Triangles — Class 10 Maths PYQs

(B) A square and a rhombus of the same area are always similar.

(C) $8 \operatorname{cm}$

(d) $40^\circ$

(d) $BC = 3 EF$

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) $2.25$ cm

(c) $7$ cm

(d) $\frac{9}{2}$

(b) $20$ cm

(a) $8$ cm

(d) $10$

(B) 2.25 cm

(C) $15.3$ cm

(D) $BC \parallel XY$

(B) $1.5$ cm

(B) $10$

(C) $8$ cm

(C) $\frac{16}{3} \text{ cm}$

(D) $3.3$

(A) $2$

(D) $4.8$ cm

(D) 1:2

(C) $3$ cm

(B) $2$ only

(A) $1.5$ cm

(A) $12.5$ cm

(B) $120$ cm

(C) $4$ cm

(C) $6$ cm

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(C) $3:5$

As $XZ \parallel BC$ Therefore $\frac{AX}{XB} = \frac{AZ}{ZC} = \frac{3}{2}$ (i)
$\triangle AXY \sim \triangle ABM$
$\Rightarrow \frac{AX}{AB} = \frac{XY}{BM}$ or $\frac{XY}{3} = \frac{3}{5}$
$\Rightarrow XY = \frac{9}{5}$ or $1.8$ cm

In $\triangle ABC$, $DE || BC$
$\therefore \frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{x}{x-2} = \frac{x+2}{x-1}$
$x(x - 1) = (x + 2)(x - 2)$
$x^2 - x = x^2 - 4 \Rightarrow x = 4$

FR = $6-3.6 = 2.4$ cm
$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$ and $\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$
Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$,∴ EF $||$ QR.

$\triangle ADE \sim \triangle ABC$
$\therefore \frac{AD}{AB} = \frac{AE}{AC}$

$\frac{AP}{AB} = \frac{1}{3}$; $\frac{AQ}{AC} = \frac{1.5}{4.5} = \frac{1}{3}$
$\angle ACP = \angle ACB$
$\Delta APQ \sim \Delta ABC$
$PQ = 1.2$ cm

Since $DE \parallel BC \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ (I) (1 Mark)
$\frac{x}{x-2} = \frac{x+2}{x-1}$ (II) (1/2 Mark)
Solving, we get $x = 4$ (III) (1/2 Mark)

XY $||$ QR
$\frac{PQ}{XQ} = \frac{PR}{YR}$ (1 Mark)
$\Rightarrow \frac{7}{3} = \frac{6.3}{YR}$ (1/2 Mark)
$\Rightarrow YR = 2.7 \text{ cm}$ (1/2 Mark)

In $\triangle BEA$, $FD \parallel EA$
$\therefore \frac{BF}{FE} = \frac{BD}{DA}$ ...(i) (I) (1 Mark)
In $\triangle BCA$, $ED \parallel CA$
$\therefore \frac{BE}{EC} = \frac{BD}{DA}$ ...(ii) (II) ($\frac{1}{2}$ Mark)
Using (i) and (ii)
$\frac{BF}{FE} = \frac{BE}{EC}$ (III) ($\frac{1}{2}$ Mark)

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. (Correct Statement: $1$ mark)
Given: In $\Delta ABC, DE \parallel BC$
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$
Construction: Draw $DM \perp AC, EN \perp AB$, join $BE$ and $CD$ (Given + To prove + Construction + Figure: $1$ mark)
Proof: $\frac{ar(\Delta ADE)}{ar(\Delta DBE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \dots (i)$ [$1$ mark]
$\frac{ar(\Delta ADE)}{ar(\Delta ECD)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \dots (ii)$ [$1$ mark]
as $\Delta DBE$ and $\Delta DCE$ lie on the same base $DE$ and between same parallels $BC$ and $DE$
$\therefore ar(\Delta DBE) = ar(\Delta ECD)$ or $\frac{ar(\Delta ADE)}{ar(\Delta DBE)} = \frac{ar(\Delta ADE)}{ar(\Delta ECD)} \dots (iii)$ [$\frac{1}{2}$ mark]
From $(i), (ii)$ and $(iii)$, we get $\frac{AD}{DB} = \frac{AE}{EC}$ [$\frac{1}{2}$ mark]

Correct Given, to prove, figure, construction
Correct proof

Correct figure, given, to prove, construction
Correct proof
Correct statement of converse of given statement

Correct statement of converse of Basic Proportionality Theorem.
In $\Delta ADC, EG || DC \implies \frac{AE}{ED} = \frac{AG}{GC} = \frac{2}{3}$
In $\Delta ABC, GF || AB \implies \frac{AG}{GC} = \frac{BF}{FC} = \frac{2}{3}$
$\Delta AEG \sim \Delta ADC$
$\implies \frac{AE}{AD} = \frac{AG}{AC} = \frac{EG}{DC} \implies \frac{2}{5} = \frac{EG}{DC} \implies EG = \frac{2}{5} \times 15 = 6$ cm
Similarly, $\Delta CFG \sim \Delta CBA$ and $\frac{FC}{BF} = \frac{3}{2} \implies \frac{FC}{BC} = \frac{GF}{AB} = \frac{3}{5} \implies GF = \frac{3}{5} \times 10 = 6$ cm
$EF = EG + GF = 6 + 6 = 12$ cm

