Question
Construct an isosceles triangle using the given data: Altitude $AD = 4\ cm$ and vertex $\angle A = 90^\circ $
✓
Answer
Altitude $AD = 4\ cm$ 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.
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.
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