**1** **Draw a line segment of length $7.6 cm$ and divide it in the ratio $5\colon 8.$ Measure the two parts. Give the justification of the construction.**

A line segment of length $7.6 cm$ can be divided in the ratio of $5\colon 8 $ as follows.$\\$ Step 1. Draw line segment $AB$ of $7.6 cm$ and draw a ray $AX$ making an acute angle with line segment $AB$.$\\$ Step 2. Locate $13 (= 5 + 8)$ points, $A_ 1 , A_ 2 , A _3 , A _4 ........ A _13 $, on $AX$ such that $ AA _1 = A _1 A_ 2 = A_ 2 A _3$ and so on.$\\$ Step 3. Join $BA 13 .$$\\$ Step 4. Through the point $A_ 5$ , draw a line parallel to $BA_ 13 $(by making an angle equal to $\angle AA _13 B$) at $A _5$ intersecting $AB$ at point $C.$$\\$ $C$ is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8. The lengths of $AC$ and $CB$ can be measured. It comes out to $2.9 cm$ and $4.7 cm$ respectively.$\\$ $Justification$$\\$ The construction can be justified by proving that$\\$ $ \dfrac{AC}{CB}=\dfrac{5}{8}$$\\$ By construction, we have $A_ 5 C \parallel A _13 B.$ By applying Basic proportionality theorem for the triangle $AA_ 13 B$, we obtain$\\$ $\dfrac{AC}{CB}=\dfrac{AA_5}{A_5 A_13} \dots (1)$$\\$ From the figure, it can be observed that $AA _5$ and $A_ 5 A _13$ contain $5$ and $8$ equal divisions of line segments respectively.$\\$ $\therefore \dfrac{AA_5}{A_5A_13} =\dfrac{5}{8} \dots (2)$$\\$ On comparing equations (1) and (2), we obtain$\\$ $ \dfrac{AC}{CB}=\dfrac{5}{8}$$\\$ This justifies the construction.

**2** **Construct a triangle of sides $4 cm, 5 cm $ and $6 cm$ and then a triangle similar to it whose sides are $\dfrac{2}{3 }$ of the corresponding sides of the first triangle. Give the justification of the construction**

The construction can be justified by proving that$\\$ $ AB'=\dfrac{2}{3}AB,B'C'=\dfrac{2}{3}BC,AC'=\dfrac{2}{3}AC$$\\$ By construction, we have $B’C’ \parallel BC\\ \therefore \angle AB'C' = \angle ABC $(Corresponding angles) $In \Delta AB'C' $ and $\Delta ABC,$$\\$ $\angle ABC = \angle AB'C $(Proved above)$\\$ $\angle BAC = \angle B'AC'$(Common)$\\$ $\therefore \Delta AB'C' \cong \Delta ABC$ (AA similarity criterion)$\\$ $\dfrac{AB'}{AB}=\dfrac{B'C'}{BC}=\dfrac{AC,}{AC} \dots (1)\\ In \Delta AA _2 B' \ and \ \Delta AA _3 B,\\ \angle A_ 2 AB' =\angle A_ 3 AB$ (Common)$\\$ $\angle AA _2 B' =\angle AA _3 B \ \ \ $(Corresponding angles)$\\$ $\therefore \Delta AA _2 B'$ and $\angle AA _3 B \ \ \ $(AA similarity criterion)$\\$ $ \dfrac{AB'}{AB}=\dfrac{AA_2}{AA_3}\\ =\dfrac{AB'}{AB}=\dfrac{2}{3} \dots (2)$$\\$ From equations (1) and (2), we obtain$\\$ $\dfrac{AB'}{AB}=\dfrac{B'C'}{BC}=\dfrac{AC'}{AC}=\dfrac{2}{3}\\ =AB'=\dfrac{2}{3}AB,B'C'=\dfrac{2}{3}BC,AC'=\dfrac{2}{3}AC$$\\$ This justifies the construction.