Correct statement of Basic Proportionality Theorem.
As DA $||$ CE $\implies \frac{BD}{DC} = \frac{BA}{AE}$ --- (1)
$\angle 2 = \angle 3$ & $\angle 1 = \angle 4$. As $\angle 1 = \angle 2 \implies \angle 3 = \angle 4 \implies AC = AE$ --- (2)
From (1) & (2), $\frac{BD}{DC} = \frac{BA}{AC}$

Correct Statement of BPT (1 mark). Correct figure, Given, To Prove, Construction (2 marks). Correct Proof of BPT (2 marks). NOTE* Given statement in English version is not a correct statement. Full marks may be awarded to any attempt in English medium.

(d) $10$ cm

(c) $83^\circ$

(d) $1$ cm

(a) $2$ cm

(b) $2.5$ cm

(b) $\angle B = \angle D$

(B) $27$ cm

(C) $\triangle PQR \sim \triangle CBA$

(B) $\frac{6}{7}$

(B) $3$

(B) $45^\circ$

(C) $\angle B = \angle R$

(B) $7.5$ cm

(D) Measure of $\angle A$ is $110^{\circ}$

(A) $5.4$ cm

(D) similar but not congruent

(D) $\frac{DE}{PQ} = \frac{EF}{RP}$

(C) $8 \operatorname{cm}$

(C) $150^\circ$

(B) $21.5$ cm

(d) $16$ cm (1 Mark)

(a) $\Delta RPQ$ and $\Delta XZY$ (1 Mark)

(C) QR : QP (1 Mark)

(C) 12 cm

(D) $3.6$ cm (1 Mark)

(A) $\angle A = \angle C$ (1 Mark)

(C) $\angle A = \angle D, \angle C = \angle F$

(A) $\frac{21}{4}$ cm

(D) $\frac{AC}{BC}=\frac{PR}{QR}$ (1 Mark)

(C) $3.5$ cm (1 Mark)

(D) $18$ cm

None of the given options is correct.
Note - One mark to be given to all students who have attempted this question.

(D) Equilateral triangles

(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

(B) $6$ cm

(C) may or may not be similar.

(B) $\triangle DEF \sim \triangle CBA$

In $\triangle ABC$, $AB = AC$ (Given)
$\therefore \angle ACB = \angle ABC$ ----- (1)
In $\triangle ABD$ and $\triangle ECF$
$\angle ADB = \angle EFC$ (each $90^\circ$)
$\angle ABD = \angle ACD$ (from (1))
$\therefore \triangle ABD \sim \triangle ECF$ (AA rule)

In $\triangle AOB$ and $\triangle COD$
$\frac{AO}{OC} = \frac{BO}{OD}$ (Given)
$\angle AOB = \angle COD$ (V.O.A.)
$\therefore \triangle AOB \sim \triangle COD$ (SAS rule) (1 Mark)
$\frac{AO}{OC} = \frac{AB}{CD}$ (C.P.S.T.)
$\frac{1}{2} = \frac{5}{CD}$
$\Rightarrow CD = 10$ cm (1 Mark)

Given $\triangle ABE \cong \triangle ACD$
$\therefore AE = AD$ or $AD = AE$ ---- (1) ($1/2$)
and $AB = AC$ ---- (2) ($1/2$)
Dividing (1) by (2), we have
$\frac{AD}{AB} = \frac{AE}{AC}$ ($1/2$)
and $\angle DAE = \angle BAC$
$\therefore \triangle ADE \sim \triangle ABC$ ($1/2$)

$\therefore \triangle AHK \sim \triangle ABC$ (given)
$$\begin{aligned}& \therefore \frac{HK}{BC} = \frac{AK}{AC} \\ & \Rightarrow \frac{6.4}{3.2} = \frac{8.0}{AC} \\ & AC = 4 \text{ cm}\end{aligned}$$

Given, $$\begin{aligned}& \triangle ABC \sim \triangle DEF \\ & \therefore \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{DF} \\ & \Rightarrow \frac{4}{6} = \frac{BC}{9} = \frac{CA}{12} \\ & \therefore BC = 6\end{aligned}$$ cm and CA = $8$ cm
Perimeter of $\triangle ABC = 4+6+8=18$ cm

In $\triangle OZX$, ZX $||$ BY (As XZ $||$ AB)
$\therefore \frac{OB}{OZ} = \frac{OY}{OX}$ --- (1)
In $\triangle OCX$, ZY $||$ CX (As YZ $||$ AC)
$\therefore \frac{OZ}{OC} = \frac{OY}{OX}$ --- (2)
Using (1) and (2), we get
$\frac{OB}{OZ} = \frac{OZ}{OC}$
$\Rightarrow OZ^2 = OB \times OC$

