# Triangles

## Class 10 NCERT

### NCERT

1   Fill in the blanks using the correct word given in brackets : $\\$ (i) All circles are ___________ . (congruent, similar) $\\$ (ii) All squares are ___________ . (similar, congruent) $\\$ (iii) All __________ triangles are similar. (isosceles, equilateral) $\\$ (iv) Two polygons of the same number of sides are similar, if : $\\$ (a) their corresponding angles are __________ and $\\$ (b) their corresponding sides are ___________. (equal, proportional)

##### Solution :

(i) Similar $\\$ (ii) Similar $\\$ (iii) Equilateral $\\$ (iv) (a) Equal $\\$ (b) Proportional $\\$

2   Give two different examples of pair of : $\\$ (i) similar figures. $\\$ (ii) non-similar figures.

##### Solution :

$\\$ (i) similar figures. $\\$

$\bullet$ Two equilateral triangles with sides $2\ cm$ and $4\ cm$

$\bullet$Two squares with sides $1\ cm$ and $2\ cm$ $\\$

(ii) non-similar figures.

$\bullet$ Trapezium and square

$\bullet$Triangle and parallelogram

3   State whether the following quadrilaterals are similar or not :

##### Solution :

Quadrilateral $PQRS$ and $ABCD$ are not similar as their corresponding sides are proportional, i.e. $1:2$, but their corresponding angles are not equal.

4   In Fig. 6.17, $(i)$ and $(ii),\ DE\ ||\ BC.$ Find $EC$ in $(i)$ and $AD$ in $(ii)$.

##### Solution :

(i)

Let $EC = x\ cm$ It is given that $DE\ ||\ BC.$ By using basic proportionality theorem, we obtain

$\dfrac{AD}{DB} = \dfrac{AE}{EC}$ $\\$ $\dfrac{1.5}{3} = \dfrac{1}{x}$ $\\$ $x = \dfrac{3 \times 1}{1.5}$ $\\$ $x = 2$ $\\$ $\therefore EC = 2 m$

(ii)

Let $AD = x\ cm$ $\\$ It is given that $DE\ ||\ BC$. $\\$ By using basic proportionality theorem, we obtain

$\dfrac{AD}{DB} = \dfrac{AE}{EC}$ $\\$ $\dfrac{x}{7.2} = \dfrac{1.8}{5.4}$ $\\$ $x = \dfrac{1.8 \times 7.2}{5.4}$ $\\$ $x = 24$ $\\$ $\therefore AD = 2.4\ cm$ $\\$

5   $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\Delta{PQR}.$ For each of the following cases, state whether $EF\ ||\ QR$ : $\\$ (i) $PE = 3.9\ cm,\ EQ = 3\ cm,\ PF = 3.6\ cm$ and $FR = 2.4\ cm$ $\\$ (ii) $PE = 4\ cm,\ QE = 4.5\ cm,\ PF = 8\ cm$ and $RF = 9\ cm$ $\\$ (iii) $PQ = 1.28\ cm,\ PR = 2.56\ cm,\ PE = 0.18\ cm$ and $PF = 0.36\ cm$

##### Solution :

(i)

Given that, $PE = 3.9\ cm,\ EQ = 3\ cm,\ PF = 3.6\ cm,\ FR = 2.4\ cm$ $\\$

$\dfrac{PE}{EQ} = \dfrac{3.9}{3}$ $\\$ $\dfrac{PF}{FR} = \dfrac{3.6}{2.4} = 1.5$ $\\$ Hence, $\dfrac{PE}{EQ} \ne \dfrac{PF}{FR}$ $\\$ $\therefore,\ EF$ is not paralell to $QR.$

(ii)

$PE = 4\ cm,\ QE = 4.5\ cm,\ PF = 8\ cm,\ RF = 9\ cm$ $\\$ $\dfrac{PE}{EQ} = \dfrac{4}{4.5} = \dfrac{8}{9}$ $\\$ $\dfrac{PF}{FR} = \dfrac{8}{9}$ $\\$ Hence, $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$ $\\$ Therefore , $EF$ is parallel to $QR$

(iii)

$PQ = 1.28\ cm,\ PR = 2.56\ cm,\ PE = 0.18\ cm,\ PF = 0.36\ cm$ $\\$ $\dfrac{PE}{PQ} = \dfrac{0.18}{1.28} = \dfrac{18}{128} = \dfrac{9}{64}$ $\\$ $\dfrac{PF}{PR} = \dfrac{0.36}{2.56} = \dfrac{9}{64}$ $\\$ Hence, $\dfrac{PE}{PQ} = \dfrac{PF}{PR}$ $\\$ Therefore, $EF$ is parallel to $QR$.

6   In Fig. 6.18, if $LM\ ||\ CB$ and $LN\ ||\ CD,$ prove that $\\$ $\dfrac{AM}{AB} = \dfrac{AN}{AD}$

##### Solution :

In the given figure, $LM\ ||\ CB$ $\\$ By using basic proportionality theorem, we obtain $\\$ $\dfrac{AM}{AB} = \dfrac{AL}{AC} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i)$ $\\$ Similarly, $LN\ ||\ CD$ $\\$ $\therefore \dfrac{AN}{AB} = \dfrac{AL}{AC} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)$ $\\$ From (i) and (ii), we obtain $\\$ $\therefore \dfrac{AM}{AB} = \dfrac{AN}{AD}$ $\\$

7   In Fig. 6.19, $DE\ ||\ AC$ and $DF\ ||\ AE.$ Prove that $\\$ $\dfrac{BF}{FE} = \dfrac{BE}{EC}$

##### Solution :

In $\Delta{ABC},\ DE\ ||\ AC$ $\\$ $\therefore \dfrac{BD}{DA} = \dfrac{BE}{EC} \ \ \ \ \ \ \ \ \ \ \ \ (Basic\ Proportionality\ Theorem) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i)$

In $\Delta{BAE},\ DF\ ||\ AE$ $\\$ $\therefore \dfrac{BD}{DA} = \dfrac{BF}{FE} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Basic\ Proportionality\ Theorem) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)$ $\\$ From (i) and (ii), we obtain $\\$ $\dfrac{BE}{EC} = \dfrac{BF}{FE}$

8   In Fig. 6.20, $DE\ ||\ OQ$ and $DF\ ||\ OR$. Show that $EF\ ||\ QR.$

##### Solution :

In $POQ,\ DE\ ||\ OQ$ $\\$ $\therefore \dfrac{PE}{EQ} = \dfrac{PD}{DO} \ \ \ \ \ \ \ \ \ \ \ \ (Basic\ Proportionality\ Theorem) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i)$ $\\$

(I) $\Delta{POR},\ CF || OR$ $\\$ $\therefore \dfrac{PF}{FR} = \dfrac{PD}{DO} \ \ \ \ \ \ \ \ \ \ \ \ \ (Basic\ Proportionally\ Theorem) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)$ $\\$ From (i) and (ii), we obtain $\\$ $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$ $\\$ $\therefore EF\ ||\ QR \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Converse\ of\ basic\ proportionality\ theorem)$

9   In Fig. 6.21, $A,\ B$ and $C$ are points on $OP,\ OQ$ and $OR$ respectively such that $AB\ ||\ PQ$ and $AC\ ||\ PR.$ Show that $BC\ ||\ QR.$

##### Solution :

In $POQ,\ AB\ ||\ PQ$ $\\$ $\therefore \dfrac{OA}{AP} = \dfrac{OB}{BQ} \ \ \ \ \ \ \ \ \ \ \ \ \ (Basic\ Proportionality\ Theorem) \ \ \ \ \ \ \ \ \ \ \ (i)$ $\\$

In $\Delta{POR},\ AC\ ||\ PR$ $\\$ $\therefore \dfrac{OA}{AP} = \dfrac{OC}{CR} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (By\ basic\ proportinality\ theorem) \ \ \ \ \ \ \ \ \ \ \ (ii)$ $\\$ From (i) and (ii), we obtain $\\$ $\dfrac{OB}{BQ} = \dfrac{OC}{OR}$ $\\$ $\therefore BC\ ||\ QR \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (By\ the\ converse\ of\ basic\ proportionality\ theorem)$ $\\$