Step 1. Draw a line segment $AB = 4 cm.$ Taking point $A$ as centre, draw an arc of $5 cm$ radius.$\\$ Similarly, taking point $B$ as its centre, draw an arc of $6 cm $radius. These arcs will intersect each other at point $C$. Now, $AC = 5 cm$ and $BC = 6 cm$ and $\Delta ABC$ is the required triangle.$\\$ Step 2. Draw a ray $AX$ making an acute angle with line $AB$ on the opposite side of vertex$C.$$\\$ Step 3. Locate $3$ points $A_ 1 , A_ 2 , A_ 3$ (as 3 is greater between 2 and 3) on line $AX$ such that $AA_ 1 = A _1 A_ 2 = A _2 A _3 .$$\\$ Step 4. Join $BA 3$ and draw a line through $A_ 2$ parallel to $BA_ 3$ to intersect $AB$ at point $B'.$$\\$ Step 5. Draw a line through $B'$ parallel to the line $BC$ to intersect $AC$ at $C'.$$\\$ $\Delta AB'C' $ is the required triangle.

**3** **Construct a triangle with sides $5 cm, 6 cm $ and $7 cm$ and then another triangle whose sides are $\dfrac{7}{5}$ of the corresponding sides of the first triangle. Give the justification of the construction.**

Step 1. Draw a line segment AB of $5 cm$. Taking $A$ and $B$ as centre, draw arcs of $6 cm $ and $5 cm$ radius respectively. Let these arcs intersect each other at point $C$. $\Delta ABC$ is the required triangle having length of sides as $5 cm, 6 cm,$ and $7 cm$ respectively.$\\$ Step 2. Draw a ray $AX$ making acute angle with line $AB$ on the opposite side of vertex $C.$$\\$ Step 3. Locate $7$ points, $A_ 1 , A_ 2 , A_ 3 , A _4 A _5 , A_ 6 , A_ 7 \ \ $ (as $7$ is greater between $5$ and $7$), on line $AX$ such that $AA _1 = A _1 A _2 = A _2 A _3 = A_ 3 A_ 4 = A _4 A _5 = A_ 5 A_ 6 = A_ 6 A_ 7 .$$\\$ Step 4. Join $BA_ 5$ and draw a line through $A_ 7$ parallel to $BA_ 5$ to intersect extended line segment $AB$ at point $B'.$$\\$ Step 5. Draw a line through $B'$ parallel to $BC$ intersecting the extended line segment $AC$ at $C'$. $\Delta AB'C' $ is the required triangle.$\\$ $Justification$ The construction can be justified by proving that$\\$ $AB;=\dfrac{7}{5}AB,B'C'=\dfrac{7}{5}BC,AC'=\dfrac{7}{5}AC$$\\$ In $\Delta ABC$ and $\Delta AB'C',$$\\$ $\angle ABC = \angle AB'C' \ \ $ (Corresponding angles)$\\$ $\angle BAC = \angle B'AC' \ \ \ $(Common)$\\$ $\therefore \Delta ABC \cong \Delta AB'C' \ \ \ $(AA similarity criterion)$\\$ $=\dfrac{AB}{AB'}=\dfrac{BC}{BC'}=\dfrac{AC}{AC'} \dots (1)$$\\$ In $\Delta AA _5 B$ and $\Delta AA _7 B',$$\\$ $\angle A_ 5 AB = \angle A _7 AB' \ \ \ $(Common)$\\$ $\angle AA _5 B = \angle AA _7 B' \ \ \ $ (Corresponding angles)$\\$ $\therefore \Delta AA _5 B \cong \Delta AA _7 B'\ \ \ \ $ (AA similarity criterion)$\\$ $=\dfrac{AB'}{AB}=\dfrac{AA_5}{AA_7}\\ =\dfrac{AB}{AB'}=\dfrac{5}{7} \dots (2)$$\\$ On comparing equations (1) and (2), we obtain$\\$ $ \dfrac{AB}{AB'}=\dfrac{BC}{B'C'}=\dfrac{AC}{AC'}=\dfrac{5}{7}\\ =AB'=\dfrac{7}{5}AB,B'C'=\dfrac{7}{5}BC,AC'=\dfrac{7}{5}AC$$\\$ This justifies the construction.

