E and F are points on the sides AB and AC respectively of a $\triangle ABC$ such that $\frac{AE}{EB} = \frac{AF}{FC} = \frac{1}{2}$. Which of the following relation is true ?
E and F are points on the sides PQ and PR respectively of a $\triangle PQR$. If EF $||$ QR and PE = $4$ cm, QE = $3$ cm and EF = $4$ cm, then the length of QR is :
In the figure, X and Y are two points on the sides AB and AC respectively in $\triangle ABC$, such that $AX = 3.4$ cm, $AB = 8.5$ cm, $AY = 2.6$ cm and $YC = 3.9$ cm. Which of the following relation is correct ?
In $\triangle XYZ$, $XY = 6$ cm. If M and N are two points on XY and XZ respectively such that $MN \parallel YZ$ and $XN = \frac{1}{4} XZ$, then the length of XM is :
In a $\triangle ABC$, a line DE is drawn parallel to BC to intersect AB at D and AC at E. If AD = $2$ cm, BD = $3$ cm and DE = $4$ cm, then the length of BC (in cm) is :
In the given figure, in $\triangle ABC$, $DE \parallel BC$. If $AD = 2.4 \text{ cm}$, $DB = 4 \text{ cm}$ and $AE = 2 \text{ cm}$, then the length of $AC$ is :
A line $l$ intersects the sides PQ and PR of a $\triangle PQR$ at L and M respectively such that LM $||$ QR. If PL = $5.7$ cm, PQ = $15.2$ cm and MR = $5.5$ cm, then the length of PM (in cm) is :
ABCD is a trapezium with AB $||$ DC and the diagonals intersect at O. If AO = $(2x + 1)$ cm, OC = $(5x – 7)$ cm, DO = $(7x – 5)$ cm and OB = $(7x + 1)$ cm, then the value of $x$ is :
E and F are points on the sides AB and AC respectively of a $\Delta ABC$ such that $\frac{AE}{EB} = \frac{AF}{FC} = \frac{1}{2}$. Which of the following relation is true ?
E and F are points on the sides PQ and PR respectively of a $\triangle PQR$. If PE = $3.9$ cm, EQ = $3$ cm, PF = $3.6$ cm and PR = $6$ cm, find whether EF $||$ QR.
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Sol. 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 $\frac{PE}{EQ} \neq \frac{PF}{FR}$, therefore EF $||$ QR.
In the adjoining figure, $AP = 1$ cm, $BP = 2$ cm, $AQ = 1.5$ cm and $AC = 4.5$ cm. Prove that $\Delta APQ \sim \Delta ABC$. Hence find the length of $PQ$, if $BC = 3.6$ cm.
If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, prove that the other two sides are divided in the same ratio.
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Correct Given, to prove, figure, construction Correct proof
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
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Correct given, to prove, figure and construction Correct Proof
In trapezium PQRS, $PQ \parallel SR$ and $SR = 2 PQ$. A line segment FE drawn parallel to PQ intersects PS at F and QR at E such that $\frac{QE}{ER} = \frac{3}{4}$. Diagonal QS intersects FE at G. Prove that $\frac{FE}{PQ} = \frac{10}{7}$.
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GE $\parallel$ PQ and PQ $\parallel$ SR $\Rightarrow$ GE $\parallel$ SR $\therefore \triangle QGE \sim \triangle QSR$ $\frac{GE}{SR} = \frac{QE}{QR} = \frac{3}{7}$ Given, $SR = 2 PQ$ $\Rightarrow \frac{GE}{PQ} = \frac{6}{7}$ ... (i) Now, FG $\parallel$ PQ $\therefore \triangle SFG \sim \triangle SPQ$ $\Rightarrow \frac{FG}{PQ} = \frac{SF}{SP}$ Also, $\frac{SF}{SP} = \frac{RE}{RQ} = \frac{4}{7}$ $\Rightarrow \frac{FG}{PQ} = \frac{4}{7}$ ... (ii) Adding (i) and (ii), $\frac{FE}{PQ} = \frac{6}{7} + \frac{4}{7} = \frac{10}{7}$
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
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Sol. Correct figure, given, to prove and construction Correct proof
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
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Correct Given, to prove, figure, construction Correct proof
If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points then it divides the two sides in the same ratio. Prove it. Also, state the converse of the above statement.