In $\triangle ACD$ and $\triangle BCA$
$\angle ADC = \angle BAC$
$\angle ACD = \angle BCA$
$\therefore \triangle ACD \sim \triangle BCA$
So, $\frac{CA}{CB} = \frac{CD}{CA}$
$\Rightarrow CA^2 = CD.CB$

Given OA.OB = OC.OD
$\frac{OA}{OC} = \frac{OD}{OB}$
$\& \angle AOD = \angle COB$
$\therefore \triangle AOD \sim \triangle COB$
So, $\angle D = \angle B$ and $\angle A = \angle C$

Given $\frac{PS}{SQ} = \frac{PT}{TR}$
$\Rightarrow ST || QR$ ($1$)
$\therefore \angle PST = \angle PQR$ ($1/2$)
and given, $\angle PST = \angle PRQ$
So, $\angle PQR = \angle PRQ$
$\therefore \triangle PQR$ is an isosceles triangle. ($1/2$)

Let AB be the lamp post and CD be the boy $1.5$ m tall.
For correct figure
Let the length of shadow be $x$ m
Speed of boy = $2.5$ m/sec
$\therefore$ Distance covered in $3$ seconds = $7.5$ m
Now, $\triangle ABE \sim \triangle CDE$
$\frac{CD}{AB} = \frac{DE}{BE}$
$\frac{1.5}{12} = \frac{x}{7.5+x}$
Solving, we get $x = \frac{15}{14}$ or $1.07$ approx.
Hence length of shadow is $1.07$ m

$\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow DE \parallel BC$
$\angle BDE + \angle DBC = \angle CED + \angle BCE$
$\angle DBC = \angle BCE$ or $\angle B = \angle C$
Thus $\Delta ABC$ is an isosceles triangle.

$\triangle ABC \sim \triangle XYZ \implies \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$ (1 Mark)
$\frac{4}{x} = \frac{6}{7.2} = \frac{y}{6}$ (1/2 Mark + 1/2 Mark)
Solving, we get $x = 4.8$ cm, $y = 5$ cm (1/2 Mark + 1/2 Mark)

$\triangle AHK \sim \triangle ABC$
$\frac{AK}{AC} = \frac{HK}{BC}$ (1 Mark)
$\frac{10}{AC} = \frac{7}{3.5}$ (1/2 Mark)
$\Rightarrow AC = 5$ cm (1/2 Mark)

Correct figure (1/2 Mark)
$\triangle AOB \sim \triangle COD$ (1 Mark)
$\therefore \angle OAB = \angle OCD$
As alternate angles are equal, so DC $\|\|$ AB (1/2 Mark)
Therefore, ABCD is a trapezium.
Alternate solution:
Correct figure. (1/2 Mark)
Draw EO $\|\|$ AB to intersect AD at E.
In $\triangle ABD$, EO $\|\|$ AB
$\therefore \frac{AE}{ED} = \frac{BO}{OD}$ (1/2 Mark)
Given, $\frac{AO}{OC} = \frac{BO}{OD}$
So, $\frac{AE}{ED} = \frac{AO}{OC}$ (1/2 Mark)
$\therefore EO\|\| DC$
So, DC $\|\|$ AB (1/2 Mark)
Therefore, ABCD is a trapezium.

$\triangle STP \sim \triangle RTQ$ (1 Mark)
$\frac{ST}{TP} = \frac{RT}{TQ} \Rightarrow ST \times TQ = PT \times TR$ (1 Mark)

$\triangle ADC \sim \triangle BAC$ (1 Mark)
$\frac{DC}{AC} = \frac{AC}{BC}$ (1/2 Mark)
$\Rightarrow AC^2 = DC \times BC$ or $CA^2 = CB \times CD$ (1/2 Mark)

DEFG is a square. $\therefore$ GF || BE (I) (1/2 Mark)
$\angle GBD = \angle AGF$ (II) (1/2 Mark)
$\therefore \triangle AGF \sim \triangle BDG$ (III) (1/2 Mark)
$\Rightarrow \frac{AG}{DB} = \frac{AF}{DG}$ (IV) (1/2 Mark)
$\Rightarrow AG \times DG = AF \times DB$

$AB \parallel DE \Rightarrow \angle DEF = \angle ABC$ (I Mark)
$AC \parallel DF \Rightarrow \angle DFE = \angle ACB$
Hence $\triangle ABC \sim \triangle DEF$
$\frac{BC}{EF} = \frac{AB}{DE} \Rightarrow \frac{10}{20} = \frac{7}{DE}$ (II Mark)
$DE = 14$ cm (III Mark)

Let the length of corresponding side of the second triangle be $x$ cm
$\frac{9}{x} = \frac{22}{33} \Rightarrow x = 13.5$
$\therefore$ The length of corresponding side of the second triangle $= 13.5$ cm

Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$
Since AD and PM are medians, $BC = 2BD$ and $QR = 2QM$
So, $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$
Therefore, $\triangle ABD \sim \triangle PQM$ (SSS similarity criterion)
$\Rightarrow \angle B = \angle Q$ (Corresponding angles of similar triangles)
Now, in $\triangle ABC$ and $\triangle PQR$
$\frac{AB}{PQ} = \frac{BC}{QR}$ (Given)
$\angle B = \angle Q$ (Proved above)
Therefore, $\triangle ABC \sim \triangle PQR$ (SAS similarity criterion)

$\triangle EFG \sim \triangle ADG$
$\Rightarrow \frac{EF}{AD} = \frac{FG}{DG}$
quad (i)
$\triangle EFC \sim \triangle BDC$
$\Rightarrow \frac{EF}{BD} = \frac{CF}{CD}$
$\Rightarrow \frac{EF}{AD} = \frac{CF}{CD}$
quad $\{BD = AD\}$
quad (ii)
Using (i) and (ii)
$\frac{FG}{DG} = \frac{CF}{CD}$

In $\triangle PQR$ and $\triangle SPR$
$\angle QPR = \angle PSR$ (Given) (1 Mark)
$\angle R = \angle R$ (common) (1 Mark)
$\Rightarrow \triangle PQR \sim \triangle SPR$ (AA similarity)
$\therefore \frac{QR}{PR} = \frac{PR}{SR}$ (1 Mark)
or $\frac{QR}{RP} = \frac{RP}{RS}$

(a)$\triangle CQB \sim \triangle CPA$
$\Rightarrow \frac{b}{a + b} = \frac{z}{x}$ (i)
Also $\triangle AQB \sim \triangle ARC$
$\Rightarrow \frac{a}{a + b} = \frac{z}{y}$ (ii)
from (i) and (ii) $\frac{z}{x} + \frac{z}{y} = \frac{a + b}{a + b} = 1$
$\Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{z}$

(i) $\triangle ABC \sim \triangle PQR$
$\angle A=\angle P$
and $\frac{AB}{PQ} = \frac{AC}{PR}$
$\Rightarrow \frac{2AD}{2PS} = \frac{AC}{PR}$
$\Rightarrow \frac{AD}{PS} = \frac{AC}{PR}$ and $\angle A = \angle P$
Therefore $\triangle ADC \sim \triangle PSR$
(ii)Hence $\frac{AD}{PS} = \frac{AC}{PR}$
$\Rightarrow AD \times PR = AC \times PS$

In $\Delta ABC$, D is a point on side BC such that $\angle ADC = \angle BAC$
In $\Delta CBA$ and $\Delta CDA$
$\angle C = \angle C$ (common)
$\angle BAC = \angle ADC$ (given)
$\therefore \Delta CBA \sim \Delta CAD$ (By AA similarity)
$\therefore$ their corresponding sides are proportional
$\frac{CB}{CA} = \frac{CA}{CD} \Rightarrow CA^2 = CB. CD$

AD and AM are medians of $\Delta ABC$ and $\Delta PQR$ respectively.
$\Delta ABC \sim \Delta PQR$
$\therefore \frac{AB}{PQ} = \frac{BC}{QR}$
$\frac{AB}{PQ} = \frac{2BD}{2QM}$
$\frac{AB}{PQ} = \frac{BD}{QM}$
Also $\angle B = \angle Q$ ($\Delta ABC \sim \Delta PQR$)
$\Rightarrow \Delta ABD \sim \Delta PQM$ (SAS similarly)
$\Rightarrow \frac{AB}{PQ} = \frac{AD}{PM}$

(i) As D and E are the mid-points of RQ and RP respectively.
By mid-point theorem, $ED \parallel PQ$ and $ED = \frac{1}{2} PQ$ ... (1)
$\Rightarrow \triangle POQ \sim \triangle DOE$
(ii) Using part (i), $\frac{PO}{OD} = \frac{PQ}{ED}$
Using (1), $PO = 2 OD$
(iii) Using part (ii), $PO = 2 OD = 2(PD – PO)$
$\Rightarrow 3PO = 2PD$
$\Rightarrow PO = \frac{2}{3} PD$

$\triangle PAC \sim \triangle QBC$
$\therefore \frac{x}{y} = \frac{AC}{BC}$ or $\frac{y}{x} = \frac{BC}{AC}$ ----- (i)
$\triangle RCA \sim \triangle QBA$
$\therefore \frac{z}{y} = \frac{AC}{AB}$ or $\frac{y}{z} = \frac{AB}{AC}$ ----- (ii)
Adding (i) and (ii)
$\frac{y}{x} + \frac{y}{z} = \frac{BC+AB}{AC}$
$\Rightarrow y\left(\frac{1}{x} + \frac{1}{z}\right) = \frac{AC}{AC} = 1$
$\Rightarrow \frac{1}{x} + \frac{1}{z} = \frac{1}{y}$

