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

(b) $20$ cm

(a) $8$ cm

(d) $10$

(B) 2.25 cm

(C) $8$ cm

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

(D) $3.3$

(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) $45^\circ$

(d) $16$ cm (1 Mark)

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

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

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

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)

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 $\angle DAE = \angle BAC$
$\therefore \triangle ADE \sim \triangle ABC$ ($1/2$)

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

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)
$\frac{FG}{DG} = \frac{CF}{CD}$

(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), $PO = 2 OD = 2(PD – PO)$
$\Rightarrow 3PO = 2PD$
$\Rightarrow PO = \frac{2}{3} PD$

$\Delta ABC \sim \Delta DAC$ (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).

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)

$$\begin{aligned}& \triangle OBE \sim \triangle ODA \\ & \frac{OB}{OD} = \frac{BE}{AD} \\ & \Rightarrow \frac{1}{2} = \frac{BE}{BC}\end{aligned}$$ (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), $\frac{PM}{MS} = \frac{QN}{NR}$

$\triangle ALE \sim \triangle CLB$
$\Rightarrow \frac{AL}{CL} = \frac{EL}{BL}$
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)
$\frac{EL}{BL} = \frac{2CM}{CM}$
$\Rightarrow EL = 2BL$

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