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Correct figure, given, to prove, construction Correct proof Correct statement of converse of given statement
State the converse of basic proportionality theorem. Also find $\frac{BF}{FC}$ in the following figure, given that $AB || DC || EF$ and $\frac{AE}{ED} = \frac{2}{3}$. Also, find the length of EF if $AB = 10$ cm and $DC = 15$ cm.
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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
State the basic proportionality theorem. Use the theorem to do the following : In $\Delta ABC$, AD is the angle bisector of angle A. BA is produced to E such that CE $||$ AD. Prove that $\frac{BD}{DC} = \frac{BA}{AC}$.
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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), $\frac{BD}{DC} = \frac{BA}{AC}$
If a line drawn parallel to one side of triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to third side. State and prove the converse of the above statement.
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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.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.
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Correct Given, To prove and construction Correct proof
The perimeters of two similar triangles are $42$ cm and $35$ cm respectively. If one side of the first triangle is $12$ cm, then the corresponding side of the second triangle is :
If $\triangle PQR \sim \triangle ABC$; $PQ = 6$ cm, $AB = 8$ cm and the perimeter of $\triangle ABC$ is $36$ cm, then the perimeter of $\triangle PQR$ is
In $\triangle ABC$, $DE \parallel BC$ (as shown in the figure). If $AD = 2 \text{ cm}$, $BD = 3 \text{ cm}$, $BC = 7.5 \text{ cm}$, then the length of $DE$ (in cm) is:
The perimeters of two similar triangles are $25$ cm and $15$ cm respectively. If one side of the first triangle is $9$ cm, then the length of the corresponding side of the second triangle is :
If in two triangles $\triangle DEF$ and $\triangle PQR$, $\angle D = \angle Q$ and $\angle R = \angle E$, then which of the following is textbf{not} true ?
In the given figure, in $\triangle ABC$, $AD \perp BC$ and $\angle BAC = 90^{\circ}$. If $BC = 16 \operatorname{cm}$ and $DC = 4 \operatorname{cm}$, then the value of $x$ is :
In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD $\perp$ BC and EF $\perp$ AC, prove that $\triangle ABD \sim \triangle ECF$.
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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 a $\triangle ABC$, D and E are points on the sides AB and AC respectively such that BD = CE. If $\angle B = \angle C$, then show that DE $||$ BC.
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In $\triangle ABC$, $$\begin{aligned}& \angle B = \angle C \\ & \Rightarrow AC = AB \dots (1) \\ & Given, BD = CE \dots (2) \\ & Subtract (2) from (1), we have \\ & AD = AE \dots (3) \\ & (3) \div (2)\end{aligned}$$, we have $$\begin{aligned}& \frac{AD}{BD} = \frac{AE}{CE} \\ & Therefore, DE \parallel BC\end{aligned}$$.
In the given figure, Z is a point on the side BC of $\triangle ABC$ such that XZ $||$ AB and YZ $||$ AC. If XY and CB produced meet at O, then prove that $ZO^2 = OB \times OC$.
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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 the given figure, $\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}}$ and $\angle 1 = \angle 2$, show that $\triangle$ PQS $\sim \triangle$ TQR.
A $1.5$ m tall boy is walking away from the base of a lamp post which is $12$ m high, at the speed of $2.5$ m/sec. Find the length of his shadow after $3$ seconds.
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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
AD and PS are angle bisectors of $\angle A$ and $\angle P$ of triangles ABC and PQR. If $\Delta ABC \sim \Delta PQR$, prove that $\Delta ACD \sim \Delta PRS$.
In $\Delta ABC$ and $\Delta PQR$, AD and PS are altitudes such that $\Delta ABD \sim \Delta PQS$ and $\Delta ACD \sim \Delta PRS$. Prove that $\Delta ABC \sim \Delta PQR$.