Correct figure
AD$^2$ = BD $\times$ DC
$\frac{AD}{DC} = \frac{BD}{AD}$
$\triangle$ADB $\sim \triangle$CDA
$\angle$BAD = $\angle$ACD & $\angle$DAC = $\angle$DBA
In $\triangle$ABC
$\angle$ABC + $\angle$CAB + $\angle$BCA = $180^{\circ}$
$\angle$ABC + $\angle$BAD + $\angle$CAD + $\angle$BCA = $180^{\circ}$
$\angle$BAD + $\angle$CAD = $90^{\circ}$
$\angle$BAC = $90^{\circ}$

Correct figure
In $\triangle ABE$ and $$\begin{aligned}& \triangle CFB \\ & \angle EAB = \angle BCF \\ & \angle AEB = \angle CBF \\ & \Rightarrow \triangle ABE \sim \triangle CFB\end{aligned}$$

Correct figure
$$\begin{aligned}& \therefore \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM} \\ & \therefore \frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \\ & \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}-------(i) \\ & \Rightarrow \triangle ABD \sim \triangle PQM \\ & \Rightarrow \angle B = \angle Q_ -----(ii) \\ & \text{In } \triangle ABC \text{ and } \triangle PQR \\ & \frac{AB}{PQ} = \frac{BC}{QR} \\ & \angle B = \angle Q \\ & \therefore \triangle ABC \sim \triangle PQR\end{aligned}$$

$\triangle FEC = \triangle GDB$
Therefore, $\angle 3 = \angle 4$
In $\triangle ABC$,
$\angle 3 = \angle 4$
$\therefore AB = AC ............(i)$
In $\triangle ADE$, $\angle 1 = \angle 2$
$AD = AE .............(ii)$
Dividing (ii) by (i)
$\frac{AD}{AB} = \frac{AE}{AC}$
$\Rightarrow DE \parallel BC$
$\angle 1 = \angle 3$ and $\angle 2 = \angle 4$
$\therefore \triangle ADE \sim \triangle ABC$

Correct figure, given, to prove and construction
Correct proof
In $\triangle ABC$, LM $||$ CB
$\frac{AM}{MB} = \frac{AL}{LC}$ --- (1)
In $\triangle ADC$, LN $||$ CD
$\frac{AN}{ND} = \frac{AL}{LC}$ --- (2)
from (1) and (2), we have
$\frac{AM}{MB} = \frac{AN}{ND}$

$\triangle ABQ \sim \triangle ACR$
$\frac{AB}{AC} = \frac{QB}{RC} = \frac{y}{z}$ ... (i)
Similarly, $\triangle CBQ \sim \triangle CAP$
$\frac{BC}{AC} = \frac{QB}{PA} = \frac{y}{x}$ ... (ii)
On adding (i) & (ii), we get
$\frac{AB}{AC} + \frac{BC}{AC} = \frac{y}{z} + \frac{y}{x}$
$\frac{AB+BC}{AC} = y(\frac{1}{z} + \frac{1}{x})$
$\frac{AC}{AC} = y(\frac{1}{z} + \frac{1}{x})$
$1 = y(\frac{1}{z} + \frac{1}{x})$
$\therefore \frac{1}{y} = \frac{1}{x} + \frac{1}{z}$

As, $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{3}{5}$
$\Rightarrow \triangle ABC \sim \triangle PQR$
$\angle C = \angle R$
(i) In $\triangle ADC$ and $\triangle PSR$,
$\angle ADC = \angle PSR$
and $\angle C = \angle R$
$\therefore \triangle ADC \sim \triangle PSR$
(ii) $\frac{AD}{PS} = \frac{AC}{PR} = \frac{3}{5}$
$\Rightarrow \frac{4}{PS} = \frac{3}{5}$
$\Rightarrow PS = \frac{20}{3}$ cm
(iii) $\frac{\text{ar (}\triangle ABC)}{\text{ar (}\triangle PQR)} = \frac{\frac{1}{2}\times BC\times AD}{\frac{1}{2}\times QR\times PS}$
$= \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$
$\therefore$ ar ($\triangle ABC$): ar ($\triangle PQR$) = $9 : 25$

Correct statement
Join AF intersecting line $m$ at G
In $\triangle ACF$, BG $||$ CF
$\Rightarrow \frac{AB}{BC} = \frac{AG}{GF}$ ...(i)
In $\triangle FDA$, GE $||$ AD
$\Rightarrow \frac{EF}{DE} = \frac{GF}{AG}$ or $\frac{DE}{EF} = \frac{AG}{GF}$ ...(ii)
From, (i) and (ii), we get $\frac{AB}{BC} = \frac{DE}{EF}$

$\Delta ABC \sim \Delta DAC$ (1 mark). Similarly, $\Delta ABC \sim \Delta DBA$ ($\frac{1}{2}$ mark). From equations ① and ②, $\Delta DAC \sim \Delta DBA$ or $\Delta ADB \sim \Delta CDA$ (1 mark). $\frac{AD}{CD} = \frac{BD}{AD}$ ($\frac{1}{2}$ mark). $AD^2 = BD \times CD = 8 \times 2$ ($\frac{1}{2} + 1$ marks). $\therefore AD = 4$ cm ($\frac{1}{2}$ mark).

