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#### If S is a point on side PQ of a triangle PQR such that PS = QS = RS, then$(A)$$PR .QR = RS^{2}$         $(B)$$QS^{2} + RS^{2} = QR^{2}$$(C)$ $PR^{2} + QR^{2} = PQ^{2}$    $(D)$ $PS^{2} + RS^{2} = PR^{2}$

Given: In triangle PQR

$PS = QS = RS$

$\Rightarrow PS = RS$

$\therefore \angle 1=\angle 2 ...(1)$

(Q If a right angle triangle have equal length of base and height then their acute angles are also equal)

$\text{Similarly}$ $\angle 3=\angle 4 ...(2)$

( Q corresponding angles of equal sides are equal)

$\text{In }\Delta PQR$

$\angle P+\angle Q+\angle R=180^{o}$

$\because$ sum of angles of a triangle is 180°

$\angle 2+\angle 4+\angle 1+\angle 3=180^{o} ....(3)$

Using equation (1) and (2) in (3)

$\angle 1+\angle 3+\angle 1+\angle 3=180$

$2\left ( \angle 1+\angle 3 \right )=180^{o}$

$\Rightarrow \angle R=90^{o}$

$\text{Here }\angle R=90^{o}\therefore \Delta PQR$ $\text{is right angle triangle }$

$PR^{2} + QR^{2} = PQ^{2}$

#### $\text{If }\Delta ABC\sim \Delta QRP,\frac{ar(ABC)}{ar(PQR)}=\frac{9}{4},AB=18\; cm$  and BC = 15 cm then PR is equal to ?$\text{(A)}$ $10\; cm$$\text{(B)}$ $12\; cm$$\text{(C)}$ $\frac{20}{3}\; cm$$\text{(D)}$ $8\; cm$

$\Delta ABC \sim \Delta QRP \; \text {and}\; AB = 18 \; cm, BC = 15 \; cm$

$\therefore \frac{AC}{PQ}=\frac{AB}{QR}=\frac{BC}{RP}$

$\frac{ar(ABC)}{ar\left ( PQR \right )}=\frac{BC^{2}}{RP^{2}}\; \; \; \; \; \; \; \; ....(1)$

[Q If triangles are similar, their areas are proportional to the squares of the corresponding sides.]

$\frac{ar(ABC)}{ar\left ( PQR \right )}=\frac{9}{4}$

[Using equation (1)]

$\Rightarrow \frac{BC^{2}}{RP^{2}}=\frac{9}{4}$

$\frac{(15)^{2}}{RP^{2}}=\frac{9}{4}$

$RP^{2}=\frac{225\times 4}{9}=100$

$RP=\sqrt{100}$

$RP=10\; cm$

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#### $\text{If in triangles ABC and DEF},$,$\frac{AB}{DE}=\frac{BC}{FD},$ $\text{then they will be similar when},$$\text{(A)}$ $\angle B=\angle E$$\text{(B)}$ $\angle A=\angle D$$\text{(C)}$ $\angle B=\angle D$$\text{(D)}$ $\angle A=\angle D$

Given :

$\frac{AB}{DE}=\frac{BC}{FD}$

Here corresponding sides of given triangles ABC and DEF are equal in ratio, therefore, the triangles are similar

Also, we know that if triangles are similar then their corresponding angles are equal

$\Rightarrow \angle A=\angle E$

$\angle B=\angle D$

$\angle C=\angle F$

Hence option C is correct.

#### $\text{It is given that}$ $\Delta ABC\sim \Delta DFE,\angle A=30^{o},\angle C=50^{o},AB=5\; cm,AC=8\; cm$ and $DF=7.5\; cm.$ $\text{Then, the following is true:}$$(A)$$DE=12\; cm,\angle E=50^{o}$$(B)$ $DE=12\; cm,\angle F=100^{o}$$(C)$ $EF=12\; cm,\angle D=100^{o}$$(D)$ $EF=12\; cm,\angle D=30^{o}$

