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Maths 3 Marks Question

P-2 Similarity · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 68 questions.

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Question 13 Marks
Prove The Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.
Answer
Given : In $\triangle \mathrm{ABC}$ line $l \|$ line $\mathrm{BC}$ and line $l$ intersects $\mathrm{AB}$ and $A C$ in point $P$ and $Q$ respectively To prove : $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$

Image

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Question 23 Marks
In figure 1.74, PM = 10 cm A(Δ PQS) = 100 sq.cm A (ΔQRS) = 110 sq.cm then find NR.
Answer
We know that, Area of triangle $=\frac{1}{2} \times$ base $\times$ height
$\Rightarrow \frac{ A (\triangle PQS )}{ A (\Delta QRS )}=\frac{\frac{1}{2} \times QS \times PM }{\frac{1}{2} \times QS \times NR } $
$\Rightarrow \frac{100}{110}=\frac{ PM }{ NR } $
$\Rightarrow \frac{100}{110}=\frac{10}{ NR } $
$\Rightarrow NR =11 cm$
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Question 33 Marks
In Δ ABC, B - D – C and BD = 7, BC = 20 then find following ratios.

$(1)\frac{ A (\Delta ABD )}{ A (\Delta ADC )}$
$(2)\frac{ A (\Delta ABD )}{ A (\Delta ABC )}$
$(3)\frac{ A (\Delta ADC )}{ A (\Delta ABC )}$
Answer
Theorem: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.(1) $\frac{ A (\triangle ABD )}{ A (\triangle ADC )}=\frac{ BD ^2}{ DC ^2}$
$=\frac{ BD ^2}{( BC - BD )^2}$
$=\frac{7^2}{(20-7)^2}$
$=\frac{7^2}{13^2}$
$(2) \frac{ A (\Delta ABD )}{ A (\Delta ABC )}=\frac{ BD ^2}{ BC ^2}$
$=\frac{ BD ^2}{ BC ^2}$
$=\frac{7^2}{20^2}$
$(3) \frac{ A (\Delta ADC )}{ A (\Delta ABC )}=\frac{ DC ^2}{ BC ^2}$
$=\frac{\left( BC ^2- BD \right)^2}{ BC ^2}$
$=\frac{(20-7)^2}{20^2}$
$=\frac{13^2}{20^2}$
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Question 43 Marks
Δ ABC and ΔDEF are equilateral triangles. If A(ΔABC) : A(ΔDEF) = 1 : 2 and AB = 4, find DE.
Answer
We know that, all the angles of an equilateral triangles are equal, i.e., 60°.⇒ Δ ABC~Δ DEF ……(By AAA Similarity Test)
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle DEF )}=\frac{ AB ^2}{ DE ^2}$
And, $\frac{ A (\triangle ABC )}{ A (\triangle DEF )}=\frac{1}{2}$ (Given)
$\Rightarrow \frac{ AB ^2}{ DE ^2}=\frac{1}{2}$
$\Rightarrow D E^2=2 \times 4^2(\because A B=4)$
$\Rightarrow D E=\sqrt{ } 32$
$\Rightarrow D E=4 \sqrt{ } 2$
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Question 53 Marks
ΔLMN ~ ΔPQR, 9 × A(ΔPQR) = 16 × A (ΔLMN). If QR = 20 then find MN.
Answer
∵ Δ ABC~Δ PQR⇒ Given that, 9× A(Δ ABC) = 16× A(Δ PQR)
$\Rightarrow \frac{ A (\Delta PQR )}{ A (\Delta LMN )}=\frac{16}{9}$
And, $\frac{ A (\triangle PQR )}{ A (\Delta LMN )}=\frac{ QR ^2}{ MN ^2}$
$\Rightarrow \frac{ QR ^2}{ MN ^2}=\frac{16}{9} $
$\Rightarrow \frac{20^2}{ MN ^2}=\frac{16}{9} $
$\Rightarrow MN ^2=\frac{400 \times 9}{16}$
⇒ MN = 15
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Question 63 Marks
Given : In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ
Answer
Given that, $A R=5 A P$ and $A S=5 A Q$
$\Rightarrow \frac{ AR }{ AP }=5\dots(1)$
And $\frac{ AS }{ AQ }=5\dots(2)$.
$\Rightarrow \frac{ AR }{ AP }=\frac{ AS }{ AQ }$
And, $\angle SAR \cong \angle QAP$ $\qquad$ (opposite angles)
$\Rightarrow \triangle SAR \sim \triangle QAP$ $\qquad$ (SAS Test of similarity)
$\Rightarrow \frac{ AS }{ AQ }=\frac{ AR }{ AP }=\frac{ SR }{ QP }$ (corresponding sides are proportional)
But, $\frac{ AS }{ AQ }=\frac{ AR }{ AP }=5$
$\Rightarrow \frac{ SR }{ QP }=5$
$\Rightarrow S R=5 P Q$
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Question 73 Marks
Are the triangles in figure 1.56 similar? If yes, by which test?
Answer
In $\triangle P Q R$ and $\triangle L M N \frac{P Q}{L M}=\frac{6}{3}=2$
And $\frac{ QR }{ MN }=\frac{8}{4}=2$
And $\frac{ PR }{ LN }=\frac{10}{5}=2$
$\Rightarrow \frac{P Q}{L M}=\frac{Q R}{M N}=\frac{P R}{L N}=2$
⇒ ΔPQR~ΔLMN …………(By SSS Similarity Test)
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Question 83 Marks
In figure 1.55, ∠ABC = 75°, ∠EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.
Answer
With one- to-one correspondence ABC ↔ EDC∵ ∠ABC ≅ ∠EDC = 75°
∠ACB ≅ ∠ECD (Is common in both the triangles ABC and EDC)