10   Using $Theorem\ 6.1$, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class $IX$).$\\$

##### Solution :

Consider the given figure in which $PQ$ is a line segment drawn through the mid-point $P$ of line $AB$, such that $PQ\ ||\ BC$ $\\$ By using basic proportionality theorem, we obtain $\\$ $\dfrac{AQ}{QC} = \dfrac{AP}{PB}$ $\\$ $\dfrac{AQ}{QC} = \dfrac{1}{1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ ($P$ is the mid-point of $AB.\ \therefore AP = PB$) $\\$ $\Rightarrow AQ = QC$ $\\$ Or, $Q$ is the mid-point of $AC.$

11   Using $Theorem\ 6.2$, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class $IX$).

##### Solution :

Consider the given figure in which $PQ$ is a line segment joining the mid-points $P$ and $Q$ of line $AB$ and $AC$ respectively. $\\$ i.e., $AP = PB$ and $AQ = QC$ $\\$ It can be observed that $\\$ $\dfrac{AP}{PB} = \dfrac{1}{1}$ $\\$ And $\dfrac{AQ}{QC} = \dfrac{1}{1}$ $\\$ $\therefore \dfrac{QP}{PB} = \dfrac{AQ}{QC}$ $\\$ Hence, by using basic proportionality theorem, we obtain $PQ\ ||\ BC$ $\\$

12   $ABCD$ is a trapezium in which $AB\ ||\ DC$ and its diagonals intersect each other at the point $O$. Show that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$

##### Solution :

Draw a line $EF$ through point $O$, such that $EF\ ||\ CD$ $\\$ In $\Delta{ADC},\ EO\ ||\ CD$ $\\$ By using basic proportionality theorem, we obtain $\\$ $\dfrac{AE}{ED} = \dfrac{AO}{OC}$ $\\$ So, by using basic proportionality theorem, we obtain $\\$ $\dfrac{ED}{AE} = \dfrac{OD}{BO}$ $\\$ $\Rightarrow \dfrac{AE}{ED} = \dfrac{BO}{OD} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$ $\\$ $\dfrac{AO}{OC} = \dfrac{BO}{OD}$ $\\$ $\Rightarrow \dfrac{AO}{BO} = \dfrac{OC}{OD}$ $\\$

13   The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\\$ $\dfrac{AO}{BO} = \dfrac{CO}{DO}$ Show that $ABCD$ is a trapezium.

##### Solution :

Let us consider the following figure for the given question. $\\$

Draw a line $OE\ ||\ AB$ $\\$

In $\dfrac{ABD},\ OE\ ||\ AB$ $\\$ By using basic proportionality theorem, we obtain $\\$ $\dfrac{AE}{ED} = \dfrac{BO}{OD} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i)$ $\\$ However, it is given that $\\$ $\dfrac{AO}{OC} = \dfrac{OB}{OD} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)$ $\\$ From equations (i) and (ii), we obtain $\\$ $\dfrac{AE}{ED} = \dfrac{AO}{OC}$ $\\$ $\Rightarrow EO\ ||\ DC$ [ By the converese of basic proportinality theorem] $\\$ $\Rightarrow AB\ ||\ OE\ ||\ DC$ $\\$ $\Rightarrow AB || CD$ $\\$ $\therefore ABCD$ is a trapezium.

