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

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.$$\\$

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$$\\$

**2** **Construct an angle of $45^o$ at the initial point of a given ray and justify the construction.**

The below given steps will be followed to construct an angle of $45^o$.$\\$ (i) Take the given ray $PQ.$ Draw an arc of some radius taking point $P$ as its centre, which intersects $PQ$ at $R.$$\\$ (ii) Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.$\\$ (iii) Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).$\\$ (iv) Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U$.$\\$ (v) Join $PU$. Let it intersect the arc at point $V.$$\\$ (vi) From $R$ and $V$, draw arcs with radius more than $\dfrac{1}{2} RV$ to intersect each other at $W$. Join $PW.$ $PW$ is the required ray making $45^o$ with $PQ$$\\$

$\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)$15^o$**

(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.$$\\$ (7) Let it intersect the arc at $X$. Taking $X$ and $R$ as centre and radius more than $\dfrac{1}{2}RX$ draw arcs to intersect each other at $Y$.$\\$ Joint $PY$ which is the required ray making $22\dfrac{1}{2}^o$ with the given ray $PQ$$\\$

(iii)$15^o$$\\$ The below given steps will be followed to construct an angle of $15^o$.$\\$ $\text{Step I:}$ Draw the given ray $PQ.$ Taking $P$ as centre and with some radius, draw an arc of a circle which intersects $PQ$ at $R.$$\\$ $\text{Step II:}$ 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 $PT$.$\\$ $\text{Step IV:}$ Let it intersect the arc at $U$. Taking $U$ and $R$ as centre and with radius more than $\dfrac{1}{2} RU,$ draw an arc to intersect each other at $V$. Join $PV$ which is the required ray making $15^o$ with the given ray $PQ.$

**4** **Construct the following angles and verify by measuring them by a protractor:$\\$ (i)$75^o$$\\$ (ii)$105^o$$\\$ (iii)$135^o$**

(i) $75^o$$\\$ The below given steps will be followed to construct an angle of $75^o$$\\$. (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 $S$ and $V$ as centre, draw arcs with radius more than $\dfrac{1}{2}SV$.Let those intersect each other at $W$. Join $PW$ which is the required ray making $75^o$ 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.**

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 $AB + AC = 13 cm.$**

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 $PR - PQ = 2 cm$**

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 $XY + YZ + ZX = 11 cm.$**

The below given steps will be followed to construct the required triangle.$\\$ $\text{Step I:}$ Draw a line segment $AB$ of $11 cm.$ $\\$ (As $XY + YZ + ZX = 11 cm$)$\\$ $\text{Step II:}$ Construct an angle, $\angle PAB,$ of $30^o$ at point $A$ and an angle, $\angle QBA,$ of $90^o $ at point $B.$ $\\$ $\text{Step III:}$ Bisect $\angle PAB$ and $\angle QBA.$ Let these bisectors intersect each other at point $X.$$\\$ $\text{Step IV:}$ Draw perpendicular bisector $ST$ of $AX$ and $UV$ of $BX.$$\\$ $\text{Step V:}$ Let $ST$ intersect $AB$ at $Y$ and $UV$ intersect $AB$ at $Z.$$\\$ Join $XY, XZ.$ $\\$ $\Delta XYZ$ is the required triangle.$\\$

**10** **Construct a right triangle whose base is $12 cm$ and sum of its hypotenuse and other side is $18 cm.$**

The below given steps will be followed to construct the required triangle.$\\$ $\text{Step I:}$ Draw line segment $AB$ of $12 cm.$ Draw a ray $AX$ making $90^o$ with $AB.$$\\$ $\text{Step II:}$ Cut a line segment $AD$ of $18 cm$ (as the sum of the other two sides is $18$) from ray $AX.$$\\$ $\text{Step III:}$ Join $DB$ and make an angle $DBY$ equal to $ADB.$$\\$ $\text{Step IV:}$ Let $BY$ intersect $ AX$ at $C.$ Join $AC, BC$.$\\$ $\Delta ABC$ is the required triangle.