⇒ Δ ABC~Δ EDC ………(By AA Test)
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Question 93 Marks
In Δ PQR, PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.
Answer
$\Rightarrow$ If $\frac{ PN }{ NR }=\frac{ PM }{ MQ ^{\prime}}$, then line NM is parallel to side $R Q$.
$\therefore$ we'll check if $\frac{ PN }{ NR }=\frac{ PM }{ MQ }$.
$\Rightarrow \frac{ PN }{ NR }=\frac{ PR - NR }{ NR }$
$=\frac{20-8}{8}$
$=\frac{12}{8}$
$=\frac{3}{2}$ 
And, $\frac{P M}{M Q}=\frac{P M}{P Q-P M}$
$=\frac{15}{25-15}$
$=\frac{15}{10}$
$=\frac{3}{2}$
$\Rightarrow \frac{ PN }{ NR }=\frac{ PM }{ MQ }=\frac{3}{2}$, therefore line $NM \|$ side RQ
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Question 103 Marks
In the figure 1.44, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

Proof: IN $\triangle XDE , PQ \| DE$ $\square$
$\therefore \frac{ XP }{\square}=\frac{\square}{ QE }$
(Basic proportionality theorem)
In $\triangle XDE , QR \| EF......... \square$
$\therefore$ $\frac{\square}{\square}=\frac{\square}{\square}...............(ii)\square$
$\therefore$ $\frac{\square}{\square}=\frac{\square}{\square}...............$ from (i) and (ii)
$\therefore \text { seg PR\|seg DE .......... }$
(converse of basic proportionality theorem)
Answer
Proof: In $\triangle X D E, P Q|| D E$..... (Given)
$\therefore \frac{ XP }{ XQ }=\frac{ PD }{ DE }$
(Basic proportionality theorem)
In $\triangle X D E, Q R \| E F$ $\qquad$ (Given)
$\therefore \frac{ XR }{ RF }=\frac{ XQ }{ QE }$ $\qquad$ (II) (Basic Proportionality Theorem)
$\therefore \frac{ XP }{ PD }=\frac{ XR }{ RF }$ $\qquad$ from (I) and (II)
$\therefore$ seg PR\|Seg DE $\qquad$
(converse of basic proportionality theorem)
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Question 113 Marks
Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠OPR.
Answer
Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.Therefore, we’ll find the ratio for all the triangle.Hence, for
$\text { (1) } \frac{Q M}{M R}=\frac{3.5}{1.5} $
$=2.33$
And $\frac{ QP }{ PR }=\frac{7}{3}$
$=2.33 $
$\Rightarrow \frac{ QM }{ MR }=\frac{ QP }{ PR }$
$\Rightarrow \ln (1)$, ray PM is a bisector.
$\text { (2) } \frac{ RM }{ MQ }=\frac{6}{8} $
$=0.75$
And $\frac{ RP }{ PQ }=\frac{7}{10}$
$=0.7 $
$\Rightarrow \frac{ RM }{ MQ } \neq \frac{ RP }{ PQ }$
$\Rightarrow \ln (2)$, ray PM is not a bisector.
$\text { (3) } \frac{ RM }{ MQ }=\frac{4}{3.6} $
$=1.1$
And $\frac{R P}{P Q}=\frac{10}{9}$
$=1.11$
$\Rightarrow \frac{ RM }{ MQ }=\frac{ RP }{ PQ }$
$\Rightarrow \ln (3)$, ray PM is a bisector.
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Question 123 Marks
In fig 1.80, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.


Activity : $2 AX =3 BX \therefore \frac{ AX }{ BX }=$ $\frac{ \square }{ \square }$
$\frac{ AX + BX }{ BX }=\frac{\square+\square}{\square}$ $\qquad$ by componendo.
$\frac{ AB }{ BX }=\frac{\square}{\square}=(1)$
$\Delta BCA \sim \Delta BYX \cdots \cdots \cdots \square\text { test of similarity. }$
$\therefore \frac{ BA }{ BX }=\frac{ AC }{ XY } \cdots \cdots \cdots . . \text { corresponding sides of similar triangles. }$
$\therefore \frac{\square}{\square \square}=\frac{ AC }{9} \therefore AC =\square . . . \text { from (l) }$
Answer
ACTIVITY: $2 A X=3 B X \therefore \frac{A X}{B X}=\frac{3 A X+B X}{2}=\frac{3+2}{2} \ldots \ldots$ (By Componendo) $\frac{ AB }{ BX }=\frac{5}{3} \ldots \ldots .(1)$
$\triangle B C A \sim \triangle B Y X \ldots \ldots$. (AA test of similarity).
$\therefore \frac{ BA }{ BX }=\frac{ AC }{ XY } \ldots \ldots \ldots$ (corresponding sides of similar triangles).
$\frac{5}{3}=\frac{A C}{9} \therefore A C=15 \ldots \ldots \text { from }(1)$
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Question 133 Marks
In adjoining figure PQ ⊥ BC, AD ⊥ BC then find following ratios.

