 # Constructions

## Class 10 NCERT

### NCERT

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}$$\\ This justifies the construction. 2 Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \dfrac{2}{3 } of the corresponding sides of the first triangle. Give the justification of the construction ##### Solution : The construction can be justified by proving that\\ AB'=\dfrac{2}{3}AB,B'C'=\dfrac{2}{3}BC,AC'=\dfrac{2}{3}AC$$\\$ 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$$\\ This justifies the construction. Step 1. Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius.\\ Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and \Delta ABC is the required triangle.\\ Step 2. Draw a ray AX making an acute angle with line AB on the opposite side of vertexC.$$\\$ 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'.$$\\ \Delta AB'C' is the required triangle. 3 Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \dfrac{7}{5} of the corresponding sides of the first triangle. Give the justification of the construction. ##### Solution : Step 1. Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively. Let these arcs intersect each other at point C. \Delta ABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.\\ Step 2. Draw a ray AX making acute angle with line AB on the opposite side of vertex C.$$\\$ 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.

##### Solution :

$Justification$ 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 .$$\\ Step 5. Join BA _2 and draw a line through A_ 3 parallel to BA_ 2 to intersect extended line segment ABat point B'.$$\\$ 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$$\\ This justifies the construction. 6 Draw a triangle ABC with side BC = 7 cm, \angle B = 45^o, \angle A = 105^o. Then, construct a triangle whose sides are 4/3 times the corresponding side of \Delta ABC. Give the justification of the construction. ##### Solution : \angle B = 45^o, \angle A = 105^o$$\\$ 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$$\\ This justifies the construction. 7 Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. then construct another triangle whose sides are \dfrac{5}{3} times the corresponding sides of the given triangle. Give the justification of the construction. ##### Solution : It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other. The required triangle can be drawn as follows.\\ Step 1. Draw a line segment AB = 4 cm. Draw a ray SA making 90^o with it.\\ Step 2. Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC.\\ \Delta ABC is the required triangle.\\ Step 3. Draw a ray AX making an acute angle with AB, opposite to vertex C.\\ Step 4. Locate 5 points (as 5 is greater in 5 and 3), A_ 1 , A _2 , A_ 3 , A_ 4 , A_ 5 , online segment AX such that AA_ 1 = A _1 A _2 = A _2 A_ 3 = A _3 A _4 = A _4 A_ 5 .$$\\$ 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$$\\ Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle. 9 Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation. Give the justification of the construction. ##### Solution : Tangents on the given circle can be drawn as follows.\\ Step 1. Draw a circle of 4 cm radius with centre as O on the given plane.\\ Step 2. Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle and join OP.\\ Step 3. Bisect OP. Let M be the mid-point of PO.\\ Step 4. Taking M as its centre and MO as its radius, draw a circle. Let it intersect the given circle at the points Q and R$$\\$. 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$$\\ Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle 10 Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q. Give the justification of the construction. ##### Solution : The tangent can be constructed on the given circle as follows.\\ Step 1. Taking any point O on the given plane as centre, draw a circle of 3 cm radius.\\ Step 2. Take one of its diameters, PQ, and extend it on both sides. Locate two points on this diameter such that OR = OS = 7 cm$$\\$ 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$$\\ Since OV is the radius of the circle, RV has to be a tangent of the circle. Similarly, OW, OX, and OY are the tangents of the circle 11 Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60^o. Give the justification of the construction. ##### Solution : The tangents can be constructed in the following manner:\\ Step 1. Draw a circle of radius 5 cm and with centre as O.\\ Step 2. Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.\\ Step 3. Draw a radius OB, making an angle of 120^o (180^o -60^o) with OA.$$\\$ 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

##### Solution :

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.

##### Solution :

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.

##### Solution :

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.