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Correct figure $\Delta ABD \sim \Delta PQS \implies \angle B = \angle Q$ --- (1) $\Delta ACD \sim \Delta PRS \implies \angle C = \angle R$ --- (2) From (1) & (2), we get $\Delta ABC \sim \Delta PQR$
Sides AB and BC and the median AD of a triangle AВС are respectively proportional to the sides PQ and QR and the median PM of $\triangle PQR$. Show that $\triangle ABC \sim \triangle PQR$.
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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)
In the given figure, $CD$ is the perpendicular bisector of $AB$. $EF$ is perpendicular to $CD$. $AE$ intersects $CD$ at $G$. Prove that $\frac{CF}{CD} = \frac{FG}{DG}$.
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$\triangle EFG \sim \triangle ADG$ $\Rightarrow \frac{EF}{AD} = \frac{FG}{DG}$ 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 the given figure, $E$ is a point on the side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, then prove that $\triangle ABD \sim \triangle ECF$.
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$ABC$ is an isosceles triangle $\therefore AB = AC \Rightarrow \angle B = \angle C$ In $\triangle ABD$ and $\triangle ECF$, $\angle ADB = \angle EFC$ $\angle ABD = \angle ECF$ $\therefore \triangle ABD \sim \triangle ECF$
PA, QB and RC are each perpendicular to AC. If AP = $x$, QB = $z$, RC = $Y$, AB = $a$ and BC = $b$, then prove that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$
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(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) $\frac{z}{x} + \frac{z}{y} = \frac{a + b}{a + b} = 1$ $\Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{z}$
In the given figure, CD and RS are respectively the medians of $\triangle ABC$ and $\triangle PQR$. If $\triangle ABC \sim \triangle PQR$ then prove that: (i) $\triangle ADC\sim\triangle PSR$ (ii) $AD \times PR = AC \times PS$
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(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 $\frac{AD}{PS} = \frac{AC}{PR}$ $\Rightarrow AD \times PR = AC \times PS$
Case Study – 2 On a road leading to a school, there is a triangle (ABC) shaped board on the road. It is divided into two parts by a line DE, which is parallel to BC. On the upper part, it is written ‘DRIVE SLOW' and on the lower part it is written, ‘SCHOOL AHEAD'. Based on the information given above, answer the following questions : (i) If AD = $3$ cm, BD = $5$ cm and AE = $4$ cm, then find the length of AC. (ii) If $\angle$ ADE = $50^{\circ}$ and $\angle$ DAE = $45^{\circ}$, then find $\angle$ ACB. (iii) (a) If AD = $4$ cm and BD = $6$ cm, then find $\frac{DE}{BC}$. OR (iii) (b) If AD = $3$ cm, BD = $6$ cm and AE = $5$ cm, then find $\frac{AB}{AC}$.
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(i) $\frac{3}{3+5} = \frac{4}{AC}$ $\Rightarrow AC = \frac{32}{3}$ cm (ii) $\angle ACB = \angle AED = 180^{\circ} – (50^{\circ} + 45^{\circ}) = 85^{\circ}$ (iii) (a) $\triangle ADE \sim \triangle ABC$ $\therefore \frac{DE}{BC} = \frac{AD}{AB} = \frac{4}{10}$ or $\frac{2}{5}$ OR (b) DE $||$ BC $\therefore \frac{3}{6} = \frac{5}{EC} \Rightarrow EC = 10$ $\frac{AB}{AC} = \frac{3+6}{5+10} = \frac{9}{15}$ or $\frac{3}{5}$
Sides $AB$ and $AC$ and median $AM$ of a $\triangle ABC$ are proportional to sides $DE$ and $DF$ and median $DN$ of another $\triangle DEF$. Show that $\triangle ABC \sim \triangle DEF$
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1 mark for figure Extend $AM$ to $A'$ so that $AM = A'M$ and $DN$ to $D'$ so that $DN = D'N$. Join $A'C$ and $D'F$. $$\begin{aligned}& \triangle AMB \sim \triangle A'MC \\ & \Rightarrow AB = A'C\end{aligned}$$. Similarly, $$\begin{aligned}& DE = D'F \\ & \text{Given } \frac{AB}{DE} = \frac{AC}{DF} = \frac{AM}{DN} \\ & \Rightarrow \frac{AC}{DF} = \frac{A'C}{D'F} = \frac{AA'/2}{DD'/2} \\ & \therefore \triangle AA'C \sim \triangle DD'F \\ & \therefore \angle 1 = \angle 2 \\ & \text{Similarly, } \angle 3 = \angle 4 \\ & \Rightarrow \angle 1 + \angle 3 = \angle A = \angle 2 + \angle 4 = \angle D \\ & \text{Hence } A ABC \sim \triangle DEF \text{ (SAS)}\end{aligned}$$
Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\triangle PQR$. Show that $\triangle ABC \sim \triangle PQR$.