(i) $\frac{ A (\Delta PQB )}{ A (\triangle PBC )}$
(ii) $\frac{ A (\Delta PBC )}{ A (\Delta ABC )}$
(iii) $\frac{ A (\Delta ABC )}{ A (\triangle ADC )}$
(iv) $\frac{ A (\Delta ADC )}{ A (\Delta PQC )}$
Answer
We know that area of triangle $=\frac{1}{2} \times$ Base $\times \operatorname{Height}( i ) \frac{ A (\triangle PQB )}{ A (\triangle PBC )}=\frac{ BQ }{ BC }$
(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)
(ii) $\frac{A(\triangle P B C)}{A(\triangle ABC )}=\frac{P Q}{A D}$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
(iii) $\frac{A(\triangle A B C)}{A(\triangle A D C)}=\frac{B C}{D C}$
(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)
(iv) $\frac{A(\triangle A D C)}{A(\triangle P Q C)}=\frac{\frac{1}{2} \times A D \times D C}{\frac{1}{2} \times P Q \times Q C}$
$=\frac{A D \times D C}{P Q \times Q C}$
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Question 143 Marks
In adjoining figure, AP ⊥ BC, AD || BC, then find A(ΔABC) : A (ΔBCD)
Answer

We can re-draw the fig.1.15(as shown above) where we add DO which will be height of $\triangle B C D$.
Now, $\frac{ A (\triangle ABC )}{ A (\triangle BCD )}=\frac{ AP }{ DO }$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
$\Rightarrow \frac{A(\triangle ABC )}{ A ( BBCD )}=\frac{ AP }{ DO } $
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle BCD )}=\frac{1}{1}$
( $\because$ the distance between the two parallel lines is always equal $\Rightarrow A P=D O$ )
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle BCD )}=1: 1$
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Question 153 Marks
In adjoining figure 1.14 seg PS ⊥ seg RQ , seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.
Answer

Considering, Area of (ΔPQR) with base QR⇒ PS will be the Height
Now, consider the Area of (ΔPQR) with base PR
⇒ QT will be the Height
∵ , the triangle is the same

⇒ the area will be the same irrespective of the base taken.
And we know that area of triangle = $\frac{1}{2}$ × Base× Height
⇒ $\frac{1}{2}$×QR×PS
= $\frac{1}{2}$×PR×QT
⇒ $\frac{1}{2}$×6×6
= $\frac{1}{2}$×12×QT
⇒ QT = 3
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Question 183 Marks
In △ ABC point △ on side BC is such that △C = 6, BC = 15. Find A(△ ABD) : A(△ ABC) and A(△ ABD) : A(△ ADC) .
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Question 193 Marks
In $\triangle ABC , PQ$ is a line segment intesecting $A B$ at point $P$ and $A C$ at point $Q$. $P Q \| B C$. If $P Q$ divides $\triangle A B C$ into two equal parts equal in area, find BP : AB.
Answer
$\frac{\sqrt{2}-1}{\sqrt{2}}$
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Question 203 Marks
In the adjoining figure, seg DE || side BC. If DE: BC=3:5, then find $A (\triangle ADE ): A (\triangle DBCE )$  
Image
Answer
$9: 16$
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Question 213 Marks
In $\Delta$RES, RE = 15, SE = 10. In $\Delta$PEA, PE = 8, AE = 12. Prove that $\Delta$RES ~ $\Delta$AEP
Image
Answer
self
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Question 223 Marks
$D$ is a point on side $B C$ of $\triangle A B C$ such that, $\angle A D C=\angle B A C$. Show that $A C^2=B C \times D C$.
Answer
self
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Question 233 Marks
E is a point on side CB, C-B-E, In $\triangle A B C$ AB = AC. If seg AD BC, B-D-C and seg EF $\perp$ side AC, A-F-C. Prove that $\triangle ABD \sim \triangle ECF$. 
Image
Answer
self
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Question 243 Marks
In $\triangle A B C, \angle B=90^{\circ}$, seg $DE \perp$ side AC . $A D=6, A B=12$, $A C=18$, then find $A E$.
Image
Answer
4 units
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Question 253 Marks
In $\triangle ABC , AB =5, BC =6, AC =7 . \Delta PQR \sim \triangle ABC$. Perimeter of $\triangle P Q R$ is 360 . Find $P Q, Q R$ and $P R$.
Answer
100 units, 120 units, 140 units
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Question 263 Marks
In the adjoining figure, seg PA, seg QB, seg RC and seg SD are $\perp$ to line $l$ AB = 6, BC = 9, CD = 12 and PS = 36, then find PQ, QR and RS.
Image
Answer
$PQ =8, QR =12, RS =16$
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Question 283 Marks
In $\Delta$PQR, ray QS bisects $\angle$PQR. P-S-R. Show that $\frac{ A (\Delta PQS )}{ A (\Delta QRS )}=\frac{ PQ }{ QR }$.
Image
Answer
self
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Question 293 Marks
Point D and E are the points on sides AB and AC such athat AB = 5.6, AD = 1.4, AC = 7.2 and AE = 1.8. Show that DE || BC.
Image
Answer
self
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Question 303 Marks
$\square ABCD$ is a trapezium in which $AB \| DC$ and its diagonals intersect each other at point $O$. Show that $A O: B O=C O: D O$.
Answer
self
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Question 323 Marks
In the adjoining figure, seg DH $\perp$ seg EF, seg GK $\perp$ seg EF. If DH = 12 cm, GK = 20 cm and A(ADEF) = 300°cm², then find (i) EF (ii) A( $\triangle GEF)$
Image
Answer
(i) 50 cm (ii) $500 cm^2$
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Question 333 Marks
In the adjoining figure, $R P: P K=3: 2$, then find the value of
(i) $A (\triangle T R P): A (\triangle T P K)$
(ii) $A (\triangle T R K): A (\triangle TPK )$
(iii) $A (\Delta T R P)$ : A( $\triangle T R K$
Image
Answer
(i) $3: 2$ (ii) $5: 2$ (iii) $3: 5$
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Question 343 Marks
Prove The Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.
Answer
Given : In $\triangle \mathrm{ABC}$ line $l \|$ line $\mathrm{BC}$ and line $l$ intersects $\mathrm{AB}$ and $A C$ in point $P$ and $Q$ respectively To prove : $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$

