Mathematics Part ii Solutions Solutions for Class 10 Maths Chapter 1 - Similarity

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Page No 5:

Question 1:

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.

Answer:

Let ABC and PQR be two right triangles with AB ⊥ BC and PQ ⊥ QR.
Given:
BC = 9, AB = 5, PQ = 6 and QR = 10.
$∴ \frac{A (△ ABC)}{A (△ PQR)} = \frac{AB \times BC}{PQ \times QR} = \frac{5 \times 9}{6 \times 10} = \frac{3}{4}$

Page No 6:

Question 2:

In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find $\frac{A (∆ ABC)}{A (∆ ADB)} .$

Answer:

Given:
BC = 4
AD = 8
$∴ \frac{A (△ ABC)}{A (△ ADB)} = \frac{AB \times BC}{AB \times AD} = \frac{BC}{AD} [∵ BC = 4 and AD = 8] = \frac{4}{8}$
$= \frac{1}{2}$

Page No 6:

Question 3:

In adjoining figure, seg PS ⊥ seg RQ seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then Find QT.

Answer:

Given:
PS ⊥ RQ
QT ⊥ PR
RQ = 6, PS = 6 and PR = 12
With base PR and height QT, $A (△ PQR) = \frac{1}{2} \times PR \times QT$
With base QR and height PS, $A (△ PQR) = \frac{1}{2} \times QR \times PS$
$∴ \frac{A (△ PQR)}{A (△ PQR)} = \frac{\frac{1}{2} \times PR \times QT}{\frac{1}{2} \times QR \times PS} \Rightarrow 1 = \frac{PR \times QT}{QR \times PS} \Rightarrow PR \times QT = QR \times PS$
$\Rightarrow QT = \frac{QR \times PS}{PR} = \frac{6 \times 6}{12} = 3$
Hence, the measure of side QT is 3 units.

Page No 6:

Question 4:

In adjoining figure, AP ⊥ BC, AD || BC, then Find A(âˆ†ABC) : A(âˆ†BCD).

Answer:

Given:
AP ⊥ BC
AD || BC
$∴ \frac{A (△ ABC)}{A (△ BCD)} = \frac{AP \times BC}{AP \times BC} = \frac{1}{1}$
Hence, the ratio of A(âˆ†ABC) and A(âˆ†BCD) is 1 : 1.

Page No 6:

Question 5:

In adjoining figure PQ ⊥ BC, AD⊥ BC then find following ratios.

(i) $\frac{A (∆ PQB)}{A (∆ PBC)}$

(ii) $\frac{A (∆ PBC)}{A (∆ ABC)}$

(iii) $\frac{A (∆ ABC)}{A (∆ ADC)}$

(iv) $\frac{A (∆ ADC)}{A (∆ PQC)}$

Answer:

(i)
$\frac{A (△ PQB)}{A (△ PBC)} = \frac{PQ \times BQ}{PQ \times BC} = \frac{BQ}{BC}$
(ii)
$\frac{A (△ PBC)}{A (△ ABC)} = \frac{PQ \times BC}{AD \times BC} = \frac{PQ}{AD}$
(iii)
$\frac{A (△ ABC)}{A (△ ADC)} = \frac{AD \times BC}{AD \times DC} = \frac{BC}{DC}$
(iv)
$\frac{A (△ ADC)}{A (△ PQC)} = \frac{AD \times DC}{PQ \times QC}$

Page No 13:

Question 1:

Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.
(1)

(2)

(3)

Answer:

(1)
$In △ QMP, \frac{QM}{QP} = \frac{3.5}{7} = \frac{1}{2}$
$In △ MRP, \frac{MR}{RP} = \frac{1.5}{3} = \frac{1}{2}$
$∴ \frac{QM}{QP} = \frac{MR}{RP}$
By converse of angle bisector theorem, ray PM is the bisector of ∠QPR.

(2)
$In △ QMP, \frac{QM}{QP} = \frac{8}{10} = \frac{4}{5}$
$In △ MRP, \frac{MR}{RP} = \frac{6}{7}$
$∴ \frac{QM}{QP} \neq \frac{MR}{RP}$
Therefore, ray PM is not the the bisector of ∠QPR.
(3)
$In △ QMP, \frac{QM}{QP} = \frac{3.6}{9} = \frac{2}{5}$
$In △ MRP, \frac{MR}{RP} = \frac{4}{10} = \frac{2}{5}$
$∴ \frac{QM}{QP} = \frac{MR}{RP}$
By converse of angle bisector theorem, ray PM is the bisector of ∠QPR.

