Constructions

Class 9 NCERT Maths

NCERT

1   Construct an angle of $90^o$ at the initial point of a given ray and justify the construction.

Solution :

The below given steps will be followed to construct an angle of $90^o$.$\\$ (i) Take the given ray $PQ$. Draw an arc of some radius taking point $P$ as its centre, which intersects $PQ$ at $R.$$\\ (ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.\\ (iii) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).\\ (iv) Taking S and T as centre, draw an arc of same radius to intersect each other at U.$$\\$ (v) Join $PU$, which is the required ray making $90^0$ with the given ray $PQ.$$\\ \textbf{Justification of Construction:}$$\\$ We can justify the construction, if we can prove $\angle UPQ = 90^o.$$\\ For this, join PS and PT.$$\\$

$\textbf{Justification of Construction:}$$\\ We can justify the construction, if we can prove \angle WPQ = 45^o.\\ For this, join PS and PT.\\ We have, \angle SPQ = \angle TPS = 60^o. In (iii) and (iv) steps of this construction, PU was drawn as the bisector of \angle TPS.$$\\$ $\therefore \angle UPS=\dfrac{1}{2}\angle TPS=\dfrac{1}{2}*60^o=30^o$$\\ Also , \angle UPQ=\angle SPQ+\angle UPS$$\\$ $=60^o+30^o$$\\ =90^o$$\\$ In step (vi) of this construction, PW was constructed as the bisector of $\angle UPQ.$$\\ \therefore \angle WPQ=\dfrac{1}{2}\angle UPQ=\dfrac{90^o}{2}=45^o 3 Construct the angles of the following measurements:\\ (i)30^o$$\\$ (ii)$22\dfrac{1}{2}^o$$\\ (iii)15^o Solution : (i) 30^o$$\\$ The below given steps will be followed to construct an angle of $30^o$$\\. \text{Step I:} Draw the given ray PQ. Taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.$$\\$ $\text{Step II:}$ Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point $S.$$\\ \text{Step III:} Taking R and S as centre and with radius more than \dfrac{1}{2}RS draw arcs to intersect each other at T. Join PT which is the required ray making 30^o with the given ray PQ.$$\\$

(ii)$22\dfrac{1}{2}^o$$\\ The below given steps will be followed to construct an angle of 22 \dfrac{1}{2}^o$$\\$ (1) Take the given ray $PQ$. Draw an arc of some radius, taking point $P$ as its centre, which intersects $PQ$ at $R.$$\\ (2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.$$\\$ (3) Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).$\\$ (4) Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U.$$\\ (5) Join PU. Let it intersect the arc at point V.$$\\$ (6)From $R$ and $V$, draw arcs with radius more than$\dfrac{1}{2}RV$ to intersect each other at $W$. Join $PW.$$\\ (7) Let it intersect the arc at X. Taking X and R as centre and radius more than \dfrac{1}{2}RX draw arcs to intersect each other at Y.\\ Joint PY which is the required ray making 22\dfrac{1}{2}^o with the given ray PQ$$\\$