Image

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Question 353 Marks
Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof.
i. The areas of two triangles of equal height are proportional to their bases.
ii. Every point on the bisector of an angle is equidistant from the sides of the angle.
Image
Given: In $\triangle CAB$,ray AD bisects ∠A.
To prove: $\frac{A B}{A C}=\frac{B D}{D C}$
Construction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Answer
Proof:
In ∆ABC,
Point D is on angle bisector of ∠A. [Given]
∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]
$\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B \times D M}{A C \times D N}$ [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
$\therefore \frac{A(\triangle A B D)}{A(\triangle A C D)}=\frac{A B}{A C}$(ii) [From (i)]
Also, ∆ABD and ∆ACD have equal height.
$\therefore \frac{ A (\triangle ABD )}{ A (\triangle ACD )}=\frac{ BD }{ CD }$(iii) [Triangles having equal height]
$\therefore \frac{ AB }{ AC }=\frac{ BD }{ DC }$ [From (ii) and (iii)]
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Question 363 Marks
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.
AB = ⬜ cm, BC = ⬜ cm,
AD = ⬜ cm, DC = ⬜ cm
iv. Find ratios $\frac{A B}{B C}$ and $\frac{A D}{D C}$
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above.
Verify that the ratios are equal.
Image
$\frac{ AB }{ BC }=\frac{4}{4}=1$$\quad$$\quad$(i)
$\frac{ AD }{ DC }=\frac{2}{2}=1$$\quad$$\quad$(ii)
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$...[From (i) and (ii)]
Answer
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.
AB = 4 cm, BC = 4 cm,
AD = 2 cm, DC = 2 cm
iv. Find ratios $\frac{A B}{B C}$ and $\frac{A D}{D C}$
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above.
Verify that the ratios are equal.
Image
$\frac{ AB }{ BC }=\frac{4}{4}=1$$\quad$$\quad$(i)
$\frac{ AD }{ DC }=\frac{2}{2}=1$$\quad$$\quad$(ii)
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$...[From (i) and (ii)]
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Question 373 Marks
In the adjoining figure, BP ⊥ AC, CQ ⊥ AB, A – P – C, A – Q – B, then prove that ∆APB and ∆AQC are similar.
Image
Answer
In $\triangle APB$ and $\triangle AQC$,
$\angle APB =90^{\circ}$...(1)
$\angle AQC =90^{\circ}$ ...(2)
$\therefore \quad \angle APB \cong \angle AQC$ [From (i) and (ii)]
$\angle PAB \cong \angle QAC$ [Common angle]
$\therefore \quad \triangle APB \sim \triangle AQC$ [AA test of similarity]
2. SAS test for similarity of triangles:
For a given correspondence, if two pairs of corresponding sides are in the same proportion and the angle between them is congruent, then the two triangles are similar.
ImageIn the given figure, if $\frac{A B}{P Q}=\frac{B C}{Q R}$, and $\angle B \cong \angle Q$, then $\triangle ABC \sim \triangle PQR$
3. SSS test for similarity of triangles:
For a given correspondence, if three sides of one triangle are in proportion with the corresponding three sides of the another triangle, then the two triangles are similar.
Image
In the given figure, if $\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R^{\prime}}$ then $\triangle ABC \sim \triangle PQR$
Properties of similar triangles:
1. Reflexivity : $\triangle ABC \sim \triangle ABC$
2. If $\triangle ABC \sim \triangle DEF$, then $\triangle DEF \sim \triangle ABC$.
3. Transitivity : If $\triangle ABC \sim \triangle DEF$ and $\triangle DEF \sim \triangle GHI$, then $\triangle ABC \sim$ΔGHI.
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Question 383 Marks
i. Draw three parallel lines.
ii. Label them as l, m, n.
iii. Draw transversals $t_1$ and $t_2$.
Iv. AB and BC are intercepts on transversal $t _1$.
v. PQ and QR are intercepts on transversal $t _2$.
vi. Find ratios $\frac{A B}{B C}$ and $\frac{P Q}{Q R}$. ou will find that they are almost equal. Verify that they are equal.
Answer