â€‹

Page No 13:

Question 2:

In âˆ†PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Answer:

Given:
PM = 15,
PQ = 25,
PR = 20 and
NR = 8
Now, PN = PR − NR
= 20 − 8
= 12
Also, MQ = PQ − PM
= 25 − 15
= 10
$In △ PRQ, \frac{PR}{NR} = \frac{12}{8} = \frac{3}{2}$
$Also, \frac{PM}{MQ} = \frac{15}{10} = \frac{3}{2}$
$∴ \frac{PR}{NR} = \frac{PM}{MQ}$
By converse of basic proportionality theorem, NM is parallel to side RQ or NM || RQ.

Page No 14:

Question 3:

In âˆ†MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then Find QP.

Answer:

$In △ PNM, \frac{QM}{QP} = \frac{MN}{PN} (By angle bisector theorem) \Rightarrow \frac{2.5}{QP} = \frac{5}{7}$
$\Rightarrow QP = \frac{2.5 \times 7}{5} = 3.5$
Hence, the measure of QP is 3.5.

Page No 14:

Question 4:

Measures of some angles in the figure are given. Prove that $\frac{AP}{PB} = \frac{AQ}{QC}$

Answer:

Given:
∠APQ = 60^âˆ˜
∠ABC = 60^âˆ˜
Since, the corresponding angles ∠APQ and â€‹∠APC are equal.
Hence, line PQ || BC.
$In △ ABC, PQ ∥ B C \frac{AP}{PB} = \frac{AQ}{QC} (By Basic proportionality theorem)$

Page No 14:

Question 5:

In trapezium ABCD, side AB || side PQ || side âˆ†C, AP = 15, PD = 12, QC = 14, Find BQ.

Answer:

Construction: Join BD intersecting PQ at X.

In â–³ABD, PX || AB
$\frac{DP}{PA} = \frac{DX}{XB} . . . (1) (By Basic proportionality theorem)$
In â–³BDC, XQ||DC
$\frac{DX}{XB} = \frac{CQ}{QB} . . . (2) (By Basic proportionality theorem)$
From (1) and (2), we get
$\frac{DP}{PA} = \frac{CQ}{QB} \Rightarrow \frac{12}{15} = \frac{14}{QB} \Rightarrow QB = 17.5$

Page No 14:

Question 6:

Find QP using given information in the figure.

Answer:

$In △ PNM, \frac{QM}{QP} = \frac{MN}{PN} (By angle bisector theorem) \Rightarrow \frac{14}{QP} = \frac{25}{40}$
$\Rightarrow QP = \frac{14 \times 40}{25} = 22.4$
Hence, the measure of QP is 22.4.

Page No 14:

Question 7:

In the given figure, if AB || CD || FE then Find x and AE.

Answer:

Construction: Join AFintersecting CD at X.

In â–³ABF, DX || AB
$\frac{FD}{DB} = \frac{FX}{XA} . . . (1) (By Basic proportionality theorem)$
In â–³AEF, XC||FE
$\frac{FX}{XA} = \frac{EC}{CA} . . . (2) (By Basic proportionality theorem)$
From (1) and (2), we get
$\frac{FD}{DB} = \frac{EC}{CA} \Rightarrow \frac{4}{8} = \frac{x}{12} \Rightarrow x = 6$
Now, AE = AC + CE
= 12 + 6
= 18

Page No 15:

Question 8:

In âˆ†LMN, ray MT bisects ∠LMN If LM = 6, MN = 10, TN = 8, then Find LT.

Answer:

$In △ LNM, \frac{LT}{NT} = \frac{LM}{NM} (By angle bisector theorem) \Rightarrow \frac{LT}{8} = \frac{6}{10}$
$\Rightarrow LT = \frac{8 \times 6}{10} = 4.8$
Hence, the measure of LT is 4.8.

Page No 15:

Question 9:

In âˆ†ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.

Answer:

In â–³ABC, ∠ABD = ∠DBC
$\frac{AD}{DC} = \frac{AB}{CB} (By angle bisector theorem) \Rightarrow \frac{x - 2}{x + 2} = \frac{x}{x + 5} \Rightarrow x^{2} + 3 x - 10 = x^{2} + 2 x$
$\Rightarrow 3 x - 2 x = 10 \Rightarrow x = 10$

Page No 15:

Question 10:

In the given figure, 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.