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In $\triangle ABC$ and $\triangle PQR$ $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$ $\frac{AB}{PQ} = \frac{2 BD}{2 QM} = \frac{AD}{PM}$ ($\therefore$ D is midpoint of BC and M is midpoint of QR) $\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM} \Rightarrow \triangle ABD \sim \triangle PQM$ $\Rightarrow \angle B = \angle Q$ -(i) Now, In $\triangle ABC$ and $\triangle PQR$ $\frac{AB}{PQ} = \frac{BC}{QR}$ (given) $\angle B = \angle Q$ from (i) $\therefore \triangle ABC \sim \triangle PQR$
D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$, prove that $CA^2 = CB.CD$
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Sol. 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 $\frac{CB}{CA} = \frac{CA}{CD} \Rightarrow CA^2 = CB. CD$
In the given figure, two medians PD and QE of $\triangle PQR$ meet each other at O. Prove that : (i) $\triangle POQ \sim \triangle DOE$ (ii) $PO = 2OD$ (iii) $PO = \frac{2}{3} PD$
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(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), $PO = 2 OD = 2(PD – PO)$ $\Rightarrow 3PO = 2PD$ $\Rightarrow PO = \frac{2}{3} PD$
In the given figure PA, QB and RC are each perpendicular to AC. If AP = $x$, BQ = $y$ and CR = $z$, then prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$.
Sides $AB$ and $AC$ and median $AD$ to $\triangle ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \triangle PQR$.
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Correct figure Produce $AD$ to $E$ such that $AD = DE$ and join $EC$ Produce $PM$ to $N$ such that $PM = MN$ and join $NR$ $\triangle ADB \cong \triangle EDC$ $\therefore AB = EC$ Similarly, $PQ=NR$ Since, $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$ $\Rightarrow \frac{EC}{NR} = \frac{AC}{PR} = \frac{AE}{PN}$ $\therefore \triangle AEC \sim \triangle PNR$ $\Rightarrow \angle 1 = \angle 2$ Similarly, $\angle 3 = \angle 4$ Hence $\angle 1 + \angle 3 = \angle 2 + \angle 4$ or $\angle A = \angle P$ Also, $\frac{AB}{PQ} = \frac{AC}{PR}$ $\therefore \triangle ABC \sim \triangle PQR$
Sides $AB$, $BC$ and the median $AD$ of $\triangle ABC$ are respectively proportional to sides $PQ$, $QR$ and the median $PM$ of another $\triangle PQR$. Prove that $\triangle ABC \sim \triangle PQR$.
Sides AB, BC and the median AD of $\triangle ABC$ are respectively proportional to sides PQ, QR and the median PM of another $\triangle PQR$. Prove that $\triangle ABC \sim \triangle PQR$.
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 $\triangle ABC \sim \triangle PQR$.