Image
Here, $AB =1.5 cm, BC =2.1 cm$,
$PQ =1.7 cm, QR =2.3 cm$
$\frac{ AB }{ BC }=\frac{1.5}{2.1}=0.714 \approx 0.7$
$\frac{ PQ }{ QR }=\frac{1.7}{2.3}=0.739 \approx 0.7$
∴$\frac{ AB }{ BC }=\frac{ PQ }{ QR }$
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Question 393 Marks
In the adjoining figure, seg PA, seg QB, seg RC and seg SD are $\perp$ to line $l$ AB = 6, BC = 9, CD = 12 and PS = 36, then find PQ, QR and RS.
Image
Answer
seg PA, seg QB, seg RC and seg SD are perpendicular to line AD.
∴ seg PA || seg QB || seg RC || seg SD (i) [Lines perpendicular to the same line are parallel to each other]
Let the value of PQ be x and that of QR be y.
PS = PQ + QS [P – Q – S]
∴ 280 – x + QS
∴ QS = 280 – x (ii)
Now, seg PA || seg QB || seg SD [From (i)]
$\therefore \frac{A B}{B D}=\frac{P Q}{Q S}$[Property of three parallel lines and their transversals]
$\therefore \frac{A B}{B C+C D}=\frac{P Q}{Q S}[B- C - D ]$
$\therefore \frac{60}{70+80}=\frac{x}{280-x}$
$\therefore \frac{60}{150}=\frac{x}{280-x}$
$\therefore \frac{2}{5}=\frac{x}{280-x}$
∴ 5x = 2 (280 – x)
∴ 5x = 560 – 2x
∴ 7x = 560
$\therefore x=\frac{560}{7}=80$
∴ PQ = 80 units
QS = 280 – x [From (ii)]
= 280 – 80
= 200 units
But, QS = QR + RS [Q – R – S]
∴ 200 = y + RS
∴ RS = 200 – y (ii)
Now, seg QB || seg RC || seg SD [From (i)]
$\therefore \frac{B C}{C D}=\frac{Q R}{R S}$ [Property of three parallel lines and their transversals]
$\therefore \frac{70}{80}=\frac{y}{200-y}$
$\therefore \frac{7}{8}=\frac{y}{200-y}$
$\therefore 8 y=7(200-y)$
$\therefore 8 y=1400-7 y$
$\therefore 15 y=1400$
$\therefore y=\frac{1400}{15}=\frac{280}{3}$
$\therefore QR =\frac{280}{3}$ units
RS = 200 – 7 [From (iii)]
$=200-\frac{280}{3}$
$=\frac{200 \times 3-280}{3}$
$=\frac{600-280}{3}$
$\therefore RS =\frac{320}{3}$units
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Question 403 Marks
In Δ ABC, B - D – C and BD = 7, BC = 20 then find following ratios.
Image
(1) $\frac{ A (\triangle ABD )}{ A (\triangle ADC )}$
(2) $\frac{ A (\triangle ABD )}{ A (\triangle ABC )}$
(3) $\frac{ A (\triangle ADC )}{ A (\triangle ABC )}$
Answer
Theorem: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.
(1) $\frac{ A (\triangle ABD )}{ A (\triangle ADC )}=\frac{ BD ^2}{ DC ^2}$
$\begin{array}{l}
=\frac{ BD ^2}{( BC - BD )^2} \\
=\frac{7^2}{(20-7)^2} \\
=\frac{7^2}{13^2} \\
\text { (2) } \frac{ A (\triangle ABD )}{ A (\triangle ABC )}=\frac{ BD ^2}{ BC ^2}
\end{array}$
$\begin{aligned}
= & \frac{ BD ^2}{ BC ^2} \\
= & \frac{7^2}{20^2}
\end{aligned}$
(3) $\frac{ A (\triangle ADC )}{ A (\triangle ABC )}=\frac{ DC ^2}{ BC ^2}$
$\begin{array}{l}
=\frac{( BC - BD )^2}{ BC ^2} \\
=\frac{(20-7)^2}{20^2} \\
=\frac{13^2}{20^2}
\end{array}$
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Question 423 Marks
Δ ABC and ΔDEF are equilateral triangles. If A(ΔABC) : A(ΔDEF) = 1 : 2 and AB = 4, find DE.
Answer
We know that, all the angles of an equilateral triangles are equal, i.e., $60^{\circ}$.
$\begin{array}{l}
\Rightarrow \triangle ABC \sim \triangle DEF \\
\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle DEF )}=\frac{ AB ^2}{ DE ^2} \\
\text { And, } \frac{ A (\triangle ABC )}{ A (\triangle DEF )}=\frac{1}{2} \text { (Given) } \\
\Rightarrow \frac{ AB ^2}{ DE ^2}=\frac{1}{2} \\
\Rightarrow D E^2=2 \times 4^2(\because A B=4) \\
\Rightarrow D E=\sqrt{32} \\
\Rightarrow D E=4 \sqrt{2} \\
\end{array}$
(By AAA Similarity Test)
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Question 433 Marks
ΔLMN ~ ΔPQR, 9 × A(ΔPQR) = 16 × A (ΔLMN). If QR = 20 then find MN.
Answer

