Class 10 NCERT Triangles

NCERT

Class 10 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)

(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.

(i) Two equilateral triangles with sides $1\ cm$ and $2\ cm$

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

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(ii) Trapezium and square

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Triangle and parallelogram

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3.   State whether the following quadrilaterals are similar or not :

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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 $(ii)$.

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(i)

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

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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$ $\\$ $\therefore AD = 24\ cm$ $\\$

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 $PF = 0.36\ cm$

(i)

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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)

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

(iii)

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$PQ = 1.28\ cm,\ PR = 2.56\ cm,\ PE = 0.18\ cm,\ PF = 0.36\ cm$ $\\$ $\dfrac{PE}{PQ} = \dfrac{0.18}{1.28} = \dfrac{18}{128} = \dfrac{9}{64}$ $\\$ $\dfrac{PF}{PR} = \dfrac{0.36}{2.56} = \dfrac{9}{64}$ $\\$ 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 $\\$ $\dfrac{AM}{AB} = \dfrac{AN}{AD}$

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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 $\\$ $\dfrac{BF}{FE} = \dfrac{BE}{EC}$

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

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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 $EF\ ||\ QR.$

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

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

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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 $BC\ ||\ QR.$

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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 $\\$ $\dfrac{OB}{BQ} = \dfrac{OC}{OR}$ $\\$ $\therefore BC\ ||\ QR \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (By\ the\ converse\ of\ basic\ proportionality\ theorem)$ $\\$

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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 $IX$).$\\$

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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 $IX$).

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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 $\dfrac{AO}{BO} = \dfrac{CO}{DO}$

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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.

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

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

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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 :

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(i) $\angle{A} = \angle{P} = 60°$ $\\$ $\angle{B} = \angle{Q} = 80°$ $\\$ $\angle{C} = \angle{R} = 40°$ The $\dfrac{AB}{QR} = \dfrac{BC}{RP} = \dfrac{CA}{PQ}\ PQR$ [By $AAA$ similarity criterion] $\\$ (ii) $\\$ $\therefore \Delta{ABC}$ ~ $\Delta{QRP}$ [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 as the corresponding sides are not proportional. $\\$ (v)The given triangles are not similar as the corresponding sides are not proportional. $\\$ (vi) In $\Delta{DEF}$, $\\$ $\angle{D} + \angle{E} + \angle{F} = 180^o$ $\\$ (Sum of the measures of the angles of a triangle is $180^o$.) $\\$

$70^o + 80^o + \angle{F} = 180^o$ $\\$ $\angle{F} = 30^o$ $\\$ 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 $\angle{OAB}$.

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$DOB$ 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 $\dfrac{OA}{OC} = \dfrac{OB}{OD}$

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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 $\Delta{PQS}$ ~ $\Delta{TQR}$.

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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 $\Delta{RPQ}$ ~ $\Delta{RTS.}$

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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 $\Delta{ADE} \sim \Delta{ABC}$

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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) $\Delta{PDC}$ ~ $\Delta{BEC}$

(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 $\Delta{ABE}$ ~ $\Delta{CFB.}$

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) $\dfrac{CA}{BC} = \dfrac{BC}{MP}$

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) $\Delta{DCA} \sim \Delta{HGF}$ $\\$

diagram

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) $\\$ $\Rightarrow \dfrac{CD}{GH} = \dfrac{AC}{FG}$ $\\$ In $\angle{DCB}$ 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) 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)

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 $\Delta{ABD} \sim \Delta{ECF}.$

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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 $\Delta{ABC} \sim \Delta{PQR}.$

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Median divides the opposite side. $\\$ $\therefore BD = \dfrac{BC}{?}$ and $QM = \dfrac{QR}{?}$ $\\$ 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) $\\$ $\therefore \Delta{ABC} \sim \Delta{PQR}$ (By $SAS$ similarity criterion) $\\$ $\dfrac{AB}{Q} = \dfrac{BC}{QR}$

26.   $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle{ADC} = \angle{BAC}$. Show that $CA^2 = CB.CD.$

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. $\\$ $\therefore \dfrac{CA}{CB} = \dfrac{CD}{CA}$ $\\$ $\Rightarrow CA^2 = CA \times CD$ $\\$

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 $\Delta{ABC} \sim \Delta{PQR}.$

Answer

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Diagram

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 $L.$ $\\$

Diagram

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.