14   State which pairs of triangles in $Fig.\ 6.34$ are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

(i) $\angle{A} = \angle{P} = 60°$ $\\$ $\angle{B} = \angle{Q} = 80°$ $\\$ $\angle{C} = \angle{R} = 40°$ Therefore ,$\Delta ABC \sim \Delta PQR$ [By $AAA$ similarity criterion] $\\$ $\dfrac{AB}{QR} = \dfrac{BC}{RP} = \dfrac{CA}{PQ}$$\\ (ii)As corresponding sides are proportional \\ \therefore \Delta{ABC} ~ \Delta{QRP} [By SSS similarity criterion] \\ (iii)The given triangles are not similar as the corresponding sides are not proportional. \\ (iv)The given triangles are not similar By SAS similarity criteria \therefore \Delta MNL \sim \Delta QPR \\ (v)The given triangles are not similar as the correspondingangle is not contained by the two corresponding sides \\ (vi) In \Delta{DEF}, \\ \angle{D} + \angle{E} + \angle{F} = 180^o \\ (Sum of the measures of the angles of a triangle is 180^o.) \\ \angle{P} + 80^o + 30^o = 180^o \\ \angle{P} = 70^o \\ 70^o + 80^o + \angle{F} = 180^o \\ \angle{F} = 30^o \\ Similarly, in \Delta{PQR}, \\ \angle{P} + \angle{Q} + \angle{R} = 180^o \\ (Sum of the measures of the angles of a triangle is 180^o.) \\ \angle{P} + 80^o + 30^o = 180^o \\ \angle{P} = 70^o \\ In \Delta{DEF} and \Delta{PQR}, \\ \angle{D} = \angle{P} (Each 70°) \\ \angle{E} = \angle{Q} (Each 80°) \\ \angle{F} = \angle{R} (Each 30°)\\ \therefore \Delta{DEF} ~ \Delta{PQR} [By AAA similarity criterion] 15 In Fig. 6.35, \Delta{ODC} ~ \Delta{OBA}, \angle{BOC = 125°} and \angle{CDO = 70°}. Find \angle{DOC}, \angle{DCO} and \angle{OAB}. ##### Solution : DOB is a straight line. \\ \therefore \angle{DOC} + \angle{COB} = 180^o \\ \Rightarrow \angle{DOC} = 180^o - 125^o \\ = 55^o \\ In \Delta{DOC}, \\ \angle{DCO} + \angle{CDO} + \angle{DOC} = 180^o \\ (Sum of the measures of the angles of a triangle is 180^o.) \\ \Rightarrow \angle{DCO} + 70^o ++ 55^o = 180^o \\ \Rightarrow \angle{DCO} = 55^o \\ It is given that \Delta{ODC} ~ \Delta{OBA}. \\ \Rightarrow \angle{OAB} = \angle{OCD} [Corresponding angles are equal in similar triangles.] \\ \Rightarrow \angle{OAB} = 55^o 16 Diagonals AC and BD of a trapezium ABCD with AB\ ||\ DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \dfrac{OA}{OC} = \dfrac{OB}{OD} ##### Solution : In \Delta{DOC} and \Delta{BOA}, \\ \angle{CDO} = \angle{ABO} [Alternate interior angles as AB\ ||\ CD] \\ \angle{DCO} = \angle{BAO} [Alternate interior angles as AB\ ||\ CD] \\ \angle{DOC} = \angle{BOA} [Vertically opposite angles] \\ \therefore \Delta{DOC} ~ \Delta{BOA} [AAA similarity criterion] \\ \therefore \dfrac{DO}{BO} = \dfrac{OC}{OA} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [Correspnding sides are proportional] \\ \Rightarrow \dfrac{OA}{OC} = \dfrac{OB}{OD} 17 In Fig. 6.36, \dfrac{QR}{QS} = \dfrac{QT}{PR} and \angle{1} = \angle{2}. Show that \Delta{PQS} ~ \Delta{TQR}. ##### Solution : In \Delta{PQR},\ \angle{PQR} = \angle{PRQ} \\ \therefore PQ = PR \\ (i) Given, \\ \dfrac{QR}{QS} = \dfrac{QT}{PR} Using (i), we obtain \\ \dfrac{QR}{QS} = \dfrac{QT}{QP} \ \ \ \ \ \ \ \ \ \ \ \ (ii) \\ In \Delta{PQS} and \Delta{TQR}, \\ \dfrac{QR}{QS} = \dfrac{QT}{QP} \ \ \ \ \ \ \ \ \ \ [Using (ii)] \\ \angle{Q} = \angle{Q} \\ \therefore \Delta{PQS} ~ \Delta{TQR} \ \ \ \ \ \ \ \ \ [SAS similarity criterion] 18 S and T are points on sides PR and QR of \Delta{PQR} such that \angle{P} = \angle{RTS}. Show that \Delta{RPQ} ~ \Delta{RTS.} ##### Solution : In \Delta{RPQ} and \Delta{RST}, \\ \angle{RTS} = \angle{QPS} (Given) \\ \angle{R} = \angle{R} (Common angle) \\ \therefore \Delta{RPQ} ~ \Delta{RTS} (By AA similarity criterion) 19 In Fig. 6.37, if \Delta{ABE} \cong \Delta{ACD}, show that \Delta{ADE} \sim \Delta{ABC} ##### Solution : It is given that \Delta{ABE} \cong \Delta{ACD}. \\ \therefore AB = AC [By CPCT] (1) \\ And, AD = AE [By CPCT] (2) \\ In \Delta{ADE} and \Delta{ABC}, \\ \dfrac{AD}{AB} = \dfrac{AE}{AC} [Dividing equation (2) and (1)] \\ \angle{A} = \angle{A} [Common angle] \\ \therefore \Delta{ADE} ~ \Delta{ABC} [By SAS similarity criterion] \\ 20 In Fig. 6.38, altitudes AD and CE of \Delta{ABC} intersect each other at the point P. Show that: \\ (i) \Delta{AEP} ~ \Delta{CDP} \\ (ii) \Delta{ABD} ~ \Delta{CBE} \\ (iii) \Delta{AEP} ~ \Delta{ADB} \\ (iv) \Delta{PDC} ~ \Delta{BEC} ##### Solution : (i) Diagram In \Delta{AEP} and \Delta{CDP}, \\ \angle{AEP} = \angle{CDP} (Each 90^o) \\ \angle{APE} = \angle{CPD} (Vertically opposite angles)\\ Hence, by using AA similarity criterion,\\ \Delta{AEP} ~ \Delta{CDP} \\ (ii) Diagram \\ In \Delta{ABD} and \Delta{CBE}, \\ \angle{ADB} = \angle{CEB} (Each 90^o) \\ \angle{ABD} = \angle{CBE} (Common) \\ Hence, by using AA similarity criterion, \\ \Delta{ABD} \sim \Delta{CBE} (iii) Diagram \\ In \Delta{AEP} and \Delta{ADB}, \\ \angle{AEP} = \angle{ADB} (Each 90^o) \\ \angle{PAE} = \angle{DAB} (Common) \\ Hence, by using AA similarity criterion, \\ \Delta{AEP} \sim \Delta{ADB} \\ (iv) Diagram \\ In \Delta{PDC} and \Delta{BEC}, \\ \angle{PDC} = \angle{BEC} (Each 90^o) \\ \angle{PCD} = \angle{BCE} (Common angle)\\ Hence, by using AA similarity criterion, \\ \Delta{PDC} ~ \Delta{BEC} \\ 21 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that \Delta{ABE} ~ \Delta{CFB.} ##### Solution : Diagram In \Delta{ABE} and \Delta{CFB}, \angle{A} = \angle{C} (Opposite angles of a parallelogram) \\ \angle{AEB} = \angle{CBF} (Alternate interior angles as AE\ ||\ BC) \\ \therefore \angle{ABE} \sim \Delta{CFB} (By AA similarity criterion) 22 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: \\ (i) \Delta{ABC} ~ \Delta{AMP} \\ (ii) \dfrac{CA}{BC} = \dfrac{BC}{MP} ##### Solution : In \Delta{ABC} and \Delta{AMP}, \\ \angle{ABC} = \angle{AMP} (Each 90^o) \\ \angle{A} = \angle{A} (Common) \\ \therefore \Delta{ABC} \sim \Delta{AMP} (By AA similarity criterion) \\ \dfrac{CA}{PA} = \dfrac{BC}{MP} \ \ \ \ \ \ \ \ \ \ \ \ (Corresponding sides of similar triangles are proportional)\\ 23 CD and GH are respectively the bisectors of \angle{ACB} and \angle{EGF} such that D and H lie on sides AB and FE of \Delta{ABC} and \Delta{EFG} respectively. If \Delta{ABC} \sim \Delta{FEG}, show that: \\ (i) \dfrac{CD}{GH} = \dfrac{AC}{FG} \\ (ii) \Delta{DCB} \sim \Delta{HGE} \\ (iii) \Delta{DCA} \sim \Delta{HGF} \\ ##### Solution : i. It is given that \Delta{ABC} \sim \Delta{FEG}. \\ \therefore \angle{A} = \angle{F},\ \angle{B} = \angle{E}, and \angle{ACB} = \angle{FGE} \\ \angle{ACB} = \angle{FGE} \\ \therefore \angle{ACD} = \angle{FGH} (Angle bisector) \\ And, \angle{DCB} = \angle{HGE} (Angle bisector) \\ In \Delta{ACD} and \Delta{FGH}, \\ \angle{A} = \angle{F} (Proved above) \\ \angle{ACD} = \angle{FGH} (Proved above) \\ \therefore \Delta{ACD} \sim \Delta{FGH} (By AA similarity criterion) \\ \Rightarrow \dfrac{CD}{GH} = \dfrac{AC}{FG} \\ ii. In \Delta {DCB} and \Delta{HGE}, \\ \angle{DCB} = \angle{HGE} (Proved above) \\ \angle{B} = \angle{E} (Proved above) \\ \therefore \Delta{DCB} \sim \Delta{HGE} (By AA similarity criterion) \\iii. In \Delta{DCA} and \Delta{HGF}, \angle{ACD} = \angle{FGH} (Proved above) \\ \angle{A} = \angle{F} (Proved above) \\ \therefore \Delta{DCA} \sim \Delta{HGF} (By AA similarity criterion) diagram 24 In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD\ \bot\ BC and EF\ \bot\ AC, prove that \Delta{ABD} \sim \Delta{ECF}. ##### Solution : It is given that ABC is an isosceles triangle. \therefore AB = AC \\ \Rightarrow \angle{ABD} = \angle{ECF} \\ In \Delta{ABD} and \Delta{ECF}, \\ \angle{ADB} = \angle{EFC} (Each 90°) \\ \angle{BAD} = \Delta{CEF} (Proved above) \\ \therefore \Delta{ABD} \sim \Delta{ECF} (By using AA similarity criterion) 25 Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of \Delta{PQR} (see Fig. 6.41). Show that \Delta{ABC} \sim \Delta{PQR}. ##### Solution : Median divides the opposite side. \\ \therefore BD = \dfrac{BC}{2} and QM = \dfrac{QR}{2} \\ Given that, \\ \dfrac{AB}{BC} = \dfrac{BC}{QR} = \dfrac{AD}{PM} \\ \Rightarrow \dfrac{AB}{PQ} = \dfrac{\frac{1}{2}BC}{\frac{1}{2}QR} = \dfrac{AD}{PM} \\ \dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM} \\ In \Delta{ABD} and \Delta{PQM}, \\ \dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM} (Proved above) \\ \therefore \Delta{ABD} \sim \Delta{PQM} (By SSS similarity criterion) \\ \Rightarrow \angle{ABD} = \angle{PQM} (Corresponding angles of similar triangles) In \Delta{ABC} and \Delta{PQR}, \\ \angle{ABD} = \angle{PQM} (Proved above) \\ \dfrac{AB}{Q} = \dfrac{BC}{QR} \therefore \Delta{ABC} \sim \Delta{PQR} (By SAS similarity criterion) \\ 26 D is a point on the side BC of a triangle ABC such that \angle{ADC} = \angle{BAC}. Show that CA^2 = CB.CD. ##### Solution : In \Delta{ADC} and \Delta{BAC}, \\ \angle{ADC} = \Delta{BAC} (Given) \\ \angle{ACD} = \angle{BCA} (Common angle) \\ \therefore \Delta{ADC} \sim \Delta{BAC} (By AA similarity criterion) \\ We know that corresponding sides of similar triangles are in proportion. \\ \therefore \dfrac{CA}{CB} = \dfrac{CD}{CA} \\ \Rightarrow CA^2 = CB \times CD \\ 27 Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \Delta{ABC} \sim \Delta{PQR}. ##### Solution : Given that, \\ \dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM} \\ Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E,\ C to E, Q to L, and R to L. \\ We know that medians divide opposite sides. Therefore, BD = DC and QM = MR \\ Also, AD = DE (By construction) \\ And, PM = ML (By construction) \\ In quadrilateral ABEC, diagonals AE and BC bisect each other at point D. \\ Therefore, quadrilateral ABEC is a parallelogram. \\ \therefore AC = BE and AB = EC (Opposite sides of a parallelogram are equal) \\ Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL,\ PQ = LR \\ It was given that, \\ \dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM} \\ \Rightarrow \dfrac{AB}{PQ} = \dfrac{BE}{QL} = \dfrac{2AD}{2PM} \\ \Rightarrow \dfrac{AB}{PQ} = \dfrac{BE}{QL} = \dfrac{AE}{PL} \\ \therefore \Delta{ABE} \sim \Delta{PQL} (By SSS similarity criterion) \\ \therefore \angle{BAE} = \angle{QPL} … (1) \\ Similarly, it can be proved that \Delta{AEC} \sim \Delta{PLR} and \angle{CAE} = \angle{RPL} … (2)\\ Adding equation (1) and (2), we obtain \\ \angle{BAE} + \angle{CAE} = \angle{QPL} + \angle{RPL} \\ \Rightarrow \angle{CAB} = \angle{RPQ} … (3)\\ In \Delta{ABC} and \Delta{PQR}, \dfrac{AB}{PQ} = \dfrac{AC}{PR} (Given) \\ \angle{CAB} = \angle{RPQ} [Using equation (3)] \\ \therefore \Delta{ABC} \sim \Delta{PQR} (By SAS similarity criterion) 28 A vertical pole of length 6\ m casts a shadow 4\ m long on the ground and at the same time a tower casts a shadow 28\ m long. Find the height of the tower. ##### Solution : Diagram Let AB and CD be a tower and a pole respectively. Let the shadow of BE and DF be the shadow of AB and CD respectively. \\ At the same time, the light rays from the sun will fall on the tower and the pole at the same angle. \\ Therefore, \angle{DCF} = \angle{BAE} And, \angle{DFC} = \angle{BEA} \\ \angle{CDF} = \angle{ABE} (Tower and pole are vertical to the ground) \\ \therefore \Delta{ABE} \sim \Delta{CDF} (AAA similarity criterion) \\ \Rightarrow \dfrac{AB}{CD} = \dfrac{BE}{DF} \\ \Rightarrow \dfrac{AB}{6\ m} = \dfrac{28}{4} \\ \Rightarrow AB = 42\ m \\ Therefore, the height of the tower will be 42 metres. \\ 29 If AD and PM are medians of triangles ABC and PQR, respectively where \Delta{ABC} \sim \Delta{PQR} , prove that \dfrac{AB}{PQ} = \dfrac{AD}{PM} ##### Solution : It is given that \Delta{ABC} \sim \Delta{PQR} \\ We know that the corresponding sides of similar triangles are in proportion. \\ \dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{BC}{QR} \ \ \ \ \ ....(1) \\ Also, \angle{A} = \angle{P},\ \angle{B} = \angle{Q},\ \angle{C} = \angle{R} … (2) \\ Since AD and PM are medians, they will divide their opposite sides. \\ \therefore BD = \dfrac{BC}{2} and QM = \dfrac{QR}{2} \ \ \ \ \ \ \ (3) \\ From equations (1) and (3), we obtain \\ \dfrac{AB}{PQ} = \dfrac{BD}{QM} \ \ \ \ \ \ .....(4) \\ In \Delta{ABD} and \Delta{PQM}, \\ \angle{B} = \angle{Q} [Using equation (2)] \\ \dfrac{AB}{PQ} = \dfrac{BD}{QM} [Using equation (4)] \\ \therefore \Delta{ABD} \sim \Delta{PQM} (By SAS similarity criterion) \\ \dfrac{AB}{PO} = \dfrac{BD}{OM} = \dfrac{AD}{PM} \\ 30 Let \Delta{ABC} \sim \Delta{DEF} and their areas be, respectively, 64\ cm^2 and 121\ cm^2. If EF = 15.4 cm, find BC. ##### Solution : It is given that \Delta{ABC} \sim \Delta{DEF}. \\ \therefore \dfrac{ar\ (\Delta{ABC})}{ar\ (\Delta{DEF})} = \Big(\dfrac{AB}{DE}\Big)^2 = \Big(\dfrac{BC}{EF}\Big)^2 = \Big(\dfrac{AC}{DF}\Big)^2 \\ Given that, \\ EF = 15.4\ cm \\ ar(\Delta{ABC}) = 64\ cm^2 \\ ar(\Delta{DEF}) = 121\ cm^2 \\ \therefore \dfrac{ar(ABC)}{ar(DEF)} = \Big(\dfrac{BC}{EF}\Big)^2 \\ \Rightarrow \Big(\dfrac{64\ cm^2}{121\ cm^2}\Big) = \dfrac{BC^2}{(15.4\ cm)^2} \\ \Rightarrow \dfrac{BC}{15.4} = \Big(\dfrac{8}{11}\Big)cm \\ \Rightarrow BC = \Big(\dfrac{8 \times 15.4}{11})cm = (8 \times 1.4)cm = 11.2\ cm 31 Diagonals of a trapezium ABCD with AB\ ||\ DC intersect each other at the point O. If AB = 2\ CD, find the ratio of the areas of triangles AOB and COD. ##### Solution : \therefore \dfrac{ar\ (\Delta{AOB})}{ar\ (\Delta{COD})} = \Big(\dfrac{AB}{CD}\Big)^2 \\ Since AB = 2\ CD, \\ \therefore \dfrac{ar\ (\Delta{AOB})}{ar\ (\Delta{COD})} = \Big(\dfrac{2\ CD}{CD}\Big)^2 = \dfrac{4}{1} = 4:1 Since AB || CD, \therefore \angle{OAB} = \angle{OCD} and \angle{OBA} = \angle{ODC} (Alternate interior angles) In \Delta{AOB} and \Delta{COD}, \angle{AOB} = \angle{COD} (Vertically opposite angles) \\ \angle{OAB} = \angle{OCD} (Alternate interior angles) \\ \angle{OBA} = \angle{ODC} (Alternate interior angles) \\ \therefore \Delta{AOB} \sim \Delta{COD} (By AAA similarity criterion)\\ \therefore \dfrac{ar\ (\Delta{AOB})}{ar\ (\Delta{COD})} = \Big(\dfrac{AB}{CD}\Big)^2 \\ Since AB = 2\ CD, \\ \therefore \dfrac{ar\ (\Delta{AOB})}{ar\ (\Delta{COD})} = \Big(\dfrac{2\ CD}{CD}\Big)^2 = \dfrac{4}{1} = 4:1 32 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \\ \dfrac{ar\ (ABC)}{ar\ (DBC)} = \dfrac{AO}{DO} ##### Solution : Let us draw two perpendiculars AP and DM on line BC. We know that area of a triangle = \dfrac{1}{2} \times\ Base\ \times\ Height \\ \therefore \dfrac{ar\ (\Delta{ABC})}{ar\ (\Delta{DBC})} = \dfrac{\frac{1}{2} BC \times AP}{\frac{1}{2} BC \times DM} = \dfrac{AP}{DM} \\ In \Delta{APO} and \Delta{DMO}, \\ \angle{APO} = \angle{DMO} (Each = 90°) \\ \angle{AOP} = \angle{DOM} (Vertically opposite angles) \\ \therefore \Delta{APO} \sim \Delta{DMO} (By AA similarity criterion) \\ \therefore \dfrac{AP}{DM} = \dfrac{AO}{DO} \\ \Rightarrow \dfrac{ar\ (\Delta{ABC})}{ar\ (\Delta{DBC})} = \dfrac{AO}{DO} 33 If the areas of two similar triangles are equal, prove that they are congruent. ##### Solution : Let us assume two similar triangles as \Delta{ABC} \sim \Delta{PQR}. \\ \dfrac{ar\ (\Delta{ABC})}{ar\ (\Delta{PQR})} = \Big(\dfrac{AB}{PQ}\Big)^2 = \Big(\dfrac{BC}{QR}\Big)^2 = \Big(\dfrac{AC}{PR}\Big)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) Given that, ar\ (\Delta{ABC}) = ar\ (\Delta{PQR}) \\ \Rightarrow \dfrac{ar\ (\Delta{ABC})}{ar\ (\Delta{PQR})} = 1 \\ Putting this value in equation (1), we obtain \\ 1 = \Big(\dfrac{AB}{PQ} \Big)^2 = \Big(\dfrac{BC}{QR}\Big)^2 = \Big(\dfrac{AC}{PR}\Big)^2 \\ \Rightarrow AB = PQ,\ BC = QR, and AC = PR \\ \therefore \Delta{ABC} \cong \Delta{PQR} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (By SSS congruence criterion) 34 D,\ E and F are respectively the mid-points of sides AB,\ BC and CA of \Delta{ABC}. Find the ratio of the areas of \Delta{DEF} and \Delta{ABC}. ##### Solution : D and E are the mid-points of \Delta{ABC}. \therefore DE\ ||\ AC and DE = \dfrac{1}{2} AC \\ In \Delta{BED} and \Delta{BCA} \\ \angle{BED} = \angle{BCA} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Corresponding angles) \\ \angle{BDE} = \angle{BAC} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Corresponding angles) \\ \angle{EBD} = \angle{CBA} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (Common angles) \\ \therefore \Delta{BED} \sim \Delta{BCA} \ \ \ \ \ \ \ \ \ \ \ \ \ (AAA similarity criterion) \dfrac{ar\ (\Delta{BED})}{ar\ (\Delta{BCA})} = \Big(\dfrac{DE}{AC}\Big)^2 \\ \Rightarrow \dfrac{ar\ (\Delta{BED})}{ar\ (\Delta{BCA})} = \dfrac{1}{4} \\ \Rightarrow ar\ (\Delta{BED}) = \dfrac{1}{4} ar( \Delta{BCA}) Similarly , ar(\Delta{CFE}) = \dfrac{1}{4} ar (\Delta{CBA}) and ar (\Delta{ADF}) = \dfrac{1}{4} ar (\Delta{ABC}) \\ Also, ar(\Delta{DEF}) = ar(\Delta{ABC}) - [ar(\Delta{BED}) + ar(\Delta{CFE}) + ar(\Delta{ADF})] \\ \Rightarrow ar(\Delta{DEF}) = ar(\Delta{ABC}) - \dfrac{3}{4}ar(\Delta{ABC}) = \dfrac{1}{4}ar(\Delta{ABC}) \\ \Rightarrow \dfrac{ar(\Delta{DEF})}{ar(\Delta{ABC})} = \dfrac{1}{4} 35 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. ##### Solution : Let us assume two similar triangles as \Delta{ABC} \sim \Delta{PQR}. Let AD and PS be the medians of these triangles. \Delta{ABC} \sim \Delta{PQR} \therefore \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\ \angle{A} = \angle{P}, \angle{B} = \angle{Q}, \angle{C} = \angle{R} \ \ \ \ \ \ \ \ \ \ \ \ \ \ … (2) \\ Since AD and PS are medians, \\ \therefore = DC = \dfrac{BC}{2} \\ And, QS = SR = \dfrac{QR}{2} \\ Equation (1) becomes \\ \dfrac{AB}{PQ} = \dfrac{BD}{QS} = \dfrac{AC}{PR} \\ In \Delta{ABD} and \Delta{PQS}, \angle{B} = \angle{Q} [Using equation (2)] \\ And, \dfrac{AB}{PQ} = \dfrac{BD}{QS} [Using equation (3)] \\ \Delta{ABD} \sim \Delta{PQS} (SAS similarity criterion) \\ Therefore, it can be said that \\ \dfrac{AB}{PQ} = \dfrac{BD}{QS} = \dfrac{AD}{PS} \ \ \ \ \ \ \ \ \ (4) \\ \dfrac{ar(\Delta{ABC})}{ar(\Delta{PQR})} = \Big(\dfrac{AB}{PQ}\Big)^2 = \Big(\dfrac{BC}{QR}\Big)^2 = \Big(\dfrac{AC}{PR}\Big)^2 \\ From equations (1) and (4), we may find that \\ \dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AC}{PR} = \dfrac{AD}{PS} \\ And hence, \\ \dfrac{ar(\Delta{ABC})}{ar(\Delta{PQR})} = \Big(\dfrac{AD}{PS}\Big)^2 36 Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. ##### Solution : Let ABCD be a square of side a. \\ Therefore, its diagonal = \sqrt{2a} \\ Two desired equilateral triangles are formed as \Delta{ABE} and \Delta{DBF}. Side of an equilateral triangle, \Delta{ABE}, described on one of its sides = a \\ Side of an equilateral triangle, \Delta{DBF}, described on one of its diagonals = \sqrt{2a} \\ We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. \dfrac{Area\ of \ \Delta{ABE}}{Area\ of \ \Delta{DBF}} = \Big(\dfrac{a}{\sqrt{2}a}\Big)^2 = \dfrac{1}{2} \\ 37 ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is ##### Solution : We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. Let side of \Delta{ABC} = x \\ Therefore, side of \Delta{BDE} = \dfrac{x}{2} \\ \dfrac{Area(\Delta{ABC})}{Area(\Delta{BDE})} = \Bigg(\dfrac{x}{\dfrac{x}{2}}\Bigg)^2 =\dfrac{4}{1} \\ Hence, the correct answer is (C). \\ 38 Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio ##### Solution : If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles. \\ It is given that the sides are in the ratio 4:9. \\ Therefore, ratio between areas of these triangles = \Bigg(\dfrac{4}{9}\Bigg)^2 =\dfrac{16}{81} \\ Hence, the correct answer is (D). 39 Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. \\ (i) 7 cm, 24 cm, 25 cm \\ (ii) 3 cm, 8 cm, 6 cm \\ (iii) 50 cm, 80 cm, 100 cm \\ (iv) 13 cm, 12 cm, 5 cm \\ ##### Solution : (i) It is given that the sides of the triangle are 7\ cm,\ 24\ cm, and 25\ cm. Squaring the lengths of these sides, we will obtain 49,\ 576, and 625.\ 49 + 576 = 625 \\ Or, 7^2 + 24^2 = 25^2 \\ The sides of the given triangle are satisfying Pythagoras theorem. \\ Therefore, it is a right triangle. \\ We know that the longest side of a right triangle is the hypotenuse. \\ Therefore, the length of the hypotenuse of this triangle is 25\ cm.$$\\$ (ii) It is given that the sides of the triangle are $3\ cm,\ 8\ cm,$ and $6\ cm$. Squaring the lengths of these sides, we will obtain $9,\ 64,$ and $36.$ However, $9 + 36 \ne 64$ $\\$ Or, $3^2 + 6^2 \ne 8^2$ $\\$ Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side. $\\$ Therefore, the given triangle is not satisfying Pythagoras theorem. $\\$ Hence, it is not a right triangle.