The angle so formed can be measured with the help of a protractor. It comes to be $75^o.$$\\ (ii) 105^o$$\\$ The below given steps will be followed to construct an angle of $105^o$$\\. (1) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersects PQ at R.\\ (2) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.\\ (3) Taking S as centre and with the same radius as before, draw an arc intersecting the arc at T (see figure).\\ (4) Taking S and T as centre, draw an arc of same radius to intersect each other at U.$$\\$ (5) Join $PU$. Let it intersect the arc at $V$. Taking $T$ and $V$ as centre, draw arcs with radius more than $\dfrac{1}{2}TV$. Let these arcs intersect each other at $W$. Join $PW$ which is the required ray making $105^o$ with the given ray $PQ.$$\\ The angle so formed can be measured with the help of a protractor. It comes to be 105^o. (iii) 135^o$$\\$ The below given steps will be followed to construct an angle of $135^o$$\\. (1) Take the given ray PQ. Extend PQ on the opposite side of Q. Draw a semi-circle of some radius taking point P as its centre, which intersects PQ at R and W.$$\\$ (2) Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.$\\$ (3) Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).$\\$ (4) Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U$.$\\$ (5)Join $PU$. Let it intersect the arc at $V$.Taking $V$ and $W$ as centre and with radius more than $\dfrac{1}{2}VW$, draw arcs to intersect each other at $X$. Join $PX$, which is the required ray making $135^o$ with the given line $PQ.$$\\ The angle so formed can be measured with the help of a protractor. It comes to be 135^o. 5 Construct an equilateral triangle, given its side and justify the construction. Solution : Let us draw an equilateral triangle of side 5 cm. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 5 cm. We also know that each angle of an equilateral triangle is 60^o.$$\\$ The below given steps will be followed to draw an equilateral triangle of $5 cm$ side.$\\$ $\text{Step I:}$ Draw a line segment $AB$ of $5 cm$ length. Draw an arc of some radius, while taking $A$ as its centre. Let it intersect $AB$ at $P.$$\\ \text{Step II:} Taking P as centre, draw an arc to intersect the previous arc at E. Join AE.$$\\$ $\text{Step III:}$ Taking $A$ as centre, draw an arc of $5 cm$ radius, which intersects extended line segment $AE$ at $C$. Join $AC$ and $BC$. $\Delta ABC$ is the required equilateral triangle of side $5 cm.$$\\ \textbf{Justification of Construction:}$$\\$ We can justify the construction by showing $ABC$ as an equilateral triangle i.e., $AB = BC = AC = 5 cm$ and $\angle A = \angle B = \angle C = 60^o$$\\. In \Delta ABC, we have AC = AB = 5 cm and \angle A = 60^o. Since AC = AB,$$\\$ $\angle B = \angle C$ (Angles opposite to equal sides of a triangle)$\\$ In $\Delta ABC,$$\\ \angle A + \angle B + \angle C = 180^o (Angle sum property of a triangle)\\ 60^o + \angle C + \angle C = 180^o$$\\$ $60^o + 2\angle C = 180^o$$\\ 2\angle C = 180^o - 60^o = 120^o$$\\$ $\therefore \angle C = 60^o$$\\ \angle B = \angle C = 60^o$$\\$ We have, $\angle A = \angle B = \angle C = 60^o ... (1)$$\\ \angle A = \angle B and \angle A = \angle C$$\\$ $\angle BC = AC$ and $BC = AB$ (Sides opposite to equal angles of a triangle)$\\$ $\angle AB = BC = AC = 5 cm ... (2)$$\\ From Equations (1) and (2), \Delta ABC is an equilateral triangle. 6 Construct a triangle ABC in which BC = 7 cm, \angle B = 75^o and AB + AC = 13 cm. Solution : The below given steps will be followed to construct the required triangle.\\ Step I: Draw a line segment BC of 7 cm. At point B, draw an angle of 75^o, say \angle XBC. \\ Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX. \\ Step III: Join DC and make an angle DCY equal to BDC.$$\\$ Step IV: Let $CY$ intersect $BX$ at $A$. $\Delta ABC$ is the required triangle.

7   Construct a triangle $ABC$ in which $BC = 8 cm, \angle B = 45^o$ and $AB - AC = 3.5 cm.$

The below given steps will be followed to draw the required triangle.$\\$ Step I: Draw the line segment $BC = 8 cm$ and at point $B$, make an angle of $45^o$, say $\angle XBC.$ $\\$ Step II: Cut the line segment $BD = 3.5 cm$ (equal to $AB - AC$) on ray $BX.$ Step III: Join $DC$ and draw the perpendicular bisector $PQ$ of $DC.$$\\ Step IV: Let it intersect BX at point A. Join AC. \Delta ABC is the required triangle. 8 Construct a triangle PQR in which QR = 6 cm, \angle Q = 60^o and PR - PQ = 2 cm Solution : The below given steps will be followed to construct the required triangle.\\ \text{Step I:} Draw line segment QR of 6 cm. At point Q, draw an angle of 60^o, say \angle XQR.$$\\$ $\text{Step II:}$ Cut a line segment $QS$ of $2 cm$ from the line segment $QT$ extended in the opposite side of line segment $XQ.$ (As $PR > PQ$ and $PR - PQ = 2 cm$). Join SR.$\\$ $\text{Step III:}$ Draw perpendicular bisector $AB$ of line segment $SR$. Let it intersect $QX$ at point $P.$ Join $PQ, PR.$ $\\$ $\Delta PQR$ is the required triangle.$\\$

9   Construct a triangle $XYZ$ in which $\angle Y = 30^o, \angle Z = 90^o$ and $XY + YZ + ZX = 11 cm.$