**4** **Construct an isosceles triangle whose base is $8 cm$ and altitude $4 cm$ and then another triangle whose side are $1\dfrac{1}{2}$ times the corresponding sides of the isosceles triangle. Give the justification of the construction.**

$Justification$ The construction can be justified by proving that$\\$ $AB'=\dfrac{2}{3}AB,B'C'=\dfrac{3}{2}BC,AC'=\dfrac{3}{2}AC$$\\$ In $\Delta ABC$ and $\Delta AB'C',\\ \angle ABC = \angle AB'C' \ \ \ \ $(Corresponding angles)$\\$ $\angle BAC = \angle B'AC' \ \ \ \ $(Common)$\\$ $\therefore \Delta ABC \cong \Delta AB'C' \ \ \ \ $(AA similarity criterion)$\\$ $\dfrac{AB}{AB'}=\dfrac{BC}{B'C'} =\dfrac{AC}{AC'} \dots(1)$$\\$ In $\Delta AA _2 B$ and $\Delta AA _3 B',$$\\$ $\angle A _2 AB = \angle A_ 3 AB' \ \ \ \ $(Common)$\\$ $\angle AA _2 B = \angle AA _3 B' \ \ \ \ $(Corresponding angles)$\\$ $\therefore \Delta AA_ 2 B \cong \Delta AA _3 B'\ \ \ \ $ (AA similarity criterion)$\\$ $\dfrac{AB}{AB'}=\dfrac{AA_2}{AA_3} =\dfrac{2}{3} \dots (2)$$\\$ On comparing equations (1) and (2), we obtain$\\$ $\dfrac{AB}{AB'}=\dfrac{BC}{B'C'}=\dfrac{AC}{AC'} =\dfrac{2}{3} \\ =AB'=\dfrac{3}{2}AB,B'C'=\dfrac{3}{2}BC,AC'=\dfrac{3}{2}AC$$\\$ This justifies the construction.

Let us assume that $\Delta ABC$ is an isosceles triangle having $CA$ and $CB$ of equal lengths, base $AB$ of $8 cm,$ and $AD$ is the altitude of $4 cm.$ A $\Delta AB'C'$ whose sides are $\dfrac{3}{2}$ times of $\Delta ABC$ can be drawn as follows. Step 1. Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point $A$ and $B$ as its centre. Let these arcs intersect each other at $O$ and $O'.$ Join $OO'$. Let $OO'$ intersect $AB$ at $D.$$\\$ Step 2. Taking $D$ as centre, draw an arc of $4 cm$ radius which cuts the extended line segment $OO'$ at point $C$. An isosceles $\Delta ABC$ is formed, having $CD$ (altitude) as $4 cm$ and $AB$ (base) as $8 cm.$$\\$ Step 3. Draw a ray $AX$ making an acute angle with line segment $AB$ on the opposite side of vertex $C.$$\\$ Step 4. Locate $3$ points (as $3$ is greater between $3$ and $2$) $A_ 1 , A_ 2 ,$ and $A_ 3$ on $AX$ such that $AA _1 = A_ 1 A _2 = A _2 A _3 .$$\\$ Step 5. Join $BA _2$ and draw a line through $A_ 3$ parallel to $BA_ 2$ to intersect extended line segment $AB$at point $B'.$$\\$ Step 6. Draw a line through $B'$ parallel to $BC$ intersecting the extended line segment $AC$ at $C'$. $\Delta AB'C'$ is the required triangle.

**5** **Draw a triangle $ABC$ with side $BC = 6 cm, AB = 5 cm $ and $\angle ABC = 60^o$. Then construct a triangle whose sides are $\dfrac{3}{4}$ of the corresponding sides of the triangle $ABC$. Give the justification of the construction.**