Answer:

Given:
Seg PQ || seg DE
seg QR || seg EF
In â–³DXE, PQ || DE
$\frac{XP}{PD} = \frac{XQ}{QE} . . . (I) (By basic proportionality theorem)$
In â–³XEF, QR || EF ....Given
$∴ \frac{XQ}{QE} = \frac{XR}{RF} . . . . . (II) (By basic proportionality theorem)$
$∴ \frac{XP}{PD} = \frac{XR}{RF} From (I) and (II)$
∴ seg PR || seg DF (Converse of basic proportional theorem)

Page No 15:

Question 11:

In âˆ†ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.

Answer:

Given:
ray BD bisects ∠ABC
ray CE bisects ∠ACB.
seg AB ≅ seg AC

In â–³ABC, ∠ABD = ∠DBC
$\frac{AD}{DC} = \frac{AB}{BC} . . . (I) (By angle bisector theorem)$
In â–³ABC, ∠BCE = ∠ACE
$\frac{AE}{EB} = \frac{AC}{BC} . . . (II) (By angle bisector theorem)$
From (I) and (II)
$\frac{AD}{DC} = \frac{AE}{EB} (∵ seg AB ≅ seg AC)$
∴ â€‹ED || BC (Converse of basic proportional theorem)

Page No 21:

Question 1:

In the given figure, ∠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:

Given:
∠ABC = 75°, ∠EDC = 75°
Now, in â–³ABC and â–³EDC
∠ABC = ∠EDC = 75° (Given)
∠C = ∠C (Common)
By AA test of similarity
â–³ABC ∼ â–³EDC

Page No 21:

Question 2:

Are the triangles in the given figure similar? If yes, by which test ?

Answer:

Given:
PQ = 6
PR = 10
QR = 8
LM = 3
LN = 5
MN = 4
Now,
$\frac{PQ}{LM} = \frac{6}{3} = 2, \frac{QR}{MN} = \frac{8}{4} = 2, \frac{RP}{NL} = \frac{10}{5} = 2$
$∴ \frac{PQ}{LM} = \frac{QR}{MN} = \frac{RP}{NL}$
By SSS test of similarity
â–³PQR ∼ â–³LMN

Page No 21:

Question 3:

As shown in the given figure, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m then how long will be the shadow of the bigger pole at the same time ?

Answer:

Given:
PR = 4
RL = 6
AC = 8
In â–³PLR and â–³ABC
∠PRL = ∠ACB (Vertically opposite angles)
∠LPR = ∠BAC (Angles made by sunlight on top are congruent)
By AA test of similarity
â–³PLR ∼ â–³ABC
$∴ \frac{PR}{AC} = \frac{LR}{BC} (Corresponding sides are proportional) \Rightarrow \frac{4}{8} = \frac{6}{x} \Rightarrow x = 12$
Hence, the length of shadow of bigger pole due to sunlight is 12 m.

Page No 21:

Question 4:

In âˆ†ABC, AP ⊥ BC, BQ ⊥ AC B– P–C, A–Q – C then prove that, âˆ†CPA ~ âˆ†CQB. If AP = 7, BQ = 8, BC = 12 then Find AC.

Answer:

Given:
AP ⊥ BC
BQ ⊥ AC
To prove: âˆ†CPA ~ âˆ†CQB
Proof: In âˆ†CPA and âˆ†CQB
∠CPA = ∠CQB = 90^âˆ˜ (Given)
∠C = ∠C (Common)
By AA test of similarity
âˆ†CPA ~ âˆ†CQB
Hence proved.
$Now, \frac{AP}{BQ} = \frac{AC}{BC} (Corresponding sides are proportional) \Rightarrow AC = \frac{AP}{BQ} \times BC = \frac{7}{8} \times 12 = 10.5$ â€‹

Page No 22:

Question 5:

Given : In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ

Answer:

Given:
side PQ || side SR
AR = 5AP,
AS = 5AQ
To prove: SR = 5PQ
Proof: In âˆ†APQ and âˆ†ARS
∠PAQ = ∠RAS (Vertically Opposite angles)
∠PQA = ∠RSA (Alternate angles, side PQ || side SR and QS is a transversal line)
By AA test of similarity
âˆ†APQ ~ âˆ†ARS
$\frac{PQ}{SR} = \frac{AP}{AR} (Corresponding sides are proportional) \Rightarrow \frac{PQ}{SR} = \frac{1}{5} (AR = 5 AP) \Rightarrow SR = 5 PQ$ â€‹
Hence proved.