$\begin{array}{l}\because \triangle ABC \sim \Delta PQR \\ \Rightarrow \text { Given that, } 9 \times A (\triangle ABC )=16 \times A (\triangle PQR ) \\ \Rightarrow \frac{ A (\triangle PQR )}{ A (\Delta LMN )}=\frac{16}{9} \\ \text { And, } \frac{ A (\Delta PQR )}{ A (\Delta LMN )}=\frac{ QR ^2}{ MN ^2} \\ \Rightarrow \frac{ QR ^2}{ MN ^2}=\frac{16}{9} \\ \Rightarrow \frac{20^2}{ MN ^2}=\frac{16}{9} \\ \Rightarrow MN ^2=\frac{400 \times 9}{16} \\ \Rightarrow MN ^2=15\end{array}$
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Question 443 Marks
Answer
Given that, $A R=5 A P$ and $A S=5 A Q$
$\Rightarrow \frac{ AR }{ AP }=5...(1)$
And $\frac{ AS }{ AQ }=5$...(2)
$\Rightarrow \frac{ AR }{ AP }=\frac{ AS }{ AQ }$
And, $\angle SAR \cong \angle QAP$ $\qquad$ (opposite angles)
$\Rightarrow \triangle SAR \sim \triangle QAP$ $\qquad$ (SAS Test of similarity) $\Rightarrow \frac{ AS }{ AQ }=\frac{ AR }{ AP }=\frac{ SR }{ QP }$ (corresponding sides are proportional)
But, $\frac{ AS }{ AQ }=\frac{ AR }{ AP }=5$
$\Rightarrow \frac{ SR }{ QP }=5$
$\Rightarrow S R=5 P Q$
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Question 473 Marks
Answer
By Converse of basic Proportionality Theorem
(Theorem: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.)
$\Rightarrow$ If $\frac{ PN }{ NR }=\frac{ PM }{ MQ }$, then line $N M$ is parallel to side $R Q$.
$\begin{array}{l}
\therefore \text { we'll check if } \frac{ PN }{ NR } \\
\Rightarrow \frac{ PN }{ NR }=\frac{ PR - NR }{ NR } \\
=\frac{20-8}{8} \\
=\frac{12}{8} \\
=\frac{3}{2}
\end{array}$
And, $\frac{ PM }{ MQ }=\frac{ PM }{ PQ - PM }$
$\begin{array}{l}
\vDash \frac{15}{25-15} \\
=\frac{\frac{15}{10}}{} \\
=\frac{3}{2} \\
\Rightarrow \frac{ PN }{ NR }=\frac{ PM }{ MQ }=\frac{3}{2} \text {, therefore line } NM \| \text { side } RQ
\end{array}$
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Question 483 Marks
Answer
Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.
Therefore, we'll find the ratio for all the triangle. Hence, for
$\begin{array}{l}
\text { (1) } \frac{Q M}{M R}=\frac{3.5}{1.5} \\
=2.33
\end{array}$
And $\frac{ QP }{ PR }=\frac{7}{3}$
$\begin{array}{l}
=2.33 \\
\Rightarrow \frac{ QM }{ MR }=\frac{ QP }{ PR }
\end{array}$
$\Rightarrow \operatorname{In}$ (1), ray PM is a bisector.
(2) $\frac{ RM }{ MQ }=\frac{6}{8}$
$=0.75$
And $\frac{ RP }{ PQ }=\frac{7}{10}$
$\begin{array}{l}
=0.7 \\
\Rightarrow \frac{ RM }{ MQ } \neq \frac{ RP }{ PQ }
\end{array}$
$\Rightarrow$ In (2), ray PM is not a bisector.
$\begin{array}{l}\text { (3) } \frac{R M}{M Q}=\frac{4}{3.6} \\ =1.1 \\ \text { And } \frac{R P}{P Q}=\frac{10}{9} \\ =1.11 \\ \Rightarrow \frac{R M}{M Q}=\frac{R P}{P Q} \\ \Rightarrow \operatorname{In}(3), \text { ray } P M \text { is a bisector. }\end{array}$
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Question 493 Marks
In adjoining figure PQ ⊥ BC, AD ⊥ BC then find following ratios.
Image
(i) $\frac{ A (\triangle PQB )}{ A (\triangle PBC )}$
(ii) $\frac{ A (\triangle PBC )}{ A (\triangle ABC )}$
(iii) $\frac{ A (\triangle ABC )}{ A (\triangle ADC )}$
(iv) $\frac{ A (\triangle ADC )}{ A (\triangle PQC )}$
Answer
We know that area of triangle $=\frac{1}{2} \times$ Base $\times$ Height
$\text { (i) } \frac{ A (\triangle PQB )}{ A (\triangle PBC )}=\frac{ BQ }{ BC }$
(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)
(ii) $\frac{ A (\triangle PBC )}{ A (\triangle ABC )}=\frac{ PQ }{ AD }$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
(iii) $\frac{ A (\triangle ABC )}{ A (\triangle ADC )}=\frac{ BC }{ DC }$
(PROPERTY:Areas of triangles with equal heights are proportional to their corresponding bases.)
$\begin{array}{l}
\text { (iv) } \frac{ A (\triangle ADC )}{ A (\triangle PQC )}=\frac{\frac{1}{2} \times A D \times D C}{\frac{1}{2} \times P Q \times Q C} \\
=\frac{ AD \times DC }{ PQ \times QC } \\
\end{array}$
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Question 503 Marks
Answer

Image
We can re-draw the fig.1.15(as shown above) where we add DO
which will be height of $\triangle B C D$.
Now, $\frac{ A (\triangle ABC )}{ A (\triangle BCD )}=\frac{ AP }{ DO }$
(PROPERTY: Areas of triangles with equal bases are proportional to their corresponding heights.)
$\begin{aligned}
\Rightarrow & \frac{A(\triangle A B C)}{A(\triangle B C D)}=\frac{A P}{D O} \\
\Rightarrow & \frac{A(\triangle A B C)}{A(\triangle B C D)}=\frac{1}{1}
\end{aligned}$
( $\because$ the distance between the two parallel lines is always equal $\Rightarrow A P=D O)$
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle BCD )}=1: 1$
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Question 513 Marks
Answer