$Justification$ The construction can be justified by proving$\\$ $AB'=\dfrac{3}{4}AB,BC'=\dfrac{3}{4}BC,A'C'=\dfrac{3}{4}AC$$\\$ In $\Delta A'BC'$ and $\delta ABC,$$\\$ $\angle A'C'B = \angle ACB \ \ \ $ (Corresponding angles)$\\$ $\angle A'BC' = \angle ABC\ \ \ \ $ (Common)$\\$ $\therefore \Delta A'BC' \cong \Delta ABC \ \ \ \ $(AA similarity criterion)$\\$ $=\dfrac{A'B}{AB}=\dfrac{BC'}{BC}=\dfrac{A'C'}{AC} \dots (1)$$\\$ In $\Delta BB _3 C' $ and $\Delta BB _4 C,$$\\$ $\angle B_ 3 BC' = \angle B_ 4 BC \ \ \ \ $(Common)$\\$ $\angle BB_ 3 C' = \angle BB _4 C \ \ \ \ $(Corresponding angles)$\\$ $\therefore \Delta BB_ 3 C' \cong \Delta BB_ 4 C\ \ \ $ (AA similarity criterion)$\\$ $=\dfrac{BC'}{BC}=\dfrac{BB_3}{BB_4}\\ \dfrac{BC'}{BC'}=\dfrac{3}{4} \dots (2)$$\\$ From equations (1) and (2), we obtain$\\$ $=\dfrac{A'B}{AB}=\dfrac{BC'}{BC}=\dfrac{A'C'}{AC}=\dfrac{3}{4}$$\\$ $ A'B=\dfrac{3}{4}AB,BC'=\dfrac{3}{4}BC,AC'=\dfrac{3}{4}AC$$\\$ This justifies the construction.

A $\Delta A'BC' $ whose sides are $\dfrac{3}{4}^{th}$of the corresponding sides of $\Delta ABC$ can be drawn as follows.$\\$ Step 1. Draw a $\Delta ABC$ with side $BC = 6 cm, AB = 5 cm $ and $\angle ABC = 60^o.$$\\$ Step 2. Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A.$ $\\$ Step 3. Locate $4$ points (as $4$ is greater in $3$ and $4$), $B_ 1 , B_ 2 , B_ 3 , B_ 4 $, on line segment $BX$.$\\$ Step 4. Join $B_ 4 C$ and draw a line through $B_ 3$ , parallel to $B _4 C$ intersecting$BC$ at $C'.$$\\$ Step 5. Draw a line through $C'$ parallel to $AC$ intersecting $AB$ at $A'$. $\Delta A'BC'$ is the required triangle.$\\$ $Justification$$\\$ The construction can be justified by proving$\\$ $AB'=\dfrac{3}{4}AB,BC'=\dfrac{3}{4}BC,A'C'=\dfrac{3}{4}AC$$\\$ In $\Delta A'BC'$ and $\delta ABC,$$\\$ $\angle A'C'B = \angle ACB \ \ \ $ (Corresponding angles)$\\$ $\angle A'BC' = \angle ABC\ \ \ \ $ (Common)$\\$ $\therefore \Delta A'BC' \cong \Delta ABC \ \ \ \ $(AA similarity criterion)$\\$ $=\dfrac{A'B}{AB}=\dfrac{BC'}{BC}=\dfrac{A'C'}{AC} \dots (1)$$\\$ In $\Delta BB _3 C' $ and $\Delta BB _4 C,$$\\$ $\angle B_ 3 BC' = \angle B_ 4 BC \ \ \ \ $(Common)$\\$ $\angle BB_ 3 C' = \angle BB _4 C \ \ \ \ $(Corresponding angles)$\\$ $\therefore \Delta BB_ 3 C' \cong \Delta BB_ 4 C\ \ \ $ (AA similarity criterion)$\\$ $=\dfrac{BC'}{BC}=\dfrac{BB_3}{BB_4}\\ \dfrac{BC'}{BC'}=\dfrac{3}{4} \dots (2)$$\\$ From equations (1) and (2), we obtain$\\$ $=\dfrac{A'B}{AB}=\dfrac{BC'}{BC}=\dfrac{A'C'}{AC}=\dfrac{3}{4}$$\\$ $ A'B=\dfrac{3}{4}AB,BC'=\dfrac{3}{4}BC,AC'=\dfrac{3}{4}AC$$\\$ This justifies the construction.