$\Delta ABC \sim \Delta DEF$

$\angle A=30^{o},\angle C=50^{o}$

$AB=5\; cm,AC=8\; cm,DF=7.5\; cm$

$\Delta ABC \sim \Delta DEF$

$\therefore \angle A=\angle D=30^{o}$

$\angle C=\angle E=100^{o}$

$\frac{AB}{DF}=\frac{BC}{FE}=\frac{AC}{DE}\; \; \; \; \; \; \; \; ....(1)$

$In$$\Delta ABC$

$\angle A+\angle B+\angle C=180^{o}$

$30+\angle B+100^{o}=180^{o}$

$\angle B=180^{o}-100^{o}-30^{o}$

$\angle B=50^{o}$

$\Rightarrow \angle B=\angle E=50^{o}$                       $\left \{ \because \Delta ABC\sim \Delta DFE \right \}$

Now equate first and the last term of equation (1) we get

$\frac{AB}{DF}=\frac{AC}{DE}$

$\frac{5}{7.5}=\frac{8}{DE}$

$DE=\frac{8 \times 75}{50}=12\; cm$

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#### $\text{It is given that}$  $\Delta ABC\sim \Delta PQR$, with $\frac{BC}{QR}=\frac{1}{3}.$$\text{Then, }$ $\frac{ar\left ( PRQ \right )}{ar\left ( BCA \right )}$ $\text{is equal to }$(a) $9$(b) $3$(c) $\frac{1}{3}$(d) $\frac{1}{9}$

Given : $\Delta ABC\sim \Delta PQR$

$\frac{BC}{QR}=\frac{1}{3}$

It is given that $\Delta ABC\sim \Delta PQR$

$\therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\; \; \; \; \; \; ...(1)$       $\therefore \frac{ar(PRQ)}{ar(BCA)}=\frac{PR^{2}}{AC^{2}}$

[Q If triangles are similar their areas are proportional to the squares of the corresponding sides.]

$\Rightarrow \frac{ar(PRQ)}{ar(BCA)}=\frac{QR^{2}}{BC^{2}}\; \; \; \; \; \; \; ...(2)$ $\text{{using equation (1)}}$

$\frac{QR}{BC}=\frac{3}{1} \; \; \; \; \; \; \; \; \; \left \{ \because \frac{BC}{QR}=\frac{1}{3} \right \}$

$\text{Put}$ $\frac{QR}{BC}=\frac{3}{1}$ $\text{in (2) we get}$

$\therefore \frac{ar(PRQ)}{ar(BCA)}=\frac{3^{2}}{1^{2}}=\frac{9}{1}=9$

Hence option A is correct.

#### In triangles $ABC \; \text {and} \; DEF, \angle B = \angle E, \angle F = \angle C \; \text {and}\; AB = 3 DE$. Then, the two triangles are(A) congruent but not similar  (B) similar but not congruent(C) neither congruent nor similar      (D) congruent as well as similar

In $\Delta ABC$ and $\Delta DEF$

$\angle B=\angle E\; \text {and}\; \angle F=\angle C\; \text {also}\; AB=3DE$

In $\Delta ABC$ and $\Delta DEF$

$\angle B=\angle E$                      (given)

$\angle F=\angle C$                       (given)

$\angle A=\angle D$                       (third angle)

Therefore $\Delta ABC\sim \Delta DEF$              (by AAA similarity)

Also, it is given that $AB=3DE$

$\Rightarrow AB\neq DE$

Hence $\Delta ABC$ and $\Delta DEF$ is not congruent because congruent figures have the same shape and same size.

Therefore the two triangles are similar but not congruent.

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#### If in two triangles DEF and PQR, $\angle D = \angle Q \; \text {and} \; \angle R = \angle E$, then which of the following is not true?(a) $\frac{EF}{PR}=\frac{DF}{PQ}$(b) $\frac{DE}{PQ}=\frac{EF}{RP}$(c) $\frac{DE}{QR}=\frac{DF}{PQ}$(d) $\frac{EF}{RP}=\frac{DE}{QR}$

In $\Delta DEF$ and $\Delta PQR$

$\angle D=\angle Q$ and $\angle R=\angle E$

Now draw two triangles with the help of given conditions:-

Here $\angle D=\angle Q$ and $\angle E=\angle R$  {Given}

$\therefore \; \; \; \; \Delta DEF\sim \Delta QRP$  {by AA similarity}

$\Rightarrow \angle F=\angle P$  {corresponding angles of similar triangles}

$\therefore \frac{DF}{QP}=\frac{ED}{RQ}=\frac{FE}{PR}\; \; \; \; \; \; \; \; ....(1)$