(i) $\triangle ABC \sim \triangle PQR \implies \angle A = \angle P$ (1/2 Mark)
$\frac{AB}{PQ} = \frac{AC}{PR} = \frac{2 AM}{2 PN} = \frac{AC}{PR}$ (as $CM$ and $RN$ are the medians) (1.5 Marks)
$\therefore \triangle AMC \sim \triangle PNR$ (1/2 Mark)
(ii) $\triangle ABC \sim \triangle PQR \implies \angle B = \angle Q$ (1/2 Mark)
$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{2 MB}{2 NQ} = \frac{BC}{QR}$ (as $CM$ and $RN$ are the medians) (1.5 Marks)
$\therefore \triangle CMB \sim \triangle RNQ$ (1/2 Mark)

Draw DZ $\|\|$ BY. (1/2 Mark)
In $\triangle CBY$, DZ $\|\|$ BY
$\therefore \frac{CD}{DB} = \frac{CZ}{ZY} = 1$
Therefore, DZ $= \frac{1}{2}$ BY --- (i) (1/2 Mark)
In $\triangle ADZ$, DZ $\|\|$ XY
So, $\triangle AXY \sim \triangle ADZ$
$\therefore \frac{AX}{AD} = \frac{XY}{DZ}$ (1 Mark)
$\Rightarrow \frac{2}{5} = \frac{XY}{DZ}$ or DZ $= \frac{5}{2}$ XY --- (ii) (1/2 Mark)
Using (i) and (ii),
BY $= 5$ XY (1 Mark)
Therefore BX = BY – XY $= 4$ XY (1/2 Mark)
Alternate solution:
Draw DZ $\|\|$ BY. (1/2 Mark)
$\triangle AXY \sim \triangle ADZ$ (1/2 Mark)
$\therefore \frac{AX}{AD} = \frac{XY}{DZ} \Rightarrow \frac{2}{2+3} = \frac{XY}{DZ}$
$\Rightarrow 2$ DZ $= 5$ XY --- (i) (1 Mark)
Now, $\triangle CDZ \sim \triangle CBY$
$\therefore \frac{CD}{CB} = \frac{DZ}{BY}$ (1/2 Mark)
But CD $= \frac{1}{2}$ BC
So, $\frac{DZ}{BY} = \frac{1}{2} \Rightarrow BY = 2$ DZ --- (ii) (1 Mark)
From (i) and (ii), we get
BY $= 5$ XY (1 Mark)
$\Rightarrow (BX + XY) = 5$ XY
$\Rightarrow BX = 4$ XY (1/2 Mark)

(i) $\therefore$ Diagonals OG and BC of quadrilateral OBGC bisect each other.
$\therefore$ OBGC is a parallelogram (I Mark)
(ii) $CO \parallel GB \Rightarrow CE \parallel GB$ (II Mark)
In $\triangle AGB, OE \parallel GB \rightarrow \frac{AO}{OG} = \frac{AE}{EB}$ (III Mark)
Similarly in $\triangle AGC, \frac{AO}{OG} = \frac{AF}{FC}$ (IV Mark)
$\Rightarrow \frac{AE}{EB} = \frac{AF}{FC} \Rightarrow EF \parallel BC$ (V Mark)
(iii) In $\triangle AEF$ and $\triangle ABC$
$\angle AEF = \angle ABC$ and $\angle A$ is common.
$\therefore \triangle AEF \sim \triangle ABC$ (VI Mark)

(i) $QC \parallel AB \therefore \triangle RQC \sim \triangle RAB$ (I Mark)
$\Rightarrow \frac{QR}{AR} = \frac{QC}{AB} = \frac{1}{2}$ (II Mark)
$\Rightarrow 2QR = AR \Rightarrow Q$ is the mid point of $AR$
$\therefore AQ = QR$
(ii) $\triangle PQD \sim \triangle PAB$ (III Mark)
$\therefore \frac{QP}{AP} = \frac{DQ}{BA} = \frac{1}{2}$ (IV Mark)
$\Rightarrow AP = 2PQ$
(iii) Since $AQ = QR$
$AP + PQ = PR - PQ$ (V Mark)
$\Rightarrow AP + \frac{1}{2}AP = PR - \frac{1}{2}AP$ (VI Mark)
$\Rightarrow PR = 2AP$