**6** **Draw a triangle $ABC$ with side $BC = 7 cm, \angle B = 45^o, \angle A = 105^o$. Then, construct a triangle whose sides are $4/3$ times the corresponding side of $\Delta ABC$. Give the justification of the construction.**

$\angle B = 45^o, \angle A = 105^o$$\\$ Sum of all interior angles in a triangle is $180^o$.$\\$ $\angle A + \angle B + \angle C = 180^o\\ 105^o + 45^o + \angle C = 180^o\\ \angle C = 180^o - 150^o \\ \angle C = 30^o$$\\$ The required triangle can be drawn as follows.$\\$ Step 1. Draw a $\Delta ABC$ with side $BC = 7 cm, \angle B = 45^o, \angle C = 30^o.$$\\$ Step 2. Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.$\\$ Step 3. Locate $4$ points (as $4$ is greater in $4$ and $3$), $B_ 1 , B_ 2 , B_ 3 , B_ 4 $, on $BX$.$\\$ Step 4. Join $B_ 3 C$. Draw a line through $B_ 4$ parallel to $B_ 3 C$ intersecting extended $BC$ at $C'$.$\\$ Step 5. Through $C'$, draw a line parallel to $AC$ intersecting extended line segment at $C'$.$\\$ $\Delta A'BC'$ is the required triangle.$\\$ $Justification$$\\$ The construction can be justified by proving that$\\$ $AB'=\dfrac{4}{3}AB,BC'=\dfrac{4}{3}BC,A'C'=\dfrac{4}{3}AC$$\\$ In $\Delta ABC$ and $\Delta A'BC',$$\\$ $\angle ABC = \angle A'BC' \ \ \ \ $(Common)$\\$ $\angle ACB = \angle A'C'B \ \ \ \ $ (Corresponding angles)$\\$ $\therefore \Delta ABC \cong \Delta A'BC' \ \ \ $ (AA similarity criterion)$\\$ $\dfrac{AB}{A'B}=\dfrac{BC}{BC'}=\dfrac{AC}{A'C'} \dots (1)$$\\$ In $\Delta BB _3 C$ and $\Delta BB _4 C',$$\\$ $\angle B _3 BC = \angle B _4 BC' \ \ \ \ $ (Common)$\\$ $\angle BB_ 3 C = \angle BB_ 4 C' \ \ \ \ $ (Corresponding angles)$\\$ $\therefore BB_ 3 C \cong \angle BB_ 4 C' \ \ \ \ $ (AA similarity criterion)$\\$ $\dfrac{BC}{BC'} =\dfrac{BB_3}{BB_4}\\ \dfrac{BC}{BC'}=\dfrac{3}{4} \dots (2)$$\\$ On comparing equations (1) and (2), we obtain$\\$ $\dfrac{AB}{A'B}=\dfrac{BC}{BC'}=\dfrac{AC}{A'C'}=\dfrac{3}{4}\\ A'B=\dfrac{4}{3}AB,BC'=\dfrac{4}{3}BC,A'C'=\dfrac{4}{3}AC$$\\$ This justifies the construction.

**7** **Draw a right triangle in which the sides (other than hypotenuse) are of lengths $4 cm$ and $3 cm$. then construct another triangle whose sides are $\dfrac{5}{3}$ times the corresponding sides of the given triangle. Give the justification of the construction.**

