Mix Examples - Constructions Questions in English

(N/A) $1$. Draw a line segment $AB = 6\,cm$ as the base.
$2$. At point $B$,construct an angle of $90^\circ$ using a compass.
$3$. From the ray $BX$ (making $90^\circ$ with $AB$),cut off a line segment $BD = 12\,cm$.
$4$. Join $AD$.
$5$. Draw the perpendicular bisector of $AD$,which intersects $BD$ at point $C$.
$6$. Join $AC$. Thus,$\triangle ABC$ is the required right-angled triangle where $BC + AC = 12\,cm$ and $AB = 6\,cm$.

(N/A) Steps of construction:
$1$. Draw a line segment $QR = 4\,cm$.
$2$. At point $Q$,construct an angle $\angle XQR = 45^{\circ}$.
$3$. From the ray $QX$,cut off a line segment $QD = 8\,cm$ (which is equal to $PQ + PR$).
$4$. Join $DR$.
$5$. Construct the perpendicular bisector of $DR$,which intersects $QD$ at point $P$.
$6$. Join $PR$. Thus,$\Delta PQR$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $BC = 6\,cm$.
$2$. At point $B$,construct an angle $\angle XBC = 45^{\circ}$.
$3$. Cut a line segment $BD = 1.5\,cm$ from the ray $BX$.
$4$. Join $DC$.
$5$. Draw the perpendicular bisector of $DC$,which intersects $BX$ at a point $A$.
$6$. Join $AC$. Thus,$\Delta ABC$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $QR = 6\,cm$.
$2$. At point $Q$,construct an angle $\angle XQR = 60^{\circ}$.
$3$. Cut a line segment $QS = 1\,cm$ from the ray $QX$ such that $QS = PR - PQ$.
$4$. Join $SR$.
$5$. Draw the perpendicular bisector of $SR$,let it intersect $QX$ at point $P$.
$6$. Join $PR$. Thus,$\Delta PQR$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $XY = 14 \, cm$ which is equal to the perimeter of the triangle $(AB + BC + CA)$.
$2$. Construct $\angle LXY = 30^{\circ}$ at point $X$ and $\angle MYX = 90^{\circ}$ at point $Y$.
$3$. Bisect $\angle LXY$ and $\angle MYX$. Let the bisectors intersect at point $A$.
$4$. Draw the perpendicular bisector of $AX$ to intersect $XY$ at $B$.
$5$. Draw the perpendicular bisector of $AY$ to intersect $XY$ at $C$.
$6$. Join $AB$ and $AC$. Thus,$\Delta ABC$ is the required triangle.

(N/A) Steps of construction:
$1$. Draw a line segment $AB = 3\,cm$ as the base.
$2$. At point $A$,construct an angle of $90^{\circ}$ using a compass.
$3$. From the ray $AX$ (making $90^{\circ}$ with $AB$),cut off a line segment $AD = 9\,cm$.
$4$. Join $DB$.
$5$. Construct the perpendicular bisector of $DB$,which intersects $AD$ at point $C$.
$6$. Join $CB$. Thus,$\triangle ABC$ is the required right-angled triangle.
Verification: In $\triangle ABC$,$AB = 3\,cm$,$\angle A = 90^{\circ}$,and $AC + CB = AD = 9\,cm$ (since $C$ lies on the perpendicular bisector of $DB$,$CD = CB$).