Solution-Class-10-Mathematics-2-Chapter-1-Similarity-Maharashtra Board

Let the base, height and area of the first triangle be b₁, h₁ and A₁ respectively. Let the base, height and area of the second triangle be b₂, h₂ and A₂ respectively.

b₁ = 9, h₁ = 5, b₂ = 10 and h₂ = 6.

The ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights

∴  \(\frac{A_1}{A_2}=\frac{b_1×h_1}{b_2×h_2}\)

∴  \(\frac{A_1}{A_2}=\frac{9×5}{10×6}\)

∴  \(\frac{A_1}{A_2}=\frac{3}{4}\)

∴ The ratio of the areas of the triangles is 3 : 4.

BC = 4, AD = 8  ……(Given)

ΔABC and ΔADB have same base AB.

Areas of triangles with same base are proportional to their corresponding heights.

∴ \(\frac{A(ΔABC)}{A(ΔADB)}=\frac{BC}{AD}\)

∴ \(\frac{A(ΔABC)}{A(ΔADB)}=\frac{4}{8}\)

∴ \(\frac{A(ΔABC)}{A(ΔADB)}=\frac{1}{2}\)

∴ The ratio of the areas of the triangles ΔABC and ΔADB is 1 : 2.

Solution : RQ = 6, PS = 6 and PR = 12  …….(Given)

Area of a triangle = \(\frac{1}{2}\) x base x height

∴ A(ΔPQR) = \(\frac{1}{2}\)x QR x PS

∴ A(ΔPQR) = \(\frac{1}{2}\) x 6 x 6

∴ A(ΔPQR) = 18 sq units.  ……(1)

Also, A(ΔPQR) = \(\frac{1}{2}\) x PR x QT

∴ 18 = \(\frac{1}{2}\) x 12 x QT  ….[From (1)1

∴ QT = \(\frac{18}{6}\)

∴ QT = 3

Answer is  QT = 3

ΔABC and ΔBCD have same base BC.

The vertices of ΔABC and ΔBCD lie on the parallel lines AD and BC. The distance between the two parallel lines is the height of the triangles.

Distance between parallel lines is constant,

ΔABC and ΔBCD have equal heights.

Areas of two triangles with same base and equal heights are equal.

A(ΔABC) = A(ΔBCD)

∴ \(\frac{A(ΔABC)}{A(ΔBCD)}=\frac{1}{1}\)

∴ A(ΔABC) : A(ΔBCD) = 1:1

Answer is A(ΔABC) : A(ΔBCD) = 1:1.

(i) ΔPQB and ΔPBC have same height PQ. Areas of triangles with equal heights are proportional to their corresponding bases.

\(\frac{A(ΔPQB)}{A(ΔPBC)}=\frac{QB}{BC}\)

(ii) ΔPBC and ΔABC have same base BC. Areas of triangles with same base are proportional to their corresponding heights.

\(\frac{A(ΔPBC)}{A(ΔABC)}=\frac{PQ}{AD}\)

(iii) ΔABC and ΔADC have same height AD. Areas of triangles with same height are proportional to their corresponding bases.

\(\frac{A(ΔABC)}{A(ΔADC)}=\frac{BC}{DC}\)

(iv) Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.

\(\frac{A(ΔADC)}{A(ΔPQC)}=\frac{DC×AD}{QC×PQ}\)

(1) PQ = 7, PR = 3, QM = 3.5 and MR = 1.5  ….(Given)

\(\frac{PQ}{PR}=\frac{7}{3}\) ……..(1)

  \(\frac{QM}{QR}=\frac{3.5}{1.5}=\frac{35}{15}=\frac{7}{3}\)  …...(2)

In ΔQPR,

\(\frac{PQ}{PR}=\frac{QM}{QR}\)……From (1) and (2)

∴ by converse of angle bisector theorem,

∴ ray PM bisects ∠ QPR.

(2) PR = 7, PQ = 10, RM = 6 and MQ = 8  ….(Given)

  \(\frac{PR}{PQ}=\frac{7}{10}\)……..(1)

  \(\frac{RM}{MQ}=\frac{6}{8}==\frac{3}{4}\) …...(2)

∴ From (1) and (2)

\(\frac{PR}{PQ}=\frac{RM}{MQ}\)

∴ ray PM is not bisector of ∠ RPQ.

(3) PQ = 9, PR = 10, QM = 3.6 and MR = 4  ….(Given)

  \(\frac{PQ}{PR}=\frac{9}{10}\)  ……..(1)

   \(\frac{QM}{QR}=\frac{3.6}{4}=\frac{36}{40}=\frac{9}{10}\) …...(2)

In ΔQPR,

\(\frac{PQ}{PR}=\frac{QM}{QR}\)……From (1) and (2)

∴ by converse of angle bisector theorem, ray PM bisects ∠ QPR.