It is given that sides other than hypotenuse are of lengths $4 cm$ and $3 cm$. Clearly, these will be perpendicular to each other. The required triangle can be drawn as follows.$\\$ Step 1. Draw a line segment $AB = 4 cm$. Draw a ray $SA$ making $90^o$ with it.$\\$ Step 2. Draw an arc of $3 cm$ radius while taking $A$ as its centre to intersect $SA$ at $C$. Join $BC$.$\\$ $\Delta ABC$ is the required triangle.$\\$ Step 3. Draw a ray $AX$ making an acute angle with $AB, $opposite to vertex $C$.$\\$ Step 4. Locate $5$ points (as $5$ is greater in $5$ and $3$),$ A_ 1 , A _2 , A_ 3 , A_ 4 , A_ 5 $, online segment $AX$ such that $AA_ 1 = A _1 A _2 = A _2 A_ 3 = A _3 A _4 = A _4 A_ 5 .$$\\$ Step 5. Join $A _3 B$. Draw a line through $A_ 5$ parallel to $A _3 B$ intersecting extended line segment $AB$ at $B'$.$\\$ Step 6. Through $B'$, draw a line parallel to $BC$ intersecting extended line segment $AC$ at $C'$. $\Delta AB'C'$ is the required triangle.$\\$ $Justification$$\\$ The construction can be justified by proving that$\\$ $ab'=\dfrac{5}{3}AB,B'C'=\dfrac{5}{3}BC,AC'=\dfrac{5}{3}AC$$\\$ In$ \Delta ABC$ and $\Delta AB'C',$$\\$ $\angle ABC = \angle AB'C' \ \ \ $(Corresponding angles)$\\$ $\angle BAC = \angle B'AC' \ \ \ \ $ (Common)$\\$ $\therefore \Delta ABC \cong \Delta AB'C' \ \ \ $(AA similarity criterion)$\\$ $\dfrac{AB}{AB'}=\dfrac{BC}{B'C'}=\dfrac{AC}{AC'} \dots (1)$$\\$ In $\Delta AA_ 3 B$ and $\Delta AA _5 B',$$\\$ $\angle A_ 3 AB = \angle A_ 5 AB' \ \ \ $(Common)$\\$ $\angle AA _3 B = \angle AA _5 B' \ \ \ $(Corresponding angles)$\\$ $\therefore \Delta AA_ 3 B \cong \Delta AA _5 B' \ \ \ \ $(AA similarity criterion)$\\$ $\dfrac{AB}{AB'}=\dfrac{AA_3}{AA_5}\\ \dfrac{AB}{AB'}=\dfrac{3}{5 } \dots (2)$$\\$ On comparing equations (1) and (2), we obtain$\\$ $\dfrac{AB}{AB'}=\dfrac{BC}{B'C'}=\dfrac{AC}{AC'}=\dfrac{3}{5}$$\\$ $AB'=\dfrac{5}{3}AB,B'C'=\dfrac{5}{3}BC,AC'=\dfrac{5}{3}AC$$\\$ This justifies the construction.

**8** **Draw a circle of radius $6 cm$. From a point $10 cm$ away from its centre, construct the pair of tangents to the circle and measure their lengths. Give the justification of the construction.**

A pair of tangents to the given circle can be constructed as follows.$\\$ Step 1. Taking any point $O$ on the given plane as centre, draw a circle of $6 cm$ radius. Locate a point $P$, $10 cm$ away from $O$. Join $OP$.$\\$ Step 2. Bisect $OP$. Let $M$ be the mid-point of $PO$.$\\$ Step 3. Taking $M$ as centre and $MO$ as radius, draw a circle.$\\$ Step 4. Let this circle intersect the previous circle at point $Q$ and $R$.$\\$ Step 5. Join $PQ$ and $PR. PQ$ and $PR$ are the required tangents.$\\$ The lengths of tangents $PQ$ and $PR$ are $8 cm $ each.$\\$ $Justification$$\\$ The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle (whose centre is $O$ and radius is $6 cm$). For this, join $OQ$ and $OR.$$\\$ $\angle PQO$ is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.$\\$ $\therefore \angle PQO = 90^o\\ = OQ \perp PQ$$\\$ Since $OQ$ is the radius of the circle, $PQ$ has to be a tangent of the circle. Similarly, $PR$ is a tangent of the circle.

**9** **Construct a tangent to a circle of radius $4 cm$ from a point on the concentric circle of radius $6 cm$ and measure its length. Also verify the measurement by actual calculation. Give the justification of the construction.**

