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#### 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.

Let ABC and PQR be two right triangles with AB ⊥ BC and PQ ⊥ QR.
Given:
BC = 9, AB = 5, PQ = 6 and QR = 10.
$\therefore \frac{\mathrm{A}\left(△\mathrm{ABC}\right)}{\mathrm{A}\left(△\mathrm{PQR}\right)}=\frac{\mathrm{AB}×\mathrm{BC}}{\mathrm{PQ}×\mathrm{QR}}\phantom{\rule{0ex}{0ex}}=\frac{5×9}{6×10}\phantom{\rule{0ex}{0ex}}=\frac{3}{4}$

#### Question 2:

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

Given:
BC = 4

$=\frac{1}{2}$

#### Question 3:

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

Given:
PS ⊥ RQ
QT ⊥ PR
RQ = 6, PS = 6 and PR = 12
With base PR and height QT, $\mathrm{A}\left(△\mathrm{PQR}\right)=\frac{1}{2}×\mathrm{PR}×\mathrm{QT}$
With base QR and height PS, $\mathrm{A}\left(△\mathrm{PQR}\right)=\frac{1}{2}×\mathrm{QR}×\mathrm{PS}$
$\therefore \frac{\mathrm{A}\left(△\mathrm{PQR}\right)}{\mathrm{A}\left(△\mathrm{PQR}\right)}=\frac{\frac{1}{2}×\mathrm{PR}×\mathrm{QT}}{\frac{1}{2}×\mathrm{QR}×\mathrm{PS}}\phantom{\rule{0ex}{0ex}}⇒1=\frac{\mathrm{PR}×\mathrm{QT}}{\mathrm{QR}×\mathrm{PS}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{PR}×\mathrm{QT}=\mathrm{QR}×\mathrm{PS}$
$⇒\mathrm{QT}=\frac{\mathrm{QR}×\mathrm{PS}}{\mathrm{PR}}\phantom{\rule{0ex}{0ex}}=\frac{6×6}{12}\phantom{\rule{0ex}{0ex}}=3$
Hence, the measure of side QT is 3 units.

#### Question 4:

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

Given:
AP ⊥ BC
$\therefore \frac{\mathrm{A}\left(△\mathrm{ABC}\right)}{\mathrm{A}\left(△\mathrm{BCD}\right)}=\frac{\mathrm{AP}×\mathrm{BC}}{\mathrm{AP}×\mathrm{BC}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1}$
Hence, the ratio of A(∆ABC) and A(∆BCD) is 1 : 1.

#### Question 5:

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

(i) $\frac{\mathrm{A}\left(∆\mathrm{PQB}\right)}{\mathrm{A}\left(∆\mathrm{PBC}\right)}$

(ii) $\frac{\mathrm{A}\left(∆\mathrm{PBC}\right)}{\mathrm{A}\left(∆\mathrm{ABC}\right)}$

(iii) $\frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆\mathrm{ADC}\right)}$

(iv) $\frac{\mathrm{A}\left(∆\mathrm{ADC}\right)}{\mathrm{A}\left(∆\mathrm{PQC}\right)}$

(i)
$\frac{\mathrm{A}\left(△\mathrm{PQB}\right)}{\mathrm{A}\left(△\mathrm{PBC}\right)}=\frac{\mathrm{PQ}×\mathrm{BQ}}{\mathrm{PQ}×\mathrm{BC}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{BQ}}{\mathrm{BC}}$
(ii)
$\frac{\mathrm{A}\left(△\mathrm{PBC}\right)}{\mathrm{A}\left(△\mathrm{ABC}\right)}=\frac{\mathrm{PQ}×\mathrm{BC}}{\mathrm{AD}×\mathrm{BC}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{PQ}}{\mathrm{AD}}$
(iii)
$\frac{\mathrm{A}\left(△\mathrm{ABC}\right)}{\mathrm{A}\left(△\mathrm{ADC}\right)}=\frac{\mathrm{AD}×\mathrm{BC}}{\mathrm{AD}×\mathrm{DC}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{BC}}{\mathrm{DC}}$
(iv)
$\frac{\mathrm{A}\left(△\mathrm{ADC}\right)}{\mathrm{A}\left(△\mathrm{PQC}\right)}=\frac{\mathrm{AD}×\mathrm{DC}}{\mathrm{PQ}×\mathrm{QC}}$

#### 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)

(1)

$\therefore \frac{\mathrm{QM}}{\mathrm{QP}}=\frac{\mathrm{MR}}{\mathrm{RP}}$
By converse of angle bisector theorem, ray PM is the bisector of ∠QPR.