Here option (A) $\frac{EF}{PR}=\frac{DF}{PQ}$  is true because both the terms are derived from equation (1)

Here option (B) $\frac{DE}{PQ}=\frac{EF}{RP}$ is not true because the first term is not derived from equation (1)

Here option (C) $\frac{DE}{QR}=\frac{DF}{PQ}$  is true because both the terms are derived from equation (1)

Here option D i.e. is also true because both terms are derived from equation (1)

Hence option (B) is not true.

#### In Fig., two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, $\angle APB=50^{o}$ and $\angle CDP=30^{o}$. Then, $\angle PBA$ is equal to(A) 50°            (B) 30°            (C) 60°            (D) 100°

In $\Delta APB$ and $\Delta CPD$.

$\frac{AP}{PD}=\frac{6}{5}$

$\frac{BP}{CP}=\frac{3}{2.5}=\frac{30}{25}=\frac{6}{5}$

$\angle APB=\angle CPD=50^{o}$ {vertically opposite angles}

$\therefore \Delta APB \sim \Delta DPC$  {By SAS similarity criterion}

$\therefore \angle A=\angle D=30^{o}$ {corresponding angles of similar triangles}

In $\Delta APB$

$\angle A+\angle B+\angle APB=180^{o}$ {Sum of interior angles of a triangle is 180°}

$30^{o}+\angle B+50^{o}=180^{o}$

$\angle B=180^{o}-50^{o}-30^{o}$

$\angle B=100^{o}$

$\Rightarrow \angle PBA=100^{o}$

Hence option D is correct.

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#### If in two triangles ABC and PQR, $\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}$, then(a) $\Delta PQR\sim \Delta CAB$(b) $\Delta PQR\sim \Delta ABC$(c) $\Delta CBA\sim \Delta PQR$(d) $\Delta BCA\sim \Delta PQR$

In $\Delta ABC$ and $\Delta PQR$

$\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}\; \; \; \; \; \; \; .......(1)$

(A) If $\Delta PQR \sim \Delta CAB$

Here; $\frac{CA}{PQ}=\frac{CB}{PR}=\frac{AB}{PQ}$

It matches the equation (1) Hence option A is correct.

(B) If $\Delta PQR \sim \Delta ABC$

Here; $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

It does not match equation (1)

Therefore option (B) is not correct.

(c) $\Delta CBA\sim \Delta PQR$

Here; $\frac{CB}{PQ}=\frac{BA}{QR}=\frac{CA}{PR}$

It does not match equation (1) Hence option (C) is not correct

(D) $\Delta BCA \sim \Delta PQR$

Here; $\frac{BC}{PQ}=\frac{CA}{QR}=\frac{BA}{PR}$

It does not match the equation (1). Hence option (D) is not correct.

#### If $\Delta ABC \sim \Delta EDF$ and $\Delta ABC$ is not similar to $\Delta DEF$, then which of the following is not true?(A) $BC . EF = A C. FD$       (B) $AB . EF = AC . DE$(C) $BC . DE = AB . EF$      (D) $BC . DE = AB . FD$

Given : $\Delta ABC\sim \Delta EDF$

$\therefore \frac{AB}{ED}=\frac{BC}{DF}=\frac{AC}{EF}$

Taking the first two terms we get

$\therefore \frac{AB}{ED}=\frac{BC}{DF}$

by cross multiply we get

$AB.DF=BC.ED$

Hence option D is correct

Now taking the last two terms we get

$\frac{BC}{DF}=\frac{AC}{EF}\Rightarrow BC.EF=DF.AC$ {By cross multiplication}

Hence option (A) is also correct

Now taking first and last terms, we get

$\frac{AB}{ED}=\frac{AC}{EF}\Rightarrow AB.EF=AC.ED$ {By cross multiplication}

Hence option (B) is also correct

By using congruence properties we conclude that only option C is not true.