Tangents on the given circle can be drawn as follows.$\\$ Step 1. Draw a circle of $4 cm$ radius with centre as $O$ on the given plane.$\\$ Step 2. Draw a circle of 6 cm radius taking $O$ as its centre. Locate a point $P$ on this circle and join $OP$.$\\$ Step 3. Bisect $OP$. Let $M$ be the mid-point of $PO$.$\\$ Step 4. Taking $M$ as its centre and $MO$ as its radius, draw a circle. Let it intersect the given circle at the points $Q$ and $R$$\\$. Step 5. Join $PQ$ and $PR$. $PQ$ and $PR$ are the required tangents.$\\$ It can be observed that $PQ$ and $PR$ are of length $4.47 cm$ each.$\\$ In $\Delta PQO,$$\\$ Since $PQ$ is a tangent,$\\$ $\angle PQO = 90^o\\ PO = 6 cm\\ QO = 4 cm$$\\$ Applying Pythagoras theorem in $\Delta PQO$, we obtain $\\$ $PQ ^2 + QO ^2 = PQ ^2\\ PQ ^2 + (4)^ 2 = (6)^ 2\\ PQ ^2 + 16 = 36\\ PQ ^2 = 36 - 16\\ PQ^ 2 = 20\\ PQ = 2 \sqrt{ 5}\\ PQ = 4.47 cm$$\\$ $Justification$$\\$ The construction can be justified by proving that $PQ$ and $PR$ are the tangents to the circle (whose centre is $O$ and radius is $4 cm$). For this, let us join $OQ$ and $OR$.$\\$ $\angle PQO$ is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.$\\$ $\therefore \angle PQO = 90^o\\ = OQ \perp PQ$$\\$ Since $OQ$ is the radius of the circle, $PQ$ has to be a tangent of the circle. Similarly, $PR$ is a tangent of the circle

**10** **Draw a circle of radius $3 cm$. Take two points $P$ and $Q$ on one of its extended diameter each at a distance of $7 cm$ from its centre. Draw tangents to the circle from these two points $P$ and $Q$. Give the justification of the construction.**

The tangent can be constructed on the given circle as follows.$\\$ Step 1. Taking any point $O$ on the given plane as centre, draw a circle of $3 cm$ radius.$\\$ Step 2. Take one of its diameters, $PQ$, and extend it on both sides. Locate two points on this diameter such that $OR = OS = 7 cm$$\\$ Step 3. Bisect $OR$ and $OS$. Let $T$ and $U$ be the mid-points of $OR$ and $OS$ respectively.$\\$ Step 4. Taking $T$ and $U$ as its centre and with $TO$ and $UO$ as radius, draw two circles. These two circles will intersect the circle at point $V, W, X, Y$ respectively. Join $RV, RW, SX,$ and $SY$. These are the required tangents.$\\$ $Justification$$\\$ The construction can be justified by proving that $RV, RW, SY, $ and $SX$ are the tangents to the circle (whose centre is $O$ and radius is $3 cm$). For this, join $OV, OW, OX,$ and $OY.$$\\$ $\angle RVO$ is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.$\\$ $\therefore \angle RVO = 90^o\\ = OV \perp RV$$\\$ Since $OV$ is the radius of the circle, $RV$ has to be a tangent of the circle. Similarly, $OW, OX,$ and $OY$ are the tangents of the circle

**11** **Draw a pair of tangents to a circle of radius $5 cm$ which are inclined to each other at an angle of $60^o$. Give the justification of the construction.**

The tangents can be constructed in the following manner:$\\$ Step 1. Draw a circle of radius $5 cm$ and with centre as $O$.$\\$ Step 2. Take a point $A$ on the circumference of the circle and join $OA$. Draw a perpendicular to $OA$ at point $A$.$\\$ Step 3. Draw a radius $OB,$ making an angle of $120^o (180^o -60^o)$ with $OA.$$\\$ Step 4. Draw a perpendicular to $OB$ at point $B$. Let both the perpendiculars intersect at point $P. PA$ and $PB$ are the required tangents at an angle of $60^o$.$\\$ $Justification$$\\$ The construction can be justified by proving that $\angle APB = 60^o$ $\\$ By our construction $\angle OAP = 90^o\\ \angle OBP = 90^o\\ \ And \ \angle AOB = 120^o$$\\$ We know that the sum of all interior angles of a quadrilateral = $360^o$$\\$ $\angle OAP + \angle AOB + \angle OBP + \angle APB = 360^o\\ 90^o + 120^o + 90^o + \angle APB = 360^o\\ \angle APB = 60^o$$\\$ This justifies the construction.