Page No 22:

Question 6:

In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD.

Answer:

Given:
side AB || side DC
AB = 20,
DC = 6,
OB = 15
In â–³COD and â–³AOB
∠COD = ∠AOB (Vertically opposite angles)
∠CDO= ∠ABO (Alternate angles, CD ||BA and BD is a transversal line)
By AA test of similarity
â–³COD ∼ â–³AOB
$∴ \frac{CD}{AB} = \frac{OD}{OB} (Corresponding sides are proportional) \Rightarrow \frac{6}{20} = \frac{OD}{15} \Rightarrow OD = 4.5$

Page No 22:

Question 7:

â—»ABCD is a parallelogram point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.

Answer:

Given: â—»ABCD is a parallelogram
To prove: DE × BE = CE × TE
Proof: In âˆ†BET and âˆ†CED
∠BET = ∠CED (Vertically opposite angles)
∠BTE = ∠CDE (Alternate angles, AT || CD and DT is a transversal line)
By AA test of similarity
âˆ†BET ∼ âˆ†CED
$∴ \frac{BE}{CE} = \frac{ET}{ED} (Corresponding sides are proportional) \Rightarrow BE \times ED = CE \times ET$
Hence proved.

Page No 22:

Question 8:

In the given figure, seg AC and seg BD intersect each other in point P and $\frac{AP}{CP} = \frac{BP}{DP}$ . Prove that, âˆ†ABP ~ âˆ†CDP

Answer:

Given: $\frac{AP}{CP} = \frac{BP}{DP}$
To prove: âˆ†ABP ~ âˆ†CDP
Proof: In âˆ†ABP and âˆ†DCP
$\frac{AP}{CP} = \frac{BP}{DP}$ (Given)
∠P = ∠P (Common)
By SAS test of similarity
$\frac{AP}{CP} = \frac{BP}{DP}$

Page No 22:

Question 9:

In the given figure, in âˆ†ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA² = CB × CD

Answer:

Given: ∠BAC = ∠ADC
To prove: CA² = CB × CD
Proof: In âˆ†ABC and âˆ†DAC
∠BAC = ∠ADC (Given)
∠C = ∠C (Common)
By AA test of similarity
âˆ†ABC ∼ âˆ†DAC
$∴ \frac{BC}{AC} = \frac{AC}{DC} (Corresponding sides are proportional) \Rightarrow {AC}^{2} = BC \times DC$
Hence proved.

Page No 25:

Question 1:

The ratio of corresponding sides of similar triangles is 3 : 5; then Find the ratio of their areas.

Answer:

Page No 25:

Question 2:

If âˆ†ABC ~ âˆ†PQR and AB : PQ = 2 : 3, then fill in the blanks.
$\frac{A (∆ ABC)}{A (∆ PQR)} = \frac{{AB}^{2}}{} = \frac{2^{2}}{3^{2}} = \frac{}{}$

Answer:

Given:
âˆ†ABC ~ âˆ†PQR
AB : PQ = 2 : 3
According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
$∴ \frac{A (∆ ABC)}{A (∆ PQR)} = \frac{{AB}^{2}}{{PQ}^{2}} = \frac{2^{2}}{3^{2}} = \frac{4}{9}$

Page No 25:

Question 3:

If âˆ†ABC ~ âˆ†PQR, A (âˆ†ABC) = 80, A (âˆ†PQR) = 125, then fill in the blanks.
$\frac{A (∆ ABC)}{A (∆ . . . .)} = \frac{80}{125} ∴ \frac{AB}{PQ} = \frac{}{}$

Answer:

Given:
âˆ†ABC ~ âˆ†PQR
A (âˆ†ABC) = 80
A (âˆ†PQR) = 125
According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
$∴ \frac{A (∆ ABC)}{A (∆ PQR)} = \frac{{AB}^{2}}{{PQ}^{2}} \Rightarrow \frac{80}{125} = \frac{{AB}^{2}}{{PQ}^{2}} \Rightarrow \frac{16}{25} = \frac{{AB}^{2}}{{PQ}^{2}}$
$\Rightarrow \frac{4^{2}}{5^{2}} = \frac{{AB}^{2}}{{PQ}^{2}} \Rightarrow \frac{AB}{PQ} = \frac{4}{5}$
Therefore, $\frac{A (∆ ABC)}{A (∆ PQR)} = \frac{80}{125} and \frac{AB}{PQ} = \frac{4}{5}$

Page No 25:

Question 4:

âˆ†LMN ~ âˆ†PQR, 9 × A (âˆ†PQR ) = 16 × A (âˆ†LMN). If QR = 20 then Find MN.