(iii) Given that sides are $50\ cm,\ 80\ cm,$ and $100\ cm.$ $\\$ Squaring the lengths of these sides, we will obtain $2500,\ 6400,$ and $10000$. However, $2500 + 6400 \ne 10000$ $\\$ Or, $50^2 + 80^2 \ne 100^2$ $\\$ Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side. $\\$ Therefore, the given triangle is not satisfying Pythagoras theorem. $\\$ Hence, it is not a right triangle. $\\$(iv) Given that sides are $13\ cm,\ 12\ cm,$ and $5\ cm$. $\\$ Squaring the lengths of these sides, we will obtain $169,\ 144,$ and $25$. $\\$ Clearly, $144 + 25 = 169$ $\\$ Or, 12^2 + 5^2 = 13^2 $\\$ The sides of the given triangle are satisfying Pythagoras theorem. $\\$ Therefore, it is a right triangle. $\\$ We know that the longest side of a right triangle is the hypotenuse. $\\$ Therefore, the length of the hypotenuse of this triangle is $13\ cm.$ $\\$

40   $PQR$ is a triangle right angled at $P$ and $M$ is a point on $QR$ such that $PM\ \bot\ QR$. $\\$Show that $PM^2 = QM . MR.$

##### Solution :

Let $\angle{MPR} = X$ $\\$ In $\Delta{MPR}$ $\\$ $\angle{MRP} = 180^o - 90^o - x$ $\\$ $\angle{MRP} = 90^o - x$ $\\$ Similarly, in $\Delta{MPQ}$, $\\$ $\angle{MPQ} = 90^o - \angle{MPR}$ $\\$ $= 90^o - x$ $\\$ $\angle{MQP} = 180^o - 90^o - (90^o - x)$ $\\$ $\angle{MQP} = x$ $\\$ In $\Delta{QMP} = 180^o - 90^o - (90^o - x)$ $\\$ $\angle{MQP} = x$ $\\$ In $\Delta{QMP}$ and $\Delta{PMR}$, $\angle{MPQ} = \angle{MRP}$ $\\$ $\angle{PMQ} = \angle{RMP}$ $\\$ $\angle{MQP} = \angle{MPR}$ $\\$