**12** **Draw a line segment $AB$ of length $8 cm$. Taking $A$ as centre, draw a circle of radius $4 cm$ and taking $B$ as centre, draw another circle of radius $3 cm$. Construct tangents to each circle from the centre of the other circle. Give the justification of the construction**

The tangents can be constructed on the given circles as follows.$\\$ Step 1. Draw a line segment $AB$ of $8 cm$. Taking $A$ and $B$ as centre, draw two circles of $4 cm$ and $3 cm$ radius.$\\$ Step 2. Bisect the line $AB$. Let the mid-point of $AB$ be $C$. Taking $C$ as centre, draw a circle of $AC$ radius which will intersect the circles at points $P, Q, R,$ and $S$. Join $BP, BQ, AS,$ and $AR$. These are the required tangents$\\$ $Justification$ The construction can be justified by proving that $AS$ and $AR$ are the tangents of the circle (whose centre is $B$ and radius is $3 cm$) and $BP$ and $BQ$ are the tangents of the circle (whose centre is $A$ and radius is $4 cm$ ). For this, join $AP, AQ, BS,$ and $BR.$$\\$ $\angle ASB$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.$\\$ $\therefore \angle ASB = 90^o\\ =BS \perp AS$$\\$ Since $BS$ is the radius of the circle, $AS$ has to be a tangent of the circle. Similarly, $AR, BP,$ and $BQ$ are the tangents.

**13** **Let $ABC$ be a right triangle in which $AB = 6 cm, BC = 8 cm$ and $\angle B = 90^o$. $BD$ is the perpendicular from $B$ on $AC$. The circle through $B, C,$ and $D$ is drawn. Construct the tangents from $A$ to this circle. Give the justification of the construction.**

Consider the following situation. If a circle is drawn through $B, D,$ and $C, BC$ will be its diameter as$\\$ $\angle BDC$ is of measure $90^o$. The centre $E$ of this circle will be the midpoint of $BC$.$\\$ The required tangents can be constructed on the given circle as follows.$\\$ Step 1. Join $AE$ and bisect it. Let $F$ be the mid-point of $AE$.$\\$ Step 2. Taking $F$ as centre and $FE$ as its radius, draw a circle which will intersect the circle at point $B$ and $G$. Join $AG.$ $AB$ and $AG$ are the required tangents.$\\$ $Justification$$\\$ The construction can be justified by proving that $AG$ and $AB$ are the tangents to the circle. For this, join $EG.$$\\$ $\angle AGE$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle.$\\$ $\therefore \angle AGE = 90^o\\ EG \perp AG$$\\$ Since $EG$ is the radius of the circle, $AG$ has to be a tangent of the circle.$\\$ Already, $\angle B = 90^o\\ = AB \perp BE$$\\$ Since BE is the radius of the circle, AB has to be a tangent of the circle.

**14** **Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circles. Give the justification of the construction.**

The required tangents can be constructed on the given circle as follows.$\\$ Step 1. Draw a circle with the help of a bangle.$\\$ Step 2. Take a point $P$ outside this circle and take two chords $QR$ and $ST$.$\\$ Step 3. Draw perpendicular bisectors of these chords. Let them intersect each other at point $O$.$\\$ Step 4. Join $PO$ and bisect it. Let $U$ be the mid-point of $PO$. Taking $U$ as centre, draw a circle of radius $OU$, which will intersect the circle at $V$ and $W$. Join $PV$ and $PW$.$\\$ $PV$ and $PW$ are the required tangents.$\\$ $Justification$$\\$ The construction can be justified by proving that $PV$ and $P W $ are the tangents to the circle. For this, first of all, it has to be proved that $O$ is the centre of the circle. Let us join $OV$ and $OW.$$\\$ We know that perpendicular bisector of a chord passes through the centre.$\\$ Therefore, the perpendicular bisector of chords $QR$ and $ST$ pass through the centre.$\\$ It is clear that the intersection point of these perpendicular bisectors is the centre of the circle. $\angle PVO$ is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. $\therefore \angle PVO = 90^o\\ = OV \perp PV$$\\$ Since $OV$ is the radius of the circle, $PV$ has to be a tangent of the circle. Similarly, $PW$ is a tangent of the circle.