Answer:

Given:
âˆ†LMN ~ âˆ†PQR
9 × A (âˆ†PQR ) = 16 × A (âˆ†LMN)
Consider, 9 × A (âˆ†PQR ) = 16 × A (âˆ†LMN)
$\frac{A (∆ LMN)}{A (∆ PQR)} = \frac{9}{16} \Rightarrow \frac{{MN}^{2}}{{QR}^{2}} = \frac{3^{2}}{4^{2}} \Rightarrow \frac{MN}{QR} = \frac{3}{4}$
$\Rightarrow MN = \frac{3}{4} \times QR \Rightarrow MN = \frac{3}{4} \times 20 [∵ QR = 20] \Rightarrow MN = 15$

Page No 25:

Question 5:

Areas of two similar triangles are 225 sq.cm. 81 sq.cm. If a side of the smaller triangle is 12 cm, then Find corresponding side of the bigger triangle.

Answer:

According to theorem of areas of similar triangles "When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides".
$∴ \frac{Area of bigger triangle}{Area of smaller triangle} = \frac{225}{81} \Rightarrow \frac{{(Side of bigger triangle)}^{2}}{{(Side of smaller triangle)}^{2}} = \frac{15^{2}}{9^{2}} \Rightarrow \frac{Side of bigger triangle}{Side of smaller triangle} = \frac{15}{9}$ â€‹
$\Rightarrow Side of bigger triangle = \frac{15}{9} \times Side of smaller triangle \Rightarrow Side of bigger triangle = \frac{15}{9} \times 12 = 20$

Hence, the corresponding side of the bigger triangle is 20.

Page No 25:

Question 6:

âˆ†ABC and âˆ†DEF are equilateral triangles. If A(âˆ†ABC) : A (âˆ†DEF) = 1 : 2 and AB = 4, find DE.

Answer:

Consider, A(âˆ†ABC) : A (âˆ†DEF) = 1 : 2
$\Rightarrow \frac{A (∆ ABC)}{A (∆ DEF)} = \frac{1}{2} \Rightarrow \frac{{AB}^{2}}{{DE}^{2}} = \frac{1}{2} \Rightarrow {DE}^{2} = 2 {AB}^{2}$ â€‹
$\Rightarrow {DE}^{2} = 2 \times 4^{2} [∵ AB = 4] \Rightarrow DE = \sqrt{32} \Rightarrow DE = 4 \sqrt{2}$

Page No 25:

Question 7:

In the given figure 1.66, seg PQ || seg DE, A(âˆ†PQF) = 20 units, PF = 2 DP, then Find A(â—»DPQE) by completing the following activity.

Answer:

Given:
seg PQ || seg DE
A(âˆ†PQF) = 20 units
PF = 2 DP
Let us assume DP = x
∴ PF = 2x
$DF = DP + PF = x + 2 x = 3 x$
In â–³FDE and â–³FPQ
∠FDE = ∠FPQ (Corresponding angles)
∠FED = ∠FQP (Corresponding angles)
By AA test of similarity
â–³FDE ∼ â–³FPQ
$∴ \frac{A (△ FDE)}{A (△ FPQ)} = \frac{{FD}^{2}}{{FP}^{2}} = \frac{{(3 x)}^{2}}{{(2 x)}^{2}} = \frac{9}{4} A (△ FDE) = \frac{9}{4} A (△ FPQ) = \frac{9}{4} \times 20 = 45$
$∴ A (□ DPQE) = A (△ FDE) - A (△ FPQ) = 45 - 20 = 25$

Page No 26:

Question 1:

Select the appropriate alternative.
(1) In âˆ†ABC and âˆ†PQR, in a one to one correspondence $\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}$ then

(A) âˆ†PQR ~ âˆ†ABC
(B) âˆ†PQR ~ âˆ†CAB
(C) âˆ†CBA ~ âˆ†PQR
(D) âˆ†BCA ~ âˆ†PQR

(2) If in âˆ†DEF and âˆ†PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false?