$\therefore \Delta{QMP} \sim \Delta{PMR} \ \ \ \ \ \ \ \ \ \ \ \$ (By $AAA$ similarity criterion) $\\$ $\Rightarrow \dfrac{QM}{PM} = \dfrac{MP}{MR}$ $\\$ $PM^2 = QM \times MR$

41   In Fig. $6.53,\ ABD$ is a triangle right angled at $A$ and $AC\ \bot\ BD$. Show that $\\$ (i) $AB^2 = BC . BD$ $\\$ (ii) $AC^2 = BC . DC$ $\\$ (iii) $AD^2 = BD . CD$ $\\$

##### Solution :

(i) $\Delta{ABD}$ and $\Delta{CAB},$ $\\$ $\angle{DAB} = \angle{ACB} \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (Each 90^o) $\\$ $\therefore \Delta{ABD} \sim \Delta{CAB} \ \ \ \ \ \ \ \ \ \ \ \ \ \$ ($AA$ similarity criterion) $\\$ $\angle{ABD} = \angle{CBA} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (common angle) $\\$ $\Rightarrow \dfrac{AB}{CB} = \dfrac{CD}{AB}$ $\\$ $\Rightarrow AB^2 = CB \times BD$ $\\$

(ii) Let $\angle{CAB} = x$ $\\$ In $\Delta{CBA}$, $\\$ $\angle{CBA} = 180^o - 90^o - x$ $\\$ $\angle{CBA} = 90^o - x$ $\\$ Similarly, in $\Delta{CAD}$, $\\$ $\angle{CAD} = 90^o - \angle{CAB}$ $\\$ $= 90^o - x$ $\\$ $\angle{CDA} = 180^o - 90^o - (90^o - x)$ $\\$ $\angle{CDA} = x$ $\\$ In $\Delta{CBA}$ and $\Delta{CAD}$ $\\$ $\angle{CBA} = \angle{CAD}$ $\\$ $\angle{CAB} = \angle{CDA}$ $\\$ $\angle{ACB} = \angle{DCA} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (each $90^o$)$\\$ $\therefore \Delta{CBA} \sim \Delta{CAD} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$ (By $AAA$ rule) $\\$ $\Rightarrow \dfrac{AC}{DC} = \dfrac{BC}{AC}$ $\\$ $\Rightarrow AC^2= DC \times BC$$\\(iii) In \Delta{DCA} and \Delta{DAB}, \\ \angle{DCA} = \angle{DAB} (Each 90º) \\ \angle{CDA} = \angle{ADB} (Common angle) \\ \therefore \Delta{DCA} \sim \Delta{DAB} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (AA similarity criterion) \\ \Rightarrow \dfrac{DC}{DA} = \dfrac{DA}{DB} \\ \Rightarrow AD^2 = BD \times CD 42 ABC is an isosceles triangle right angled at C. Prove that AB^2 = 2AC^2. ##### Solution : Given that \Delta{ABC} is an isosceles triangle. \\ \therefore AC = CB \\ Applying Pythagoras theorem in \Delta{ABC} (i.e., right-angled at point C), we obtain \\ AC^2 + CB^2 = AB^2 \\ \Rightarrow AC^2 + AC^2 = AB^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ (AC = CB) \\ \rightarrow 2AC^2 = AB^2 43 ABC is an isosceles triangle with AC = BC. If AB^2 = 2\ AC^2, prove that ABC is a right triangle. ##### Solution : Given that,\\ AB^2 = 2AC^2 \\ \Rightarrow AB^2 = AC^2 + AC^2 \\ \Rightarrow AB^2 = AC^2 + BC^2 \ \ \ \ \ \ \ \ \ \ \ (As AC = BC) \\ The triangle is satisfying the pythagoras theorem. \\ Therefore, the given triangle is a right - angled triangle. 44 ABC is an equilateral triangle of side 2a. Find each of its altitudes. ##### Solution : Let AD be the altitude in the given equilateral triangle, \Delta{ABC}. \\ We know that altitude bisects the opposite side. \\ \therefore BD = DC = a \\ In \Delta{ABC} \\ \angle{ABC} = 90^o \\ Applying pythagoras theorem, we obtain \\ AD^2 + DB^2 = AB^2 \\ \Rightarrow AD^2 + a^2 = (2a)^2 \\ \Rightarrow AD^2 + a^2 = 4a^2 \\ \Rightarrow AD^2 = 3a^2 \\ \Rightarrow AD = a\sqrt{3} \\ In an equilateral triangle, all the altitudes are equal in length. \\ Therefore, the length of each altitude will be \sqrt{3a} . 45 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. ##### Solution : In \Delta{AOB},\ \Delta{BOC},\ \Delta{COD},\ \Delta{AOD}, \\ Applying Pythagoras theorem, we obtain \\ AB^2 = AO^2 + OB^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\ BC^2 = BO^2 + OC^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\ CD^2 = CO^2 + OD^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\ AD^2 = AO^2 + OD^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4) \\ Adding all these equations , we obtain \\ AB^2 + BC^2 + CD^2 + AD^2 = 2(AO^2 + OB^2 + OC^2 + OD^2) \\ =2 \Bigg(\Big(\dfrac{AC}{2}\Big)^2 + \Big(\dfrac{BD}{2}\Big)^2 + \Big(\dfrac{AC}{2}\Big)^2 + \Big(\dfrac{BD}{2}\Big)^2 \Bigg) \\ (Diagoinals bisect each other) \\ =2 \Bigg(\dfrac{(AC)^2}{2} + \dfrac{(BD)^2}{2}\Bigg) \\ = (AC)^2 + (BD)^2 46 In Fig. 6.54, O is a point in the interior of a triangle ABC,\ OD\ \bot\ BC,\ OE\ \bot\ AC and OF\ \bot\ AB. Show that \\ (i) OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + CE^2, \\ (ii) AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2. ##### Solution : Join OA,\ OB, and OC. \\ (i) Applying Pythagoras theorem in \Delta{AOF}, we obtain \\ OA^2 = OF^2 + AF^2 \\ Similarly, in \Delta{BOD}, \\ OB^2 = OB^2 + BD^2 \\ Similarly, in \Delta{COE} \\ OC^2 = OE^2 + EC^2 \\ Adding these equations, \\ OA^2 + OB^2 + OC^2 = OF^2 + AF^2 + OD^2 + BD^2 + OE^2 + EC^2 \\ OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2 = AF^2 + BD^2 + EC^2 \\ (ii) From the above result, \\ AF^2 + BD^2 + EC^2 = (OA^2 - OE^2) + (OC^2 - OD^2) + (OB^2 - OF^2) \\ \therefore AF^2 + BD^2 + EC^2 = AE^2 + CD^2 + BF^2 \\ 47 A ladder 10\ m long reaches a window 8\ m above the ground. Find the distance of the foot of the ladder from base of the wall. ##### Solution : Let OA be the wall and AB be the ladder. \\ Therefore, by Pythagoras theorem, \\ AB^2 = OA^2 + BO^2 \\ (10\ m)^2 = (8\ m)^2 + OB^2 \\ 100\ m^2 = 64\ m^2 + OB^2 \\ OB = 6\ m \\ Therefore, the distance of the foot of the ladder from the base of the wall is 6\ m. 48 A guy wire attached to a vertical pole of height 18\ m is 24\ m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut? ##### Solution : Let OB be the pole and AB be the wire. \\ By Pythagoras theorem, \\ AB^2 = OB^2 + OA^2 \\ (24\ m)^2 = (18\ m)^2 + OA^2 \\ OA^2 = (576 - 324)\ m^2 = 252\ m^2 \\ OA = \sqrt{252}\ m = \sqrt{6 \times 6 \times 7} m = 6 \sqrt{7}\ m \\ Therefore, the distance from the base is 6 \sqrt{7}\ m. 49 Two poles of heights 6\ m and 11\ m stand on a plane ground. If the distance between the feet of the poles is 12\ m, find the distance between their tops. ##### Solution : Let CD and AB be the poles of height 11\ m and 6\ m. \\ Therefore, CP = 11 - 6 = 5\ m \\ From the figure, it can be observed that AP = 12\ m \\ Applying Pythagoras theorem for \Delta{APC}, we obtain \\ AP^2 + PC^2 = AC^2 \\ (12\ m)^2 + (5\ m)^2 = AC^2 \\ AC^2 = (144 + 25)\ m^2 = 169\ m^2 \\ AC = 13\ m \\ Therefore, the distance between their tops is 13\ m 50 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE^2 + BD^2 = AB^2 + DE^2. ##### Solution : Applying Pythagoras theorem in \Delta{ACE}, we obtain \\ AC^2 + CE^2 = AE^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .....(1) \\ Applying Pythagoras theorem in \Delta{BCD}, we obtain \\ BC^2 + CD^2 = BD^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(2) \\ Using equation (1) and equation (2),we obtain \\ AC^2 + CE^2 + BC^2 + CD^2 = AE^2 + BD^2 \ \ \ \ \ \ \ \ \ \ .....(3) \\ Applying Pythagoras theorem in \Delta{CDE}, we obtain \\ DE^2 = CD^2 + CB^2 \\ Applying Pythagoras theorem in \Delta{ABC}, we obtain \\ AB^2 = AC^2 + CB^2 \\ Putting the values in equation (3), we obtain, \\ DE^2 + AB^2 = AE^2 + BD^2 51 The perpendicular from A on side BC of a \Delta{ABC} intersects BC at D such that DB = 3\ CD \\ (see Fig. 6.55). Prove that 2\ AB^2 = 2\ AC^2 + BC^2. ##### Solution : Applying Pythagoras theorem for \Delta{ACD}, we obtain \\ AC^2 = AD^2 + DC^2 \\ AD^2 = AC^2 - DC^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(1) \\ Applying Pythagoras theorem in \Delta{ABD}, we obtain \\ AB^2 = AD^2 + DB^2 \\ AD^2 = AB^2 - DB^2 \ \ \ \ \ \ \ \ \ \ \ \ \ ...(2) \\ From equation (1) and equation (2), we obtain \\ AC^2 - DC^2 = AB^2 - DB^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(3) \\ It is given that 3DC = DB \\ \therefore DC = \dfrac{BC}{4} and DB = \dfrac{3BC}{4} \\ Putting these values in equation (3), we obtain \\ AC^2 - \Bigg(\dfrac{BC}{4}\Bigg)^2 = AB^2 - \Bigg(\dfrac{3BC}{4}\Bigg) \\ AC^2 - \dfrac{BC^2}{16} = AB^2 - \dfrac{9BC^2}{16} \\ 16AC^2 - BC^2 = 16AB^2 - 9BC^2 \\ 16AB^2 - 16AC^2 = 8BC^2 \\ 2AB^2 = 2AC^2 + BC^2 52 In an equilateral triangle ABC,\ D is a point on side BC such that BD = \dfrac{1}{3} BC. Prove that 9\ AD^2 = 7\ AB^2. ##### Solution : Let the side of the equilateral triangle be a, and AE be the altitude of \Delta ABC.$$\\$ $\therefore BE=EC=\dfrac{BC}{2}=\dfrac{a}{2}$$\\ And, AE=\dfrac{a\sqrt{3}}{2}$$\\$ Given that,$BD=\dfrac{1}{3}BC$$\\ \therefore BD=\dfrac{a}{3}\\ DE=BE-BD=\dfrac{a}{2}-\dfrac{a}{3}=\dfrac{a}{6}$$\\$ Applying Pythagoras theorem in $\Delta ADE$, we obtain $AD^2=AE^2+DE^2\\ AD^2=(\dfrac{a\sqrt{3}}{2})^2+(\dfrac{a}{6})^2\\ =(\dfrac{3a^2}{4})+(\dfrac{a^2}{36})\\ =\dfrac{28a^2}{36}\\ =\dfrac{7}{9}AB^2\\ =9AD^2=7AB^2$