(2)

$\therefore \frac{\mathrm{QM}}{\mathrm{QP}}\ne \frac{\mathrm{MR}}{\mathrm{RP}}$
Therefore, ray PM is not the the bisector of ∠QPR.
(3)

$\therefore \frac{\mathrm{QM}}{\mathrm{QP}}=\frac{\mathrm{MR}}{\mathrm{RP}}$
By converse of angle bisector theorem, ray PM is the bisector of ∠QPR.

#### Question 2:

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

Given:
PM = 15,
PQ = 25,
PR = 20 and
NR = 8
Now, PN = PR − NR
= 20 − 8
= 12
Also, MQ = PQ − PM
= 25 − 15
= 10

$\mathrm{Also},\phantom{\rule{0ex}{0ex}}\frac{\mathrm{PM}}{\mathrm{MQ}}=\frac{15}{10}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$
$\therefore \frac{\mathrm{PR}}{\mathrm{NR}}=\frac{\mathrm{PM}}{\mathrm{MQ}}$
By converse of basic proportionality theorem, NM is parallel to side RQ or NM || RQ.

#### Question 3:

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

$⇒\mathrm{QP}=\frac{2.5×7}{5}\phantom{\rule{0ex}{0ex}}=3.5$
Hence, the measure of QP is 3.5.

#### Question 4:

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

Given:
∠APQ = 60
∠ABC = 60
Since, the corresponding angles ∠APQ and ​∠APC are equal.
Hence, line PQ || BC.

#### Question 5:

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

Construction: Join BD intersecting PQ at X.

In △ABD, PX || AB

In △BDC, XQ||DC

From (1) and (2), we get
$\frac{\mathrm{DP}}{\mathrm{PA}}=\frac{\mathrm{CQ}}{\mathrm{QB}}\phantom{\rule{0ex}{0ex}}⇒\frac{12}{15}=\frac{14}{\mathrm{QB}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{QB}=17.5$

#### Question 6:

Find QP using given information in the figure.

$⇒\mathrm{QP}=\frac{14×40}{25}\phantom{\rule{0ex}{0ex}}=22.4$
Hence, the measure of QP is 22.4.

#### Question 7:

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

Construction: Join AFintersecting CD at X.

In △ABF, DX || AB

In △AEF, XC||FE

From (1) and (2), we get
$\frac{\mathrm{FD}}{\mathrm{DB}}=\frac{\mathrm{EC}}{\mathrm{CA}}\phantom{\rule{0ex}{0ex}}⇒\frac{4}{8}=\frac{x}{12}\phantom{\rule{0ex}{0ex}}⇒x=6$
Now, AE = AC + CE
= 12 + 6
= 18

#### Question 8:

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

$⇒\mathrm{LT}=\frac{8×6}{10}\phantom{\rule{0ex}{0ex}}=4.8$
Hence, the measure of LT is 4.8.

#### Question 9:

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

In △ABC, ∠ABD = ∠DBC

$⇒3x-2x=10\phantom{\rule{0ex}{0ex}}⇒x=10$

#### 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.

Given:
Seg PQ || seg DE
seg QR || seg EF
In △DXE, PQ || DE

In △XEF, QR || EF                ....Given

∴ seg PR || seg DF        (Converse of basic proportional theorem)

#### Question 11:

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

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

In △ABC, ∠ABD = ∠DBC

In △ABC, ∠BCE = ∠ACE

From (I) and (II)

∴ ​ED || BC        (Converse of basic proportional theorem)

#### 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.

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

#### Question 2:

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

Given:
PQ = 6
PR = 10
QR = 8
LM = 3
LN = 5
MN = 4
Now,
$\frac{\mathrm{PQ}}{\mathrm{LM}}=\frac{6}{3}=2,\phantom{\rule{0ex}{0ex}}\frac{\mathrm{QR}}{\mathrm{MN}}=\frac{8}{4}=2,\phantom{\rule{0ex}{0ex}}\frac{\mathrm{RP}}{\mathrm{NL}}=\frac{10}{5}=2$
$\therefore \frac{\mathrm{PQ}}{\mathrm{LM}}=\frac{\mathrm{QR}}{\mathrm{MN}}=\frac{\mathrm{RP}}{\mathrm{NL}}$
By SSS test of similarity
△PQR ∼ △LMN

#### 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 ?

Given:
PR = 4
RL = 6
AC = 8
In △PLR and △ABC
∠PRL = ∠ACB         (Vertically opposite angles)
∠LPR = ∠BAC         (Angles made by sunlight on top are congruent)
By AA test of similarity
△PLR ∼ △ABC

Hence, the length of shadow of bigger pole due to sunlight is 12 m.