(A)

\frac{EF}{PR} = \frac{DF}{PQ}

(B)

\frac{DE}{PQ} = \frac{EF}{RP}

(C)

\frac{DE}{QR} = \frac{DF}{PQ}

(D)

\frac{EF}{RP} = \frac{DE}{QR}

(3) In âˆ†ABC and âˆ†DEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the two triangles is true ?
(A)The triangles are not congruent and not similar
(B) The triangles are similar but not congruent.
(C) The triangles are congruent and similar.
(D) None of the statements above is true.

(4) âˆ†ABC and âˆ†DEF are equilateral triangles, A (âˆ†ABC) : A (âˆ†DEF) = 1 : 2
If AB = 4 then what is length of DE?
(A)

2 \sqrt{2}

(B) 4
(C) 8
(D)

4 \sqrt{2}

(5) In the given figure, seg XY || seg BC, then which of the following statements is true?

(A)

\frac{AB}{AC} = \frac{AX}{AY}

(B)

\frac{AX}{XB} = \frac{AY}{AC}

(C)

\frac{AX}{YC} = \frac{AY}{XB}

(D)

\frac{AB}{YC} = \frac{AC}{XB}

Answer:

(1)
Given: $\frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ}$
By SSS test of similarity
âˆ†PQR ~ âˆ†CAB
Hence, the correct option is (B).

(2)
In âˆ†DEF and âˆ†PQR
∠D ≅ ∠Q
∠R ≅ ∠E
By AA test of similarity
âˆ†DEF~ âˆ†PQR
$∴ \frac{DE}{PQ} = \frac{EF}{QR} = \frac{DF}{PR} (Corresponding sides of similar triangles are proportional)$
$∴ \frac{DE}{PQ} \neq \frac{EF}{RP}$
Hence, the correct option is (B).

(3)
In âˆ†ABC and âˆ†DEF
∠B = ∠E,
∠F = ∠C
By AA test of similarity
âˆ†ABC ~ âˆ†DEF
Since, there is not any congruency criteria like AA.
Thus, âˆ†ABC and âˆ†DEF are not congruent.
Hence, the correct option is (B).

(4)
Given: âˆ†ABC and âˆ†DEF are equilateral triangles
Constrcution: Draw a perependicular from vertex A and D on AC and DF in both triangles.

In âˆ†ABX and âˆ†DEY
∠B = ∠C = 60^âˆ˜ (âˆ†ABC and âˆ†DEF are equilateral triangles)
∠AXB = ∠DYB (By construction)
By AA test of similarity
âˆ†ABX ~ âˆ†DEY
$∴ \frac{AB}{DE} = \frac{AX}{DY} (Corresponding sides of similar triangles are proportional)$
$∴ \frac{DE}{PQ} \neq \frac{EF}{RP}$

$\frac{A (△ ABC)}{A (△ DEF)} = \frac{1}{2} \Rightarrow \frac{\frac{1}{2} \times AB \times AX}{\frac{1}{2} \times DE \times DY} = \frac{1}{2} \Rightarrow \frac{{AB}^{2}}{{DE}^{2}} = \frac{1}{2} (∵ \frac{AB}{DE} = \frac{AX}{DY}) \Rightarrow {DE}^{2} = 32 \Rightarrow DE = 4 \sqrt{2}$
Hence, the correct option is (D).

(5)
Given: seg XY || seg BC
By basic proportionality theorem
$\frac{AX}{BX} = \frac{AY}{YC} \Rightarrow \frac{BX}{AX} + 1 = \frac{YC}{AY} + 1 \Rightarrow \frac{BX + AX}{AX} = \frac{YC + AY}{AY}$
$\Rightarrow \frac{AB}{AX} = \frac{AC}{AY} \Rightarrow \frac{AB}{AC} = \frac{AX}{AY}$
Hence, the correct option is (D).

Page No 27:

Question 2:

In âˆ†ABC, B – D – C and BD = 7, BC = 20 then Find following ratios.

(1) $\frac{A (∆ ABD)}{A (∆ ADC)}$

(2) $\frac{A (∆ ABD)}{A (∆ ABC)}$

(3) $\frac{A (∆ ADC)}{A (∆ ABC)}$

Answer:

Construction: Draw a perpendicular from vertex A to line BC.