53   In $ABC,\ AB = 6 \sqrt{3}\ cm,\ AC = 12\ cm$ and $BC = 6\ cm$. The angle $B$ is :

##### Solution :

Given that, $AB = 6\sqrt{3}\ cm,\ AC = 12\ cm,$ and $BC = 6\ cm$ $\\$ It can be observed that $\\$ $AB^2 = 108$ $\\$ $AC^2 = 144$ $\\$ And, $BC^2 = 36$ $\\$ $AB^2 + BC^2 = AC^2$ $\\$ The given triangle, $\Delta{ABC}$, is satisfying Pythagoras theorem. $\\$ Therefore, the triangle is a right triangle, right-angled at $B.$ $\\$ $\therefore \angle{B} = 90^o$ $\\$ Hence, the correct answer is $(C)$.

54   In the given figure, PS is the bisector of $\angle QPR of \Delta PQR$. Prove that $\dfrac{QS}{SR}=\dfrac{PQ}{PR}$.

##### Solution :

Let us draw a line segment $RT$ parallel to $SP$ which intersects extended line segment $QP$ at point $T.$$\\ Given that, PS is the angle bisector of \angle QPR.\\ \angle QPS = \angle SPR ... (1)$$\\$ By construction,$\\$ $\angle SPR = \angle PRT (As PS || TR) ... (2)\\ \angle QPS = \angle QTR (As PS || TR) ... (3)$$\\ Using these equations, we obtain\\ \angle PRT = \angle QTR\\ \therefore PT = PR$$\\$ By construction, $PS || TR$ By using basic proportionality theorem for $\Delta QTR$,$\\$ $\angle QPS = \angle SPR ... (1)$$\\ By construction,\\ \angle SPR = \angle PRT (As PS || TR) ... (2)\\ \angle QPS = \angle QTR (As PS || TR) ... (3)$$\\$ Using these equations, we obtain $\angle PRT = \angle QTR\\ \therefore PT = PR$$\\ By construction,\\ PS || TR By using basic proportionality theorem for \Delta QTR,\\ \dfrac{QS}{SR}=\dfrac{QP}{PT} \implies \dfrac{QS}{SR}=\dfrac{PQ}{PR} \ \ \ \ (PT=TR) \angle QPS = \angle SPR ... (1)$$\\$ By construction,$\\$ $\angle SPR = \angle PRT (As PS || TR) ... (2)\\ \angle QPS = \angle QTR (As PS || TR) ... (3)$$\\ Using these equations, we obtain \angle PRT = \angle QTR\\ \therefore PT = PR$$\\$ By construction,$\\$ $PS || TR$ By using basic proportionality theorem for $\Delta QTR$,$\\$ $\dfrac{QS}{SR}=\dfrac{QP}{PT} \implies \dfrac{QS}{SR}=\dfrac{PQ}{PR} \ \ \ \ (PT=TR)$

55   In the given figure,$D$ is a point on hypotenuse $AC$ of $\Delta ABC$, such that $BD \perp AC, DM \perp BC and DN \perp AB,$$\\ Prove that: (i) DM ^2 = DN.MC\\ (ii) DN ^2 = DM.AN ##### Solution : (i) Let us join DB.\\ We have, DN || CB, DM || AB, and \angle B = 90^o \therefore DMBN is a rectangle.\\ \therefore DN = MB and DM = NB$$\\$

The condition to be proved is the case when $D$ is the foot of the perpendicular drawn from $B$ to $AC$.$\\$$\therefore \angle CDB = 90^o\\ = \angle 2 + \angle 3 = 90^o ... (1)\\ In \Delta CDM, \angle 1 + \angle 2 + \angle DMC = 180^o\\ = \angle 1 + \angle 2 = 90^o ... (2)\\ In \Delta DMB,\\ \angle 3 + \angle DMB + \angle 4 = 180^o\\ = \angle 3 + \angle 4 = 90^o ... (3)$$\\$ From equation (1) and (2), we obtain$\\$ $\angle 1 = \angle 3$$\\ From equation (1) and (3), we obtain \angle 2 = \angle 4\\ In \Delta DCM and \Delta BDM,$$\\$ $\angle 1 = \angle 3$(Proved above) $\angle 2 =\angle 4$ (Proved above) $\therefore \Delta DCM \sim \Delta BDM$ (AA similarity criterion)$\\$ $= \dfrac{BM}{DM}=\dfrac{DM}{MC}\\ \dfrac{DN}{DM}=\dfrac{DM}{MC} \ \ \ (BM=DN)\\ = DM^2=DN*MC$