(i) $\triangle ABC \sim \triangle FEG$ (given)
$\angle ACB = \angle FGE$
$\frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE \Rightarrow \angle ACD = \angle FGH$ ($\frac{1}{2}$ Mark)
Also, $\angle A = \angle F$
$\therefore \triangle ACD \sim \triangle FGH$ (2 Marks)
$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$ ($\frac{1}{2}$ Mark)
(ii) In $\triangle DCB$ and $\triangle HGE$
$\angle DBC = \angle HEG$ (1 Mark)
$\angle DCB = \angle HGE$ (1 Mark)
$\therefore \triangle DCB \sim \triangle HGE$

(b) (i) $\triangle ABC \sim \triangle PQR \Rightarrow \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} = \frac{2DC}{2SR}$ (2)
$\Rightarrow \frac{AC}{PR} = \frac{DC}{SR}$ (½)
and $\angle C = \angle R$ (½)
$\therefore \triangle ADC \sim \triangle PSR$ (by SAS similarity) (1)
(ii) Since $\triangle ADC \sim \triangle PSR$, $\frac{AD}{PS} = \frac{DC}{SR}$ (1)
$\frac{AD}{PS} = \frac{BC/2}{QR/2} = \frac{BC}{QR}$

(D) trapezium

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

$$\begin{aligned}& \triangle OBE \sim \triangle ODA \\ & \frac{OB}{OD} = \frac{BE}{AD} \\ & \Rightarrow \frac{1}{2} = \frac{BE}{BC}\end{aligned}$$ (AD = BC)
$BE = 1.5$ cm $\Rightarrow BC = 3$ cm

In $\triangle AOB$ and $\triangle COD$,
$\angle OAB = \angle OCD$
$\angle OBA = \angle ODC$
Therefore, $\triangle AOB \sim \triangle COD$
$\therefore \frac{OA}{OC} = \frac{OB}{OD}$

Join PR
PQ $||$ SR and MN $||$ PQ $\Rightarrow$ MN $||$ SR
In $\triangle PSR$,
$\frac{PM}{MS} = \frac{PO}{OR}$ ... (i)
In $\triangle PQR$,
$\frac{PO}{OR} = \frac{QN}{NR}$ ... (ii)
From (i) and (ii), $\frac{PM}{MS} = \frac{QN}{NR}$

(i) In $\triangle ABD \& \triangle CBD$
$\angle 3 = \angle 4$
$\angle 1 = \angle 2$
$\therefore \triangle ABD \sim \triangle CBD$
(ii) $\triangle ABD \cong \triangle CBD$
$\therefore AB = BC$

$\triangle OAB \sim \triangle OCD \Rightarrow \frac{OD}{OB} = \frac{CD}{AB} \therefore \frac{OD}{OB} = \frac{1}{2} \text{ Therefore } \frac{CD}{AB} = \frac{1}{2} \Rightarrow AB = 2 CD$

Draw OE $||$ CD
In $\triangle DAB$, OE $||$ AB (since OE $||$ CD and AB $||$ CD)
By Basic Proportionality Theorem (BPT): $\frac{DE}{AE} = \frac{DO}{OB}$
In $\triangle ADC$, OE $||$ DC
By BPT: $\frac{AE}{DE} = \frac{AO}{OC}$
From the two ratios: $\frac{DO}{OB} = \frac{AO}{OC}$
$\Rightarrow \frac{OA}{OB} = \frac{OC}{OD}$

Let D, E, F be the mid-points of sides BC, CA, AB respectively of $\triangle ABC$.
By Mid-point Theorem, DE $||$ AB and DE $= \frac{1}{2}$ AB.
EF $||$ BC and EF $= \frac{1}{2}$ BC.
FD $||$ AC and FD $= \frac{1}{2}$ AC.
Consider $\triangle AFE$ and $\triangle ABC$.
$\frac{AF}{AB} = \frac{1}{2}$ and $\frac{AE}{AC} = \frac{1}{2}$ (F and E are mid-points)
$\angle A$ is common.
So, $\triangle AFE \sim \triangle ABC$ (SAS similarity criterion).
Similarly, $\triangle BDF \sim \triangle ABC$ and $\triangle CED \sim \triangle ABC$.
Also, DE $||$ AB, so ADEF is a parallelogram.
$\angle FDE = \angle A$ (Opposite angles of parallelogram)
$\frac{FD}{AC} = \frac{1}{2}$, $\frac{DE}{AB} = \frac{1}{2}$, $\frac{FE}{BC} = \frac{1}{2}$
So, $\triangle FDE \sim \triangle ABC$ (SSS similarity criterion).
Thus, all four triangles are similar to the original triangle.