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Correct figure Produce AD to E such that $AD = DE$ and join EC. Produce PM to L such that $PM = ML$ and join LR. $\therefore \triangle ABD \cong \triangle ECD$ $\therefore AB = EC$ Similarly, $PQ = LR$ $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$ $\frac{EC}{LR} = \frac{AC}{PR} = \frac{2AD}{2PM} = \frac{AE}{PL}$ $$\begin{aligned}& \therefore \triangle AEC \sim \triangle PLR \\ & \Rightarrow \angle 2 = \angle 4\end{aligned}$$ Similarly, $\angle 1 = \angle 3$ Adding both, $\angle BAC = \angle QPR$ $\therefore \triangle ABC \sim \triangle PQR$
The diagonal $BD$ of a parallelogram $ABCD$ intersects the line segment $AE$ at the point $F$, where $E$ is any point on the side $BC$. Prove that $DF \times EF = FB \times FA$.
Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points divides the other two sides in the same ratio. Hence, in the figure given below, prove that $\frac{AM}{MB} = \frac{AN}{ND}$ where LM $||$ CB and LN $||$ CD.
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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 $\frac{AM}{MB} = \frac{AN}{ND}$
In the given figure, PA, QB and RC are perpendicular to AC. If $PA = x$ units, $QB = y$ units and $RC = z$ units, prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$.
Sides AB and BC and median AD of triangle ABC are respectively proportional to sides PQ and QR and median PM of $\triangle PQR$. Show that $\triangle ABC \sim \triangle PQR$.
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Sol. Correct figure In $\triangle ABD$ and $\triangle PQM$ $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$ (given) $\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM}$ $\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$ $\therefore \triangle ABD \sim \triangle PQM$ $\therefore \angle B = \angle Q$ In $\triangle ABC$ and $\triangle PQR$ $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$ $\triangle ABC \sim \triangle PQR$
The corresponding sides of $\triangle ABC$ and $\triangle PQR$ are in the ratio $3 : 5$. AD$\perp$BC and PS$\perp$QR as shown in the following figures : (i) Prove that $\triangle ADC \sim \triangle PSR$ (ii) If $AD = 4$ cm, find the length of PS. (iii) Using (ii) find ar ($\triangle ABC$) : ar ($\triangle PQR$)
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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 ($\triangle PQR$) = $9 : 25$
State basic proportionality theorem. Use it to prove the following : If three parallel lines $l, m, n$ are intersected by transversals $q$ and $s$ as shown in the adjoining figure, then $\frac{AB}{BC} = \frac{DE}{EF}$.
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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 $\frac{AB}{BC} = \frac{DE}{EF}$
The corresponding sides of $\Delta ABC$ and $\Delta PQR$ are in the ratio $3 : 5$. $AD \perp BC$ and $PS \perp QR$ as shown in the following figures : (i) Prove that $\Delta ADC \sim \Delta PSR$ (ii) If $AD = 4$ cm, find the length of $PS$. (iii) Using (ii) find $ar (\Delta ABC) : ar (\Delta PQR)$
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As, $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{3}{5}$ $\implies \Delta ABC \sim \Delta PQR \implies \angle C = \angle R$ (i) In $\Delta ADC$ and $\Delta PSR, \angle ADC = \angle PSR = 90^{\circ}$ and $\angle C = \angle R \implies \Delta ADC \sim \Delta PSR$ (ii) $\frac{AD}{PS} = \frac{AC}{PR} = \frac{3}{5} \implies \frac{4}{PS} = \frac{3}{5} \implies PS = \frac{20}{3}$ cm (iii) $\frac{ar (\Delta ABC)}{ar (\Delta PQR)} = \frac{\frac{1}{2} \times BC \times AD}{\frac{1}{2} \times QR \times PS} = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \implies ar (\Delta ABC) : ar (\Delta PQR) = 9 : 25$
In the adjoining figure, $\Delta CAB$ is a right triangle, right angled at $A$ and $AD \perp BC$. Prove that $\Delta ADB \sim \Delta CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of $AD$.
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$\Delta ABC \sim \Delta DAC$ (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).