(ii) In right triangle $DBN,\\ \angle 5 + \angle 7 = 90^o ... (4)$$\\ In right triangle DAN,\\ \angle 6 + \angle 8 = 90^o... (5)$$\\$ $D$ is the foot of the perpendicular drawn from $B$ to $AC$. $\therefore \angle ADB = 90^o\\ =\angle 5 + \angle 6 = 90^o ... (6)$$\\ From equation (4) and (6), we obtain \angle 6 = \angle 7$$\\$ From equation (5) and (6), we obtain $\angle 8 = \angle 5\\ In$\Delta DNA$and$\Delta BND$,$\angle 6 = \angle 7$(Proved above)$\angle 8 = \angle 5$(Proved above)$\therefore \Delta DNA \sim \Delta BND$(AA similarity criterion) \dfrac{AN}{DN}=\dfrac{DN}{NB}\\ = DN ^2 = AN × NB = DN ^2 = AN × DM (As NB = DM)$

56   In the given figure, $ABC$ is a triangle in which $\angle ABC> 90^o$ and $AD \perp CB$ produced. Prove that $AC ^2 = AB ^2 + BC ^2 + 2BC.BD.$

##### Solution :

Applying Pythagoras theorem in $\Delta ADB$, we obtain $AB ^2 = AD ^2 + DB ^2 ... (1)$ Applying Pythagoras theorem in $\Delta ACD$, we obtain $AC^ 2 = AD^ 2 + DC^ 2\\ AC^ 2 = AD^ 2 + (DB + BC) ^2\\ AC ^2 = AD ^2 + DB ^2 + BC^ 2 + 2DB × BC\\ AC ^2 = AB^ 2 + BC^ 2 + 2DB × BC$ [Using equation (1)]

57   In the given figure, $ABC$ is a triangle in which $\angle ABC < 90^o$ and $AD \perp BC$. Prove that $AC^2 = AB^2 + BC^2 - 2BC.BD.$

##### Solution :

Applying Pythagoras theorem in $\Delta ADB$, we obtain $AD^2 + DB^2 = AB^2\\ =AD^2 =AB^2 -DB^2 ...(1)$ Applying Pythagoras theorem in $\Delta ADC$, we obtain $AD^2 + DC^2 = AC^2\\ AB^2 -BD^2 + DC^2 = AC^2$ [Using equation (1)] $AB^2 -BD^2 +(BC-BD)^2 =AC^2\\ AC^2 =AB^2 -BD^2 +BC^2 +BD^2 -2BC×BD =AB^2 +BC^2 -2BC×BD$

58   In the given figure, $AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that:$\\$ $(i)AC^2 =AD^2 +BC.DM+(\dfrac{BC^2}{2})^2$$\\ (ii)AB^2 =AD^2 =BC.DM+(\dfrac{BC}{2})^2$$\\$ $(iii)AC^2+AB^2=2AD^2+\dfrac{1}{2}BC^2$

##### Solution :

(i) Applying Pythagoras theorem in $\Delta AMD$, we obtain $AM^2 +MD^2 =AD^2 ...(1)$ Applying Pythagoras theorem in $\Delta AMC$, we obtain $AM^2 + MC^2 = AC^2\\ AM^2 +(MD+DC)^2 =AC^2\\ (AM^2 + MD^2) + DC^2 + 2MD.DC = AC^2\\ AD^2 + DC^2 + 2MD.DC = AC^2$ [Using equation (1)] Using the result, $DC =BC$, we obtain $AD^2 +(\dfrac{BC}{2})^2+2MD.(\dfrac{BC}{2})=AC^2\\ AD^2+(\dfrac{BC}{2})^2+MD*BC=AC^2$

60   In the given figure, two chords $AB$ and $CD$ intersect each other at the point $P$. prove that:$\\$$(i) \Delta APC \sim \Delta DPB\\ (ii) AP.BP = CP.DP ##### Solution : Let us join CB.\\ (i) In \Delta APC and \Delta DPB,$$\\$ $\angle APC = \angle DPB$ (Vertically opposite angles)$\\$ $\angle CAP = \angle BDP$ (Angles in the same segment for chord CB)$\\$ $\Delta APC \sim \Delta DPB$ (By AA similarity criterion)$\\$ (ii) We have already proved that$\\$ $\Delta APC \sim \Delta DPB$ We know that the corresponding sides of similar triangles are proportional. $\therefore \dfrac{AP}{DP}=\dfrac{PC}{PB}=\dfrac{CA}{BD}\\ \implies \dfrac{AP}{DP}=\dfrac{PC}{PB}\\ \therefore AP.PB=PC.DP$

61   In the given figure, two chords $AB$ and $CD$ of a circle intersect each other at the point $P$ (when produced) outside the circle. Prove that$\\$ $(i) \Delta PAC \sim \Delta PDB\\ (ii) PA.PB = PC.PD$

(i) In$\Delta PAC$ and $\Delta PDB,$ $\angle P = \angle P$ (Common) $\angle PAC = \angle PDB$ (Exterior angle of a cyclic quadrilateral is $\angle PCA = \angle PBD$ equal to the opposite interior angle)$\\$ $\therefore \Delta PAC \sim \Delta PDB$$\\ (ii)We know that the corresponding sides of similar triangles are proportional.\\ \dfrac{PA}{PD}=\dfrac{AC}{DB}=\dfrac{PC}{PB}\\ \implies \dfrac{PA}{PD}=\dfrac{PC}{PB}\\ \therefore PA.PB=PC.PD 62 In the given figure, D is a point on side BC of \Delta ABC such that \dfrac{BD}{CD}=\dfrac{AB}{AC}. Prove that AD is the bisector of \angle BAC. ##### Solution : Let us extend BA to P such that AP = AC. Join PC. It is given that,\\ \dfrac{BD}{CD}=\dfrac{AB}{AC}\\ \implies \dfrac{BD}{CD}=\dfrac{AP}{AC}$$\\$ By using the converse of basic proportionality theorem, we obtain$\\$ $AD \parallel PC$ $\implies \angle BAD=\angle APC$(Corresponding angles) ... (1)$\\$ And $\angle DAC=\angle ACP$ (Alternate interior angles) ... (2)$\\$

By construction, we have$\\$ $AP = AC\\ \implies \angle APC = \angle ACP$ ... (3) On comparing equations (1), (2), and (3), we obtain $\angle BAD = \angle APC\\ \implies AD$ is the bisector of the angle $BAC$

63   Nazima is fly fishing in a stream. The tip of her fishing rod is $1.8 m$ above the surface of the water and the fly at the end of the string rests on the water $3.6 m$ away and $2.4 m$ from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after $12$ seconds?

##### Solution :

Let $AB$ be the height of the tip of the fishing rod from the water surface. Let $BC$ be the horizontal distance of the fly from the tip of the fishing rod. Then, $AC$ is the length of the string.$\\$ $AC$ can be found by applying Pythagoras theorem in $\Delta ABC$.$\\$ $AC^2 = AB^2 + BC^2\\ AB^2 = (1.8 m)^2 + (2.4 m)^2\\ AB^2 = (3.24 + 5.76) m^2 \\ AB^2 = 9.00 m^2\\ \implies AB=\sqrt{9}=3m$$\\ Thus, the length of the string out is 3 m.\\ She pulls the string at the rate of 5 cm per second.\\ Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m$$\\$

Let the fly be at point $D$ after $12$ seconds.$\\$ Length of string out after $12$ seconds is $AD$.$\\$ $AD = AC$ - String pulled by Nazima in $12$ seconds $\\$ $= (3.00 - 0.6) m\\ = 2.4 m\\ In \Delta ADB,\\ AB^2 + BD^2 = AD^2\\ (1.8 m)^2 + BD^2 = (2.4 m)^2\\ BD^2 = (5.76 -3.24) m^2 = 2.52 m^2\\ BD = 1.587 m$$\\$ Horizontal distance of fly =$BD + 1.2 m\\ = (1.587 + 1.2) m\\ = 2.787 m\\ = 2.79 m$