$\triangle ALE \sim \triangle CLB$
$\Rightarrow \frac{AL}{CL} = \frac{EL}{BL}$
quad (i)
Also $\triangle CLM \sim \triangle ALB$
$\Rightarrow \frac{AL}{CL} = \frac{AB}{CM}$
$\Rightarrow \frac{AL}{CL} = \frac{CD}{CM}$
quad $\{AB = CD\}$
quad (ii)
Using (i) and (ii)
$\frac{EL}{BL} = \frac{2CM}{CM}$
$\Rightarrow EL = 2BL$

(i) $$\begin{aligned}& \triangle DPC \sim \triangle LPB \\ & \Rightarrow \frac{DP}{PL} = \frac{PC}{PB} = \frac{DC}{BL} \quad \text{(i)} \\ & \text{(ii) As } BC \parallel AD \\ & \therefore \triangle LPB \sim \triangle LDA \\ & \text{In } \triangle DLA, AD \parallel BP \\ & \Rightarrow \frac{LP}{DP} = \frac{LB}{AB} \\ & \Rightarrow \frac{LP}{DP} + 1 = \frac{LB}{AB} + 1 \\ & \Rightarrow \frac{DL}{DP} = \frac{AL}{AB} \\ & \Rightarrow \frac{DL}{DP} = \frac{AL}{CD} \quad (AB = CD) \\ & \text{(iii) } \frac{LP}{LD} = \frac{PB}{AD} \quad (\triangle LPB \sim \triangle LDA) \\ & \Rightarrow \frac{2}{5} = \frac{PB}{BC} \quad (AD = BC)\end{aligned}$$

In $\triangle BMC$ and $\triangle EMD$
MC = MD
$\angle CMB = \angle EMD$
$\angle MBC = \angle MED$
$\therefore \triangle BMC \cong \triangle EMD$
$\Rightarrow$ BC = DE
But AD = BC
$\therefore$ AD = DE
$\Rightarrow$ AE = $2$ BC
In $\triangle AEL \sim \triangle CBL$
$\therefore \frac{EL}{BL} = \frac{AE}{BC}$
$\frac{EL}{BL} = \frac{2BC}{BC}$
$\frac{EL}{BL} = 2$
$\Rightarrow EL = 2 BL$

Draw $AL \perp BC$ and $DM \perp BC\\$In $\Delta AOL$ and $$\begin{aligned}& \Delta DOM, \\ & \angle AOL = \angle DOM\end{aligned}$$ (vertically opposite angles)
$\angle ALO = \angle DMO$ (each $90^\circ$)
$\Delta AOL \sim \Delta DOM$ (AA similarity)
$\Rightarrow \frac{AL}{DM} = \frac{AO}{DO}$
dots (i)
$$\begin{aligned}& \frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta DBC)} = \frac{\frac{1}{2} \times BC \times AL}{\frac{1}{2} \times BC \times DM} \\ & = \frac{AL}{DM} = \frac{AO}{DO}\end{aligned}$$ [using (i)]

MP $||$ AB
$\Rightarrow \triangle QMP \sim \triangle QAB$
$\Rightarrow \frac{MP}{AB} = \frac{QP}{QB}$ ... (i)
Now, NO $||$ MP $||$ AB
$\Rightarrow \triangle CNO \sim \triangle CAB$
$\Rightarrow \frac{NO}{AB} = \frac{CO}{CB}$ ...(ii)
As MP = NO
From (i) and (ii),
$\frac{QP}{QB} = \frac{CO}{CB}$
$\frac{QB}{QP} = \frac{CB}{CO}$
$\frac{QB}{QP} - 1 = \frac{CB}{CO} - 1$
$\frac{QB-QP}{QP} = \frac{CB-CO}{CO}$
$\frac{BP}{QP} = \frac{BO}{CO}$
or $\frac{QP}{BP} = \frac{CO}{BO}$
$\therefore$ QC $||$ PO

$\frac{NE}{EO} = \frac{3}{4} \Rightarrow \frac{NE}{NO} = \frac{3}{7}$
XE $||$ PO
Therefore, $\frac{NX}{NP} = \frac{NE}{NO} = \frac{XE}{PO} = \frac{3}{7}$
$\because$ PO = $2$ MN
$\frac{XE}{MN} = \frac{6}{7}$ --- (1)
Also, $\frac{NX}{NP} = \frac{XP}{NP} = \frac{3}{7}$
Now, FX $||$ MN
$\frac{XP}{NP} = \frac{FX}{MN} = \frac{4}{7}$ --- (2)
Using (1) and (2),
$\frac{XE}{MN} + \frac{XF}{MN} = \frac{6}{7} + \frac{4}{7}$
$\frac{EF}{MN} = \frac{10}{7}$
or $7$ FE = $10$ MN

(c) $25$ m

BD = speed $\times$ time = $1.2 \times 4 = 4.8 \text{ m}$ (1 Mark)
$\triangle ABE \sim \triangle CDE$ (1/2 Mark)
$\frac{BE}{DE} = \frac{AB}{CD}$ (1/2 Mark)
$\frac{4.8 + DE}{DE} = \frac{3.6}{0.9}$ (1 Mark)
$\Rightarrow DE = 1.6 \text{ m}$ (1/2 Mark)
$\therefore$ Shadow of the girl after walking for 4 seconds is 1.6 m long

(b) $\frac{1}{2}ab$

(D) $5\sqrt{3}$ cm