(1)
$\frac{A (∆ ABD)}{A (∆ ADC)} = \frac{\frac{1}{2} \times AX \times BD}{\frac{1}{2} \times AX \times DC} = \frac{BD}{DC} = \frac{7}{13} (DC = BC - BD)$
(2)
â€‹ $\frac{A (∆ ABD)}{A (∆ ABC)} = \frac{\frac{1}{2} \times AX \times BD}{\frac{1}{2} \times AX \times BC} = \frac{BD}{BC} = \frac{7}{20}$
(3)
â€‹ $\frac{A (∆ ADC)}{A (∆ ABC)} = \frac{\frac{1}{2} \times AX \times DC}{\frac{1}{2} \times AX \times BC} = \frac{DC}{BC} = \frac{13}{20} (∵ DC = BC - BD)$

Page No 27:

Question 3:

Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?

Answer:

$\frac{Area of smaller triangle}{Area of bigger triangle} = \frac{2}{3} \Rightarrow \frac{\frac{1}{2} \times Height of smaller triangle \times Base of smaller triangle}{\frac{1}{2} \times Height of bigger triangle \times Base of bigger triangle} = \frac{2}{3} \Rightarrow \frac{6}{Base of bigger triangle} = \frac{2}{3}$
$\Rightarrow Base of bigger triangle = \frac{3}{2} \times 6 = 9$

Page No 27:

Question 4:

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then $\frac{A (∆ ABC)}{A (∆ DCB)} = ?$

Answer:

Given:
∠ABC = ∠DCB = 90°
AB = 6
DC = 8
$Now, \frac{A (∆ ABC)}{A (∆ DCB)} = \frac{\frac{1}{2} \times AB \times BC}{\frac{1}{2} \times DC \times BC} = \frac{6}{8} = \frac{3}{4}$

Page No 27:

Question 5:

In the given figure, PM = 10 cm A(âˆ†PQS) = 100 sq.cm A(âˆ†QRS) = 110 sq.cm then Find NR.

Answer:

Given:
PM = 10 cm
A(âˆ†PQS) = 100 sq.cm
A(âˆ†QRS) = 110 sq.cm
$Now, \frac{A (∆ PQS)}{A (∆ QRS)} = \frac{100}{110} \Rightarrow \frac{\frac{1}{2} \times PM \times QS}{\frac{1}{2} \times RN \times QS} = \frac{10}{11}$ â€‹
â€‹ $\Rightarrow \frac{10}{RN} = \frac{10}{11} \Rightarrow RN = 11 cm$

Page No 27:

Question 6:

âˆ†MNT ~ âˆ†QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio $\frac{A (∆ MNT)}{A (∆ QRS)}$ .

Answer:

The areas of two similar triangles are proportional to the squares of their corresponding altitudes.
$∴ \frac{A (∆ MNT)}{A (∆ QRS)} = {(\frac{5}{9})}^{2} = \frac{25}{81}$

Page No 28:

Question 7:

In the given figure, A – D– C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then Find BE.

Answer:

Given:
AD = 5,
DC = 3,
BC = 6.4
In â–³ABC, DE || AB
$∴ \frac{CD}{DA} = \frac{CE}{EB} (By basic proportionality theorem) \Rightarrow \frac{3}{5} = \frac{6.4 - x}{x} \Rightarrow 3 x = 32 - 5 x$ â€‹â€‹
$\Rightarrow 8 x = 32 \Rightarrow x = 4$

Page No 28:

Question 8:

In the given figure, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280 then Find PQ, QR and RS.

Answer:

Given:
AB = 60,
BC = 70,
CD = 80,
PS = 280
Now, AD = AB + BC + CD
= 60 + 70 + 80
= 210
By intercept theorem, we have
$\frac{PQ}{AB} = \frac{QR}{BC} = \frac{RS}{CD} = \frac{PS}{AD} \Rightarrow \frac{PQ}{60} = \frac{QR}{70} = \frac{RS}{80} = \frac{280}{210} \Rightarrow \frac{PQ}{60} = \frac{QR}{70} = \frac{RS}{80} = \frac{4}{3}$
$∴ PQ = \frac{4}{3} \times 60 = 80 QR = \frac{4}{3} \times 70 = \frac{280}{3} RS = \frac{4}{3} \times 80 = \frac{320}{3}$

Page No 28:

Question 9:

In âˆ†PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.