Assertion (A): ABCD is a trapezium with $DC \parallel AB$. E and F are points on AD and BC respectively, such that $EF \parallel AB$. Then $\frac{AE}{ED} = \frac{BF}{FC}$. Reason (R): Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Questions number $19$ and $20$ are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A): $ABCD$ is a trapezium with $DC \parallel AB$. $E$ and $F$ are points on $AD$ and $BC$ respectively, such that $EF \parallel AB$. Then $\frac{AE}{ED} = \frac{BF}{FC}$ Reason (R): Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.
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(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
In the given figure, $ABCD$ is a parallelogram. $AE$ divides the line segment $BD$ in the ratio $1 : 2$. If $BE = 1.5$cm, then find the length of $BC$.
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$$\begin{aligned}& \triangle OBE \sim \triangle ODA \\ & \frac{OB}{OD} = \frac{BE}{AD} \\ & \Rightarrow \frac{1}{2} = \frac{BE}{BC}\end{aligned}$$ (AD = BC) $BE = 1.5$ cm $\Rightarrow BC = 3$ cm
PQRS is a trapezium with PQ $||$ SR. If M and N are two points on the non-parallel sides PS and QR respectively, such that MN is parallel to PQ, then show that $\frac{PM}{MS} = \frac{QN}{NR}$.
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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), $\frac{PM}{MS} = \frac{QN}{NR}$
In the given figure, ABCD is a quadrilateral. Diagonal BD bisects $\angle B$ and $\angle D$ both. Prove that : (i) $\triangle ABD \sim \triangle CBD$ (ii) $AB = BC$
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Sol. (i) In $\triangle ABD \& \triangle CBD$ $\angle 3 = \angle 4$ $\angle 1 = \angle 2$ $\therefore \triangle ABD \sim \triangle CBD$ (ii) $\triangle ABD \cong \triangle CBD$ $\therefore AB = BC$
ABCD is a trapezium in which AB $||$DC and its diagonals AC and BD intersect at O. Show that $\frac{OA}{OB} = \frac{OC}{OD}$.
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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: $\frac{DO}{OB} = \frac{AO}{OC}$ $\Rightarrow \frac{OA}{OB} = \frac{OC}{OD}$
Prove that each of the four triangles formed by joining the mid-points of the sides of a triangle are similar to the original triangle.
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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.
$ABCD$ is a parallelogram, $P$ is a point on side $BC$ and $DP$ when produced meets $AB$ produced at $L$. Prove that (i) $\frac{DP}{PL} = \frac{DC}{BL}$ (ii) $\frac{DL}{DP} = \frac{AL}{DC}$ (iii) If $LP : PD = 2 : 3$ then find $BP : BC$
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD (produced) in E. Prove that EL = 2BL.
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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 $\triangle AEL \sim \triangle CBL$ $\therefore \frac{EL}{BL} = \frac{AE}{BC}$ $\frac{EL}{BL} = \frac{2BC}{BC}$ $\frac{EL}{BL} = 2$ $\Rightarrow EL = 2 BL$
In the given figure, $\Delta ABC$ and ADBC are on the same base BC. If AD intersects BC at O, prove that $\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta DBC)} = \frac{AO}{DO}$.
Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2BL.
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Correct fig. In $\triangle BLC$ and $\triangle ELA$ $\frac{BL}{EL} = \frac{BC}{EA} \dots(i)$ DM $||$ AB $\therefore$ AD = DE $\Rightarrow AE = 2AD$ $\Rightarrow AE = 2BC \dots(ii)$ From (i) and (ii) $EL=2BL$
In the figure, MNOP is a trapezium with, MN $||$ PO and PO = $2$ MN. A line segment FE drawn parallel to MN intersects MP at F and NO at E such that $\frac{NE}{EO} = \frac{3}{4}$. Diagonal PN intersects FE at X. Prove that $7$ FE = $10$ MN.
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$\frac{NE}{EO} = \frac{3}{4} \Rightarrow \frac{NE}{NO} = \frac{3}{7}$ 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
A vertical pole $10$ m long casts a shadow of length $5$ m on the ground. At the same time, a tower casts a shadow of length $12.5$ m on the ground. The height of the tower is:
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