#### Question 4:

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

Given:
AP ⊥ BC
BQ ⊥ AC
To prove: ∆CPA ~ ∆CQB
Proof: In ∆CPA and ∆CQB
∠CPA = ∠CQB = 90         (Given)
∠C = ∠C                             (Common)
By AA test of similarity
∆CPA ~ ∆CQB
Hence proved.

#### Question 5:

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

Given:
side PQ || side SR
AR = 5AP,
AS = 5AQ
To prove: SR = 5PQ
Proof: In ∆APQ and ∆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
∆APQ ~ ∆ARS

Hence proved.

#### 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.

Given:
side AB || side DC
AB = 20,
DC = 6,
OB = 15
In △COD and △AOB
∠COD = ∠AOB         (Vertically opposite angles)
∠CDO= ∠ABO         (Alternate angles, CD ||BA and BD is a transversal line)
By AA test of similarity
△COD ∼ △AOB

#### Question 7:

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

Given: ◻ABCD is a parallelogram
To prove: DE × BE = CE × TE
Proof: In ∆BET and ∆CED
∠BET = ∠CED         (Vertically opposite angles)
∠BTE = ∠CDE         (Alternate angles, AT || CD and DT is a transversal line)
By AA test of similarity
∆BET ∼ ∆CED

Hence proved.

#### Question 8:

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

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

#### Question 9:

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

To prove: CA2 = CB × CD
Proof: In ∆ABC and ∆DAC
∠C = ∠C                   (Common)
By AA test of similarity
∆ABC ∼ ∆DAC

Hence proved.

#### Question 1:

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

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".
Therefore, the ratio of the areas of triangles $=\frac{{3}^{2}}{{5}^{2}}$
$=\frac{9}{25}$

#### Question 2:

If ∆ABC ~ ∆PQR and AB : PQ = 2 : 3, then fill in the blanks.

Given:
∆ABC ~ ∆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".

#### Question 3:

If ∆ABC ~ ∆PQR, A (∆ABC) = 80, A (∆PQR) = 125, then fill in the blanks.

Given:
∆ABC ~ ∆PQR
A (∆ABC) = 80
A (∆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".
$\therefore \frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆\mathrm{PQR}\right)}=\frac{{\mathrm{AB}}^{2}}{{\mathrm{PQ}}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{80}{125}=\frac{{\mathrm{AB}}^{2}}{{\mathrm{PQ}}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{16}{25}=\frac{{\mathrm{AB}}^{2}}{{\mathrm{PQ}}^{2}}$
$⇒\frac{{4}^{2}}{{5}^{2}}=\frac{{\mathrm{AB}}^{2}}{{\mathrm{PQ}}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\overline{)4}}{\overline{)5}}$
Therefore,

#### Question 4:

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

Given:
∆LMN ~ ∆PQR
9 × A (∆PQR ) = 16 × A (∆LMN)
Consider, 9 × A (∆PQR ) = 16 × A (∆LMN)
$\frac{\mathrm{A}\left(∆\mathrm{LMN}\right)}{\mathrm{A}\left(∆\mathrm{PQR}\right)}=\frac{9}{16}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{MN}}^{2}}{{\mathrm{QR}}^{2}}=\frac{{3}^{2}}{{4}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{MN}}{\mathrm{QR}}=\frac{3}{4}$

#### 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.

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".