Image
Considering, Area of $(\triangle P Q R)$ with base $Q R$
$\Rightarrow PS$ will be the Height
Now, consider the Area of $(\triangle P Q R)$ with base $P R$
$\Rightarrow$ QT will be the Height
$\because$, the triangle is the same
$\Rightarrow$ the area will be the same irrespective of the base taken.
And we know that area of triangle $=\frac{1}{2} \times$ Base $\times$ Height
$\begin{aligned}
& \frac{1}{2} \times Q R \times P S \\
= & \frac{1}{2} \times P R \times Q T
\end{aligned}$
$\begin{aligned}
& \frac{1}{2} \times 6 \times 6 \\
= & \frac{1}{2} \times 12 \times Q T \\
\Rightarrow & Q T=3
\end{aligned}$
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Question 523 Marks

Image
Answer
In $\triangle A B C$ point $P$ and $D$ are on side $A C$, hence $B$ is common vertex of $\triangle ABD , \triangle BDC , \triangle ABC$ and $\triangle APB$ and their sides $AD , DC , AC$ and $AP$ are collinear. Heights of all the triangles are equal. Hence, areas of these triangles are proportinal to their bases. $AC =16, DC =9$
$\therefore A D=16-9=7$
$\therefore \frac{ A (\triangle ABD )}{ A (\triangle ABC )}=\frac{ AD }{ AC }=\frac{7}{16} \ldots \ldots$. triangles having equal heights
$\frac{ A (\triangle BDC )}{ A (\triangle ABC )}=\frac{ DC }{ AC }=\frac{9}{16} \ldots \ldots$. triangles having equal heights
$\frac{ A (\Delta ABD )}{ A (\Delta BDC )}=\frac{ AD }{ DC }=\frac{7}{9} \ldots \ldots$. triangles having equal heights
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Question 533 Marks
In △ ABC point △ on side BC is such that △C = 6, BC = 15. Find A(△ ABD) : A(△ ABC) and A(△ ABD) : A(△ ADC) .
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Question 543 Marks