PM = 15, PQ = 25, PR = 20 and NR = 8.

PM + MQ = PQ  …..(P-M-Q)

15 + MQ = 25

∴ MQ= 25 - 15,  \MQ = 10

PN + NR = PR  ….(P-N-R)

PN + 8 = 20

PN = 20 - 8,  ∴ PN = 12

\(\frac{PN}{NR}=\frac{12}{8}=\frac{3}{2}\)…… (1)

\(\frac{PM}{MQ}=\frac{15}{10}=\frac{3}{2}\)   …… (2)

In ΔPRQ,

\(\frac{PN}{NR}=\frac{PM}{MQ}\)  …..[From (1) and (2)]

∴ by converse of basic proportionality theorem, line NM || side RQ.

In ΔMNP, ray NQ bisects ∠ MNP

∴ by the theorem of angle bisector of a triangle,

\(\frac{MN}{NP}=\frac{MQ}{QP}\)

∴ \(\frac{5}{7}=\frac{2.5}{QP}\)

∴ QP = \(\frac{7×2.5}{5}\) = 3.5.

∴ Answer is QP = 3.5

∠ APQ = ∠ ABC = 60°  ….(Given)

∴ ∠ APQ ≅ ∠ ABC

seg PQ || seg BC  ….(Corresponding angles test for parallel lines)  (1)

In ΔABC,

seg PQ ll side BC  [From (1)]

∴  \(\frac{AP}{PB}=\frac{AQ}{QC}\)    … (Basic proportionality theorem)

Side AB || side PQ || Side DC  (Given)

∴ by the property of three parallel lines and their transversals,

\(\frac{AP}{PD}=\frac{BQ}{QC}\)

∴ \(\frac{15}{12}=\frac{BQ}{14}\)

∴ BQ x 12 = 15 x 14

∴ BQ = \(\frac{15×14}{12}=\frac{210}{6}=\frac{35}{2}\)

∴  BQ = 17.5

Answer BQ = 17.5

In ΔMNP,

ray NQ bisects ∠ MNP

∴ by the theorem of angle bisector of a triangle,

\(\frac{MN}{NP}=\frac{MQ}{PQ}\)

∴ \(\frac{25}{40}=\frac{14}{PQ}\)

25 x PQ = 40 x 14

∴ PQ = \(\frac{40×14}{25}=\frac{8×14}{5}=22.4\)

Answer is  PQ = 22.4.

Line AB || line CD || line EF

∴ by the property of three parallel lines and their transversals,

\(\frac{BD}{DF}=\frac{AC}{CE}\)

∴ \(\frac{8}{4}=\frac{12}{x}\)

∴ \(8 × x = 4×12\)

∴ \(x=\frac{4×12}{8}\)

∴ x = 6

AE = AC + CE ...(A-C_E)

 AE = 12 + x

 AE = 12 + 6

 AE = 18

Answer is  x = 6 and AE = 18.

In ΔLMN,

ray MT bisects ∠ LMN  (Given)

∴ by the theorem of angle bisector of a triangle,

\(\frac{LM}{MN}=\frac{LT}{TN}\)

∴ \(\frac{6}{10}=\frac{LT}{8}\)

∴ LT x 10 = 6 x 8

 LT =  = 4.8

Answer is LT = 4.8.

In ΔABC,

seg BD bisects ∠ ABC

∴ by the theorem of angle bisector of a triangle,

\(\frac{AB}{BC}=\frac{AD}{DC}\)

\(\frac{x}{x+5}=\frac{x-2}{x+2}\)

∴ x(x+2) = (x-20)(x+5)

∴ x² + 2x = x² + 5x - 2x - 10

∴ 2x = 3x - 10

∴ 2x - 3x = -10

∴ - x = -10

∴ x = 10

Answer is  The value of x is 10.

Proof : In ΔXDE, PQ || DE ..........[Given]

∴ \(\frac{XP}{[PD]}=\frac{[XQ]}{QE}\) .......... (I) (Basic proportionality theorem)

In ΔXEF, QR || EF .......... [Given]

∴  \(\frac{[XQ]}{[QE]}=\frac{[XR]}{[RF]}\)..........(II) […..]

∴ \(\frac{[XP]}{[PD]}=\frac{[XR]}{[RF]}\)  .......... from (I) and (II)

∴ seg PR || seg DF .......... (converse of basic proportionality theorem)

In ΔABC, ray BD is the bisector of ∠ ABC

∴ by the theorem of angle bisector of a triangle,

\(\frac{AB}{BC}=\frac{AD}{DC}\).………(1)

In ΔABC, ray CE is the bisector of ∠ ACB

∴ by the theorem of angle bisector of a triangle,

\(\frac{AC}{BC}=\frac{AE}{EB}\)………(2)

Seg AB ≅ seg AC  …..(Given)….(3)

∴ \(\frac{AC}{BC}=\frac{AC}{BC}\)   …..[From (1), (2) and (3)]….(4)

In ΔABC,

\(\frac{AE}{EB}=\frac{AD}{DC}\)……..[From (1), (2) and (4)]

∴ by converse of basic proportionality theorem,

seg ED ll side BC

i.e. ED ll BC.