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

#### Question 6:

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

Consider, A(∆ABC) : A (∆DEF) = 1 : 2
$⇒\frac{\mathrm{A}\left(∆\mathrm{ABC}\right)}{\mathrm{A}\left(∆\mathrm{DEF}\right)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mathrm{AB}}^{2}}{{\mathrm{DE}}^{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{DE}}^{2}=2{\mathrm{AB}}^{2}$

#### Question 7:

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

Given:
seg PQ || seg DE
A(∆PQF) = 20 units
PF = 2 DP
Let us assume DP = x
∴ PF = 2x
$\mathrm{DF}=\mathrm{DP}+\overline{)\mathrm{PF}}=\overline{)x}+\overline{)2x}=3x$
In △FDE and △FPQ
∠FDE = ∠FPQ         (Corresponding angles)
∠FED = ∠FQP         (Corresponding angles)
By AA test of similarity
△FDE ∼ △FPQ
$\therefore \frac{\mathrm{A}\left(△\mathrm{FDE}\right)}{\mathrm{A}\left(△\mathrm{FPQ}\right)}=\frac{\overline{){\mathrm{FD}}^{2}}}{\overline{){\mathrm{FP}}^{2}}}=\frac{{\left(3x\right)}^{2}}{{\left(2x\right)}^{2}}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}\mathrm{A}\left(△\mathrm{FDE}\right)=\frac{9}{4}\mathrm{A}\left(△\mathrm{FPQ}\right)=\frac{9}{4}×\overline{)20}=\overline{)45}$
$\therefore \mathrm{A}\left(\square \mathrm{DPQE}\right)=\mathrm{A}\left(△\mathrm{FDE}\right)-\mathrm{A}\left(△\mathrm{FPQ}\right)\phantom{\rule{0ex}{0ex}}=\overline{)45}-\overline{)20}\phantom{\rule{0ex}{0ex}}=\overline{)25}$

#### Question 1:

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

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

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

(A) $\frac{\mathrm{EF}}{\mathrm{PR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}$ (B) $\frac{\mathrm{DE}}{\mathrm{PQ}}=\frac{\mathrm{EF}}{\mathrm{RP}}$

(C) $\frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}$ (D) $\frac{\mathrm{EF}}{\mathrm{RP}}=\frac{\mathrm{DE}}{\mathrm{QR}}$

(3) In ∆ABC and ∆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) ∆ABC and ∆DEF are equilateral triangles, A (∆ABC) : A (∆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) $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AX}}{\mathrm{AY}}$ (B) $\frac{\mathrm{AX}}{\mathrm{XB}}=\frac{\mathrm{AY}}{\mathrm{AC}}$

(C) $\frac{\mathrm{AX}}{\mathrm{YC}}=\frac{\mathrm{AY}}{\mathrm{XB}}$ (D) $\frac{\mathrm{AB}}{\mathrm{YC}}=\frac{\mathrm{AC}}{\mathrm{XB}}$

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

(2)
In ∆DEF and ∆PQR
∠D ≅ ∠Q
∠R ≅ ∠E
By AA test of similarity
∆DEF~ ∆PQR

$\therefore \frac{\mathrm{DE}}{\mathrm{PQ}}\ne \frac{\mathrm{EF}}{\mathrm{RP}}$
Hence, the correct option is (B).

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

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

In ∆ABX and ∆DEY
∠B = ∠C = 60             (∆ABC and ∆DEF are equilateral triangles)
∠AXB = ∠DYB           (By construction)
By AA test of similarity
∆ABX ~ ∆DEY

$\therefore \frac{\mathrm{DE}}{\mathrm{PQ}}\ne \frac{\mathrm{EF}}{\mathrm{RP}}$

Hence, the correct option is (D).

(5)
Given: seg XY || seg BC
By basic proportionality theorem
$\frac{\mathrm{AX}}{\mathrm{BX}}=\frac{\mathrm{AY}}{\mathrm{YC}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BX}}{\mathrm{AX}}+1=\frac{\mathrm{YC}}{\mathrm{AY}}+1\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{BX}+\mathrm{AX}}{\mathrm{AX}}=\frac{\mathrm{YC}+\mathrm{AY}}{\mathrm{AY}}$
$⇒\frac{\mathrm{AB}}{\mathrm{AX}}=\frac{\mathrm{AC}}{\mathrm{AY}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AX}}{\mathrm{AY}}$
Hence, the correct option is (D).