Image
Answer
$ABCD$ is a parallelogram.
$\therefore AD \| BC$ and $AB \| DC$
Consider $\triangle ABC$ and $\triangle BDC$.
Both the triangles are drawn in two parallel lines. Hence the distance between the two parallel lines is the height of both triangles.
In $\triangle ABC$ and $\triangle BDC$, common base is $BC$ and heights are equal.
Hence, $A (\triangle ABC )= A (\triangle BDC )$
In $\triangle A B C$ and $\triangle A B D, A B$ is common base and and heights are equal.
$\therefore A (\triangle ABC )= A (\triangle ABD )$
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Question 553 Marks
$\square ABCD$ is a trapezium in which $AB \| DC$ and its diagonals intersect each other at point $O$. Show that $A O: B O=C O: D O$.
Answer
Since $A B \| D C$,
$\angle A O B=\angle C O D$ (vertically opposite angles),
$\angle A B O=\angle C D O$((alternate interior angles, as $A B \| D C)$.
Therefore, triangles $\triangle A O B$ and $\triangle C O D$ are similar by the AA similarity criterion.
Hence, the corresponding sides of similar triangles are proportional.
So, $\frac{A O}{B O}=\frac{C O}{D O}$
Therefore,
$A O: B O=C O: D O$
Hence, the given statement is proved.
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Question 563 Marks
Point D and E are the points on sides AB and AC such athat AB = 5.6, AD = 1.4, AC = 7.2 and AE = 1.8. Show that DE || BC.
Image
Answer
In triangle $\triangle A B C$,
$\frac{A D}{A B}=\frac{1.4}{5.6}=\frac{1}{4}$
Also,
$\frac{A E}{A C}=\frac{1.8}{7.2}=\frac{1}{4}$
Therefore,
$\frac{A D}{A B}=\frac{A E}{A C}$
Thus, point D divides side AB and point E divides side AC in the same ratio.
By the Converse of the Basic Proportionality Theorem, the line joining such points is parallel to the third side of the triangle.
Hence, $D E \| B C$.
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Question 573 Marks
In $\triangle ABC , PQ$ is a line segment intesecting $A B$ at point $P$ and $A C$ at point $Q$. $P Q \| B C$. If $P Q$ divides $\triangle A B C$ into two equal parts equal in area, find BP : AB.
Answer
$\frac{\sqrt{2}-1}{\sqrt{2}}$
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Question 583 Marks
In $\triangle A B C, \angle B=90^{\circ}$, seg $DE \perp$ side AC . $A D=6, A B=12$, $A C=18$, then find $A E$.
Image
Answer
4 units
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Question 593 Marks
In $\triangle ABC , AB =5, BC =6, AC =7 . \Delta PQR \sim \triangle ABC$. Perimeter of $\triangle P Q R$ is 360 . Find $P Q, Q R$ and $P R$.
Answer
100 units, 120 units, 140 units
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Question 603 Marks
In the adjoining figure, seg PQ || seg AB. Seg PR || seg BD. Prove that QR || AD.
Image
Answer
Since segment $P Q \| A B$, the line through point Q is parallel to side AB.
Also, since segment $P R \| B D$, the line through point R is parallel to side BD.
Quadrilateral ABDC is formed by joining AB and BD to point D.
Through point Q, a line is drawn parallel to AB, and through point R, a line is drawn parallel to BD.
When two lines are drawn parallel to two sides of a triangle, the line joining their corresponding points is parallel to the third side.
Hence, the line joining points Q and R, that is segment QR, is parallel to segment AD.
Conclusion:
Therefore,$Q R \| A D$
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Question 613 Marks
In the adjoining figure, seg ML || seg BC, seg NL || seg DC. Prove that AM : AB = AN : AD.
Image
Answer
Since segment $M L \| B C$, in triangle $\triangle A B C$, by the Basic Proportionality Theorem, the parallel line divides the sides proportionally.
Therefore, $\frac{A M}{A B}=\frac{A L}{A C}$ .....(1)
Similarly, since segment $N L \| D C$, in triangle $\triangle A D C$, by the Basic Proportionality Theorem, we get:
$\frac{A N}{A D}=\frac{A L}{A C}$ ...(2)
From equations (1) and (2),
$\frac{A M}{A B}=\frac{A N}{A D}$
Hence,
$\frac{A M}{A B}=\frac{A N}{A D}$
Therefore, the given statement is proved.
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Question 623 Marks
In the adjoining figure, seg DH $\perp$ seg EF, seg GK $\perp$ seg EF. If DH = 12 cm, GK = 20 cm and A(ADEF) = 300°cm², then find (i) EF (ii) A( $\triangle GEF)$
Image
Answer
(i) 50 cm (ii) $500 cm^2$
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Question 633 Marks
In the adjoining figure, seg DE || side BC. If DE: BC=3:5, then find $A (\triangle ADE ): A (\triangle DBCE )$  
Image
Answer
$9: 16$
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Question 643 Marks
In the adjoining figure, $R P: P K=3: 2$, then find the value of
(i) $A (\triangle T R P): A (\triangle T P K)$
(ii) $A (\triangle T R K): A (\triangle TPK )$
(iii) $A (\Delta T R P)$ : A( $\triangle T R K$
Image
Answer
(i) $3: 2$ (ii) $5: 2$ (iii) $3: 5$
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Question 653 Marks
In $\Delta$RES, RE = 15, SE = 10. In $\Delta$PEA, PE = 8, AE = 12. Prove that $\Delta$RES ~ $\Delta$AEP
Image
Answer
In $\triangle R E S$ and $\triangle A E P$,
$\frac{R E}{S E}=\frac{15}{10}=\frac{3}{2}$
Also, $\frac{A E}{P E}=\frac{12}{8}=\frac{3}{2}$
Therefore, $\frac{R E}{S E}=\frac{A E}{P E}$
Angle $\angle R E S$ and angle $\angle A E P$ are vertically opposite angles.
Hence,
$\angle R E S=\angle A E P$
Thus, two sides of $\triangle R E S$ are proportional to two sides of $\triangle A E P$. and the included angle between them is equal.
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Question 663 Marks
In $\Delta$PQR, ray QS bisects $\angle$PQR. P-S-R. Show that $\frac{ A (\Delta PQS )}{ A (\Delta QRS )}=\frac{ PQ }{ QR }$.
Image
Answer
Triangles $\triangle P Q S$ and $\triangle Q R S$ lie on the same base line PR and have the same altitude from point Q to line PR.
Hence, the ratio of the areas of triangles $s \triangle P Q S$ and $\triangle Q R S$ is equal to the ratio of their corresponding bases on line PR.
Therefore,
$\frac{A(\triangle P Q S)}{A(\triangle Q R S)}=\frac{P S}{S R}$
Since ray QS bisects $\angle P Q R$,by the Angle Bisector Theorem, the bisector divides the opposite side PR in the ratio of the sides containing the angle.
Thus,
$\frac{P S}{S R}=\frac{P Q}{Q R}$
Substituting this value in the area ratio, we get:
$\frac{A(\triangle P Q S)}{A(\triangle Q R S)}=\frac{P Q}{Q R}$
Conclusion:
Hence, $\frac{A(\triangle P Q S)}{A(\triangle Q R S)}=\frac{P Q}{Q R}$
Therefore, the given statement is proved.
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Question 673 Marks
E is a point on side CB, C-B-E, In $\triangle A B C$ AB = AC. If seg AD BC, B-D-C and seg EF $\perp$ side AC, A-F-C. Prove that $\triangle ABD \sim \triangle ECF$. 
Image
Answer
In $\triangle A B D$ and $\triangle E C F$,
$\angle A D B=90^{\circ} \quad($ since $A D \perp B C)$
$\therefore \angle A D B=\angle E F C$ ....(1)
Also, since $A B=A C, \triangle A B C$ is an isosceles triangle.
$\therefore \angle A B D=\angle B C A$ ....(2)
As points $E, B, C$ are collinear,
$\angle E C F=\angle B C A$ ....(3)
From (2) and (3),
$\angle A B D=\angle E C F$
Thus, in $\triangle A B D$ and $\triangle E C F$,
$\therefore \triangle A B D \sim \triangle E C F$( by AA similarity test)
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Question 683 Marks
$D$ is a point on side $B C$ of $\triangle A B C$ such that, $\angle A D C=\angle B A C$. Show that $A C^2=B C \times D C$.
Answer
To Prove: $A C^2=B C \times D C$
Consider $\triangle A D C$ and $\triangle B A C$.
$\angle A D C=\angle B A C$(Given)
Also,
$\angle A C D=\angle B C A \quad$ (Common angle)
Therefore,
$\triangle A D C \sim \triangle B A C$ (by AA similarity test)
Hence, the corresponding sides of similar triangles are proportional.
$\frac{A C}{B C}=\frac{D C}{A C}$
On cross-multiplying, we get
$A C^2=B C \times D C$
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