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 $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$

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)] $\\$

$\dfrac{}{} = \dfrac{}{}$ [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 $BC.$

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 $COD.$

Answer

31   None

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 $\\$ $\dfrac{ar\ (ABC)}{ar\ (DBC)} = \dfrac{AO}{DO}$

Let us draw two perpendiculars AP and DM on line BC.

diagram

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.

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 $\Delta{ABC}.$

Diagram

$D$ 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.

Answer

35   None

diagram

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.

diagram

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

Options

1   2:1

2   1:2

3   4:1

4   1:4

Answer

37   None

Diagram

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 $\Delta{BDE} = \dfrac{x}{2}$ $\\$ $\dfrac{Area(\Delta{ABC})}{Area(\Delta{BDE})} = \Bigg(\dfrac{x}{\dfrac{x}{2}}\Bigg)^2$ $\\$ 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

Options

1   2 : 3

2   4 : 9

3   81 : 16

4   16 : 81

Answer

38   None

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 $\\$

(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 $PM^2 = QM . MR.$

Answer

40   None

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) $AD^2 = BD . CD$ $\\$

Answer

41   None

(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 $AB^2 = 2AC^2.$

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.

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.

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.

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) $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2.$

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.

Let $OA$ be the wall and $AB$ be the ladder. $\\$ Therefore, by Pythagoras theorem, $\\$ $AB^2 = OA^2 + BO^2$ $\\$ $(10\ m)^2 = (8\ m)^2 + OB^2$ $\\$ $100\ m^2 = 64\ m^2 + OB^2$ $\\$ $OB = 6\ m^2$ $\\$ 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?

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.

Distance travelled by the plane flying towards north in $1\dfrac{1}{2}hrs = 1000 \times 1\dfrac{1}{2} = 1500\ km $ $\\$ Similarly, distance travelled by the plane flying towards west in $1\dfrac{1}{2}hrs = 1200 \times 1 \dfrac{1}{2} = 1800\ km$ $\\$ Let these distances be represented by $OA$ and $OB$ respectively. $\\$ Applying Pythagoras theorem, $\\$ Distance between these planes after $1\dfrac{1}{2}hrs,\ AB = \sqrt{OA^2 + OB^2}$ $\\$ $= \Bigg(\sqrt{(1500)^2 + (1800)^2}\Bigg)\ km = \Bigg(\sqrt{(2250000 + 3240000)}\Bigg)\ km $ $\\$ $= (\sqrt{5490000})\ km = (\sqrt{9 \times 610000})\ km = 300\sqrt{61}\ km$ $\\$ Therefore, the distance between these planes will be $300\sqrt{61}\ km$ after $1 \dfrac{1}{2}\ hrs$.

50.   $D$ and $E$ are points on the sides $CA$ and $CB$ respectively of a triangle $ABC$ right angled at $C$. Prove that $AE^2 + BD^2 = AB^2 + DE^2.$

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 $13\ m$

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 $2\ AB^2 = 2\ AC^2 + BC^2.$

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, $\\$ $DE^2 + AB^2 = AE^2 + BD^2$

52.   In an equilateral triangle $ABC,\ D$ is a point on side $BC$ such that $BD = \dfrac{1}{3} BC$. Prove that $9\ AD^2 = 7\ AB^2.$

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 $\\$ $AC^2 - \Bigg(\dfrac{BC}{4}\Bigg)^2 = AB^2 - \Bigg(\dfrac{3BC}{4}\Bigg)$ $\\$ $AC^2 - \dfrac{BC^2}{16} = AB^2 - \dfrac{9BC^2}{16}$ $\\$ $16AC^2 - BC^2 = 16AB^2 - 9BC^2$ $\\$ $16AB^2 - 16AC^2 = 8BC^2$ $\\$ $2AB^2 = 2AC^2 + BC^2$

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

Options

1   120°

2   60°

3   90°

4   45°

Answer

53   None

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 $(C)$.