#### Question 2:

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

(1) $\frac{\mathrm{A}\left(∆\mathrm{ABD}\right)}{\mathrm{A}\left(∆\mathrm{ADC}\right)}$

(2) $\frac{\mathrm{A}\left(∆\mathrm{ABD}\right)}{\mathrm{A}\left(∆\mathrm{ABC}\right)}$

(3) $\frac{\mathrm{A}\left(∆\mathrm{ADC}\right)}{\mathrm{A}\left(∆\mathrm{ABC}\right)}$

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

(1)

(2)
$\frac{\mathrm{A}\left(∆\mathrm{ABD}\right)}{\mathrm{A}\left(∆\mathrm{ABC}\right)}=\frac{\frac{1}{2}×\mathrm{AX}×\mathrm{BD}}{\frac{1}{2}×\mathrm{AX}×\mathrm{BC}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{BD}}{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}=\frac{7}{20}$
(3)

#### 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 ?

#### Question 4:

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then

Given:
∠ABC = ∠DCB = 90°
AB = 6
DC = 8

#### Question 5:

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

Given:
PM = 10 cm
A(∆PQS) = 100 sq.cm
A(∆QRS) = 110 sq.cm

#### Question 6:

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

The areas of two similar triangles are proportional to the squares of their corresponding altitudes.
$\therefore \frac{\mathrm{A}\left(∆\mathrm{MNT}\right)}{\mathrm{A}\left(∆\mathrm{QRS}\right)}={\left(\frac{5}{9}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{25}{81}$

#### 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.

Given:
DC = 3,
BC = 6.4
In △ABC,  DE || AB
​​
$⇒8x=32\phantom{\rule{0ex}{0ex}}⇒x=4$

#### 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.

Given:
AB = 60,
BC = 70,
CD = 80,
PS = 280
Now, AD = AB + BC + CD
= 60 + 70 + 80
= 210
By intercept theorem, we have
$\frac{\mathrm{PQ}}{\mathrm{AB}}=\frac{\mathrm{QR}}{\mathrm{BC}}=\frac{\mathrm{RS}}{\mathrm{CD}}=\frac{\mathrm{PS}}{\mathrm{AD}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PQ}}{60}=\frac{\mathrm{QR}}{70}=\frac{\mathrm{RS}}{80}=\frac{280}{210}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{PQ}}{60}=\frac{\mathrm{QR}}{70}=\frac{\mathrm{RS}}{80}=\frac{4}{3}$
$\therefore \mathrm{PQ}=\frac{4}{3}×60=80\phantom{\rule{0ex}{0ex}}\mathrm{QR}=\frac{4}{3}×70=\frac{280}{3}\phantom{\rule{0ex}{0ex}}\mathrm{RS}=\frac{4}{3}×80=\frac{320}{3}\phantom{\rule{0ex}{0ex}}$

#### Question 9:

In ∆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.

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

#### Question 10:

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

In △ABY, ∠YBX = ∠XBA

In △ACY, ∠YCX = ∠XCA

From (I) and (II)
$\frac{\mathrm{AC}}{\mathrm{CY}}=\frac{\mathrm{AB}}{\mathrm{BY}}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{AC}}{\mathrm{BC}-\mathrm{BY}}=\frac{\mathrm{AB}}{\mathrm{BY}}\phantom{\rule{0ex}{0ex}}⇒\frac{4}{6-\mathrm{BY}}=\frac{5}{\mathrm{BY}}$
$⇒4\mathrm{BY}=30-5\mathrm{BY}\phantom{\rule{0ex}{0ex}}⇒9\mathrm{BY}=30\phantom{\rule{0ex}{0ex}}⇒\mathrm{BY}=\frac{10}{3}$
From (I), we have
$\frac{\mathrm{AX}}{\mathrm{XY}}=\frac{5}{\frac{10}{3}}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$

#### Question 11:

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

Given: ▢ABCD is a parallelogram
To prove: $\frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{BP}}$
Proof: In △APD and △CPB
∠APD = ∠CPB         (Vertically opposite angles)
∠PAD = ∠PCB         (Alternate angles, AD || BC and BD is a transversal line)
By AA test of similarity
△APD ∼ △CPB

Hence proved.

#### Question 12:

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

Given:
XY || seg AC
2AX = 3BX
XY = 9
Consider, 2AX = 3BX

#### Question 13:

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

Given: ▢DEFG is a square
To prove: DE2 = BD × EC
Proof: In △GBD and △AGF
∠GDB = ∠GAF =  90°         (Given)
∠AGF = ∠GBF                    (Corresponding, GF || BC and AB is a transversal line)
By AA test of similarity
△GBD ∼ △AGF                         ...(1)
In △CFEand △AGF
∠FEC = ∠GAF =  90°         (Given)
∠FCE = ∠AGF                   (Corresponding, GF || BC and AC is a transversal line)
By AA test of similarity
△CFE ∼ △AGF                          ...(2)
From (1) and (2), we get
△CFE ∼ △GBD

Hence proved.

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