: Perimeter of triangle is $9 \ cm$, and the base angles are $60^\circ $ and $45^\circ $
AnswerPerimeter of triangle is $9 \ cm$, and the base angles are $60^\circ $ and $45^\circ $
Steps of construction:
$1$. Draw $DE = 9 \ cm.$
$2.$ Draw $DP$ and $EQ$ such that $\angle PDE = 45^\circ $ and $\angle QED = 60^\circ .$
$3.$ Draw $AD$ and $AE,$ the bisectors of angles $\text{PDE}$ and $\text{QED}$ respectively, intersecting each other at $A.$
$4.$ Draw perpendicular bisectors of $AD$ and $AE,$ intersecting $DE$ at points $B$ and $C$ respectively.
$5.$ Join $AB$ and $AC.$
Thus, $\text{ABC}$ is the required triangle. View full question & answer→Question 205 Marks
Construct a triangle using the given data: Perimeter of triangle is $6.4 \ cm$, and the base angles are $60^\circ $ and $45^\circ $
AnswerPerimeter of triangle is $6.4 \ cm$, and the base angles are $60^\circ $ and $45^\circ $
Steps of construction:
$1$. Draw $DE = 6.4 \ cm.$
$2$. Draw $DP$ and $EQ$ such that $\angle PDE = 45^\circ $ and $\angle QED = 60^\circ $
$3$. Draw $AD$ and $AE$, the bisectors of angles $\text{PDE}$ and $\text{QED}$ respectively, intersecting each other at $A.$
$4.$ Draw perpendicular bisectors of $AD$ and $AE$, intersecting $DE$ at points $B$ and $C$ respectively.
$5.$ Join $AB$ and $AC.$
Thus, $\text{ABC}$ is the required triangle. View full question & answer→Question 215 Marks
Construct a triangle using the given data: $XY - XZ = 1.5 \ cm, YZ = 3.4 \ cm$ and $\angle X = 45^\circ $
Answer$XY - XZ = 1.5 \ cm, YZ = 3.4 \ cm$ and $\angle X = 45^\circ $
Steps of Construction:
$1.$ Draw a line segment $YZ = 3.4 \ cm.$
$2$. With Y as centre, draw $\angle TYZ = 45^\circ .$
$3$. From $YT$, cut $YS = 1.5 \ cm.$
$4$. Join $S$ and $Z.$
$5$. Draw perpendicular bisector of $SZ$ which cuts $YT$ at $X$.
$6.$ Join $XZ.$
Thus, $\text{XYZ}$ is the required triangle. View full question & answer→Question 225 Marks
Construct a triangle using the given data: $AB - AC = 1.2 \ cm, BC = 6.0 \ cm$ and $\angle B = 60^\circ $
Answer$AB - AC = 1.2 \ cm, BC = 6.0 \ cm$ and $\angle B = 60^\circ $
Steps of Construction:
$1$. Draw a line segment $BC = 6 \ cm.$
$2$. With $B$ as centre, draw $\angle TBC = 60^\circ .$
$3$. From $BT$, cut $BS = 1.2 \ cm.$
$4$. Join $S$ and $C.$
$5$. Draw perpendicular bisector of $SC$ which cut $BT$ at $A$.
$6$. Join $AC.$
Thus,$\text{ABC}$ is the required triangle. View full question & answer→Question 235 Marks
Construct a triangle using the given data: $PQ - PR = 1.5 \ cm, QR = 6.0$ and $\angle Q = 45^\circ $
Answer$PQ - PR = 1.5 \ cm, QR = 6.0$ and $\angle Q = 45^\circ $
Steps of Construction:
$1$. Draw a line segment $QR = 6 \ cm.$
$2$. With $Q$ as centre, draw $\angle TQR = 45^\circ $
$3$. From $QT$, cut $QS = 1.5 \ cm.$
$4$. Join $S$ and $R$.
$5$. Draw perpendicular bisector of $SR$ which cuts $QT$ at $P.$
$6$. Join $PR.$
Thus, $\text{PQR}$ is the required triangle. View full question & answer→Question 245 Marks
Construct a triangle using the following data: $DE + EF = 10.3 \ cm, EF = 6.4 \ cm$ and $\angle E = 75^\circ $
Answer$DE + EF = 10.3 \ cm, EF = 6.4 \ cm$ and $\angle E = 75^\circ $
Steps of construction:
$1$. Draw a line segment $EF = 6.4 \ cm.$
$2.$ With $E$ as centre, construct $\angle SEF = 75^\circ $
$3.$ Cut $ET = 10.3 \ cm$ on $ES.$
$4$. Join $TF.$
$5$. Draw perpendicular bisector of $TF$ which cuts $ET$ at $D.$
$6$. Join $DF.$
Thus $\text{DEF}$ is the required triangle. View full question & answer→Question 255 Marks
Construct a triangle using the following data: $PQ + PR = 10.6 \ cm, QR = 4.8 \ cm$ and $\angle R = 45^\circ $
Answer$PQ + PR = 10.6 \ cm, QR = 4.8 \ cm$ and $\angle R = 45^\circ $
Steps of construction:
$1.$ Draw a line segment $QR = 4.8 \ cm$
$2.$ With $Q$ as centre, construct $\angle SQR = 45^\circ $
$3.$ Cut $QT = 10.6 \ cm$ on $QS.$
$4$. Join $TR.$
$5$. Draw perpendicular bisector of $TR$ which cuts $QT$ at $P.$
$6.$ Join $PR.$
Thus $\text{PQR}$ is the required triangle. View full question & answer→Question 265 Marks
Construct a triangle using the following data: $XY + YZ = 5.6 \ cm, XZ = 4.5 \ cm$ and $\angle X = 45^\circ $
Answer$XY + YZ = 5.6 \ cm, XZ = 4.5 \ cm$ and $\angle X = 45^\circ $
Steps of construction:
$1$. Draw a line segment $XZ = 4.5 \ cm$
$2$. With $X$ as centre, construct $\angle SXZ = 45^\circ $
$3$. Cut $XT = 5.6 \ cm$ on $XS.$
$4$. Join $TZ.$
$5$. Draw perpendicular bisector of $TZ$ which cuts $XT$ at $Y$.
$6.$ Join $YZ.$
Thus $\text{XYZ}$ is the required triangle. View full question & answer→