Question
Construct an isosceles triangle using the given data: Altitude $XT = 6.8\ cm$ and vertex $\angle X = 30^\circ $
✓
Answer
Altitude $XT = 6.8\ cm$ 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.
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.
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