Answer:

In â–³PMQ, ray MX is bisector of â–³PMQ.
$∴ \frac{PX}{XQ} = \frac{MQ}{MP}$ .......... (I) theorem of angle bisector.
In â–³PMR, ray MY is bisector ofâ–³PMR.
$∴ \frac{PY}{YR} = \frac{MR}{MP}$ .......... (II) theorem of angle bisector.
But $\frac{MP}{MQ} = \frac{MP}{MR}$ ......... M is the midpoint QR, hence MQ = MR.
$∴ \frac{PX}{XQ} = \frac{PY}{YR}$
∴XY || QR .......... converse of basic proportionality theorem.

Page No 29:

Question 10:

In the given fig, bisectors of ∠B and ∠C of âˆ†ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find $\frac{AX}{XY}$ .

Answer:

In â–³ABY, ∠YBX = ∠XBA
$\frac{AX}{XY} = \frac{AB}{BY} . . . (I) (By angle bisector theorem)$
In â–³ACY, ∠YCX = ∠XCA
â€‹ $\frac{AX}{XY} = \frac{AC}{CY} . . . (II) (By angle bisector theorem)$
From (I) and (II)
$\frac{AC}{CY} = \frac{AB}{BY} \Rightarrow \frac{AC}{BC - BY} = \frac{AB}{BY} \Rightarrow \frac{4}{6 - BY} = \frac{5}{BY}$
$\Rightarrow 4 BY = 30 - 5 BY \Rightarrow 9 BY = 30 \Rightarrow BY = \frac{10}{3}$
From (I), we have
â€‹ $\frac{AX}{XY} = \frac{5}{\frac{10}{3}} = \frac{3}{2}$

Page No 29:

Question 11:

In â–¢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that $\frac{AP}{PD} = \frac{PC}{BP}$

Answer:

Given: â–¢ABCD is a parallelogram
To prove: $\frac{AP}{PD} = \frac{PC}{BP}$
Proof: In â–³APD and â–³CPB
∠APD = ∠CPB (Vertically opposite angles)
∠PAD = ∠PCB (Alternate angles, AD || BC and BD is a transversal line)
By AA test of similarity
â–³APD ∼ â–³CPB
$∴ \frac{AP}{PC} = \frac{PD}{PB} (Corresponding sides are proportional) \Rightarrow \frac{AP}{PD} = \frac{PC}{PB}$
Hence proved.

Page No 29:

Question 12:

In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC.

Answer:

Given:
XY || seg AC
2AX = 3BX
XY = 9
Consider, 2AX = 3BX
$∴ \frac{AX}{BX} = \frac{3}{2} \frac{AX + BX}{BX} = \frac{3 + 2}{2} . . . . . by componendo$
$\frac{AB}{BX} = \frac{5}{2} . . . . (I) △ BCA ~ △ BYX . . . SAS test of similarity ∴ \frac{BA}{BX} = \frac{AC}{XY} . . . corresponding sides of similar triangles ∴ \frac{5}{2} = \frac{AC}{9} ∴ AC = 17.5 . . . from (I)$

Page No 29:

Question 13:

In the given figure, the vertices of square DEFG are on the sides of âˆ†ABC. ∠A = 90°. Then prove that DE² = BD × EC (Hint : Show that âˆ†GBD is similar to âˆ†CFE. Use GD = FE = DE.)

Answer:

Given: â–¢DEFG is a square
To prove: DE² = BD × EC
Proof: In â–³GBD and â–³AGF
∠GDB = ∠GAF = 90° (Given)
∠AGF = ∠GBF (Corresponding, GF || BC and AB is a transversal line)
By AA test of similarity
â–³GBD ∼ â–³AGF ...(1)
In â–³CFEand â–³AGF
∠FEC = ∠GAF = 90° (Given)
∠FCE = ∠AGF (Corresponding, GF || BC and AC is a transversal line)
By AA test of similarity
â–³CFE ∼ â–³AGF ...(2)
From (1) and (2), we get
â–³CFE ∼ â–³GBD
$∴ \frac{CE}{GD} = \frac{FE}{BD} (Corresponding sides are proportional) \Rightarrow \frac{CE}{DE} = \frac{DE}{BD} (∵ GD = FE = DE) \Rightarrow {DE}^{2} = BD \times CE$
Hence proved.

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