Question
Construct an isosceles triangle using the given data: Altitude $RM = 5\ cm$ and vertex $\angle R = 120^\circ $
✓
Answer
Altitude $RM = 5\ cm$ 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.
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.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In the given figure, $A D$ is the median on $B C$ from $A$. If $A D=8 \ cm$ and $B C=12 \ cm$, find the value of $\frac{1}{\sin ^2 x}-\frac{1}{\tan ^2 x}$
→The length breadth and height of a cuboid are in the ratio of $3: 3: 4$. Find its volume in $m^3$ if its diagonal is $5 \sqrt{34} \ cm$.
→Show that: $\frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{2}{5-\sqrt{3}}=\frac{52}{11}$
→The difference between the compound interest for 1 year, compounded half-yearly and the simple interest for 1 year on a certain sum of money at 10% per annum is ₹ 360. Find the sum.
In the following, find the values of $a$ and $b:\frac{3+\sqrt{7}}{3-\sqrt{7}}=a+b \sqrt{7}$
→In the given figure, $A B=B C$ and $A C=C D$. Show that : $\angle BAD : \angle ADB =3: 1$
→The cross-section of a tunnel perpendicular to its length is a trapezium $\text{ABCD}$ as shown in the following figure; also given that:$A M=B N ; A B=7 m ; C D=5 m$. The height of the tunnel is $2.4 m $. The tunnel is $40 m$ long. Calculate:
$(i)$ The cost of painting the internal surface of the tunnel $($excluding the floor$)$ at the rate of $Rs. 5$ per $m^2 ($sq. meter$).(ii)$ The cost of paving the floor at the rate of $Rs.18$ per $m^2.$
→$(i)$ The cost of painting the internal surface of the tunnel $($excluding the floor$)$ at the rate of $Rs. 5$ per $m^2 ($sq. meter$).(ii)$ The cost of paving the floor at the rate of $Rs.18$ per $m^2.$
In $\triangle A B C, \angle B=90^{\circ}$ and $\tan A=0.75$. If $A C=30 \ cm$, find the lengths of $A B$ and $B C$.
→In the adjoining figure, ABCD is a parallelogram. Any line through A cuts DC at a point P and BC produced at Q. Prove that :ar $(\triangle BPC )=\operatorname{ar}(\triangle DPQ )$.
→If $x=\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}$ and $y=\frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}$, find the values of $x^2-y^2+x y$
→