In ΔDCE and ΔBCA,

∠ CDE ≅ ∠ CBA  (Each measures 75°)

∠ DCE ≅ ∠ BCA  (Common angle)

∴ ∠ DCE ∼ ∠ BCA  (AA test of similarity)

\(\frac{PQ}{LM}=\frac{6}{3}=\frac{2}{1}\) …….(1)

\(\frac{QR}{MN}=\frac{8}{4}=\frac{2}{1}\)  …..(2)

\(\frac{PR}{LN}=\frac{10}{5}=\frac{2}{1}\) …….(3)

In ΔPQR and ΔLMN,

\(\frac{PQ}{LM}=\frac{QR}{MN}=\frac{PR}{LN}\)  ….[From (1), (2) and (3)1

∴ ΔPQR ∼ ΔLMN  (SSS test of similarity)

PR and AC are two poles perpendicular to the ground. QR and BC are their shadows respectively.

Because the shadows are cast at the same time

∴ ΔPQR ∼ ΔABC

\(\frac{PR}{AC}=\frac{QR}{BC}\)     …..[Corresponding sides of similar triangles are in proportion]

∴  \(\frac{4}{8}=\frac{6}{x}\)  

∴ 4 x x  = 8 x 6

∴ x = \(\frac{8×6}{4}

∴ x = 12,  ∴ BC = 12m

Answer is -The length of the shadow of the bigger pole will be 12 m.

In ΔCPA and ΔCQB,

∠ CPA ≅ ∠ CQB   …(Each measures 90°)

∠ ACP ≅ ∠ BCQ   ….(Common angle)

∴ ΔCPA ∼ ΔCQB  ….(AA test of similarity)

∴ \(\frac{AP}{BQ}=\frac{AC}{BC}\)  ….[Corresponding sides of similar triangles are in proportion]

∴  \(\frac{7}{8}=\frac{AC}{12}\)

∴ AC x 8 = 7 x 12

∴ AC = 10.5

Answer is AC = 10.5

Side PQ ll Side SR and line QS is the transversal

∠ PQS ≅ ∠ RSQ  …….(Alternate angles)

i.e. ∠ PQA ≅ ∠ RSA  ……(Q—A—S)…(1)

In ΔPQA and ΔRSA,

∠ PQA ≅ ∠ RSA  ….[From (1)]

∠ PAQ ≅ ∠ RAS  …..(Vertically opposite angles)

ΔPQA ∼ ΔRSA  (AA test of similarity)

  \(\frac{PQ}{RS}=\frac{AQ}{AS}=\frac{AP}{PR}\)….(Corresponding sides of similar triangles are in proportion)..(2)

AR = 5AP  ....(Given)..(3)

Substituting (3) in (2), We get,

\(\frac{PQ}{SR}=\frac{AQ}{AS}=\frac{AP}{5AP}\)

∴  \(\frac{PQ}{SR}=\frac{AQ}{AS}=\frac{1}{5}\)…. (4)

∴  \(\frac{PQ}{SR}=\frac{1}{5}\)…[From (4)]

∴ SR = 5PQ

Side AB || side DC and line DB is the transversal,

∠ CDB ≅ ∠ ABD  (Alternate angles)

i.e. ∠ CDO ≅ ∠ ABO  (B—O—D)  ….(1)

In ΔCOD and ΔAOB,

∠ CDO ≅ ∠ ABO  ….[From (1)]

∠ COD ≅ ∠ AOB  …..(Vertically opposite angles)

∴ ΔCOD ∼ ΔAOB  (AA test of similarity)

\(\frac{OD}{OB}=\frac{DC}{AB}   …(Corresponding sides of similar triangles are in proportion)

∴ \(\frac{OD}{15}=\frac{6}{20} 

OD x 20 = 15 x 6

∴ OD = \(\frac{15×6}{20} = \frac{90}{20}\)

∴ OD = 4.5

Answers is OD = 4.5.

Proof : Seg AB || seg CD  …(Opposite sides of parallelogram are parallel)

i.e. seg AT || seg CD  (A—B—T)

Line TD is the transversal,

∴ ∠ ATD ≅ ∠ CDT  (Alternate angles)

i.e. ∠ BTE ≅ ∠ CDE  (A—B—T, T—E—D) ...(1)

In ΔBET and ΔCED

∠ BTE ≅ ∠ CDE  [From (1)1

∠ BET ≅ ∠ CED  (Vertically opposite angles)

∴ ΔBET ∼ ΔCED  (AA test of similarity)

∴  \(\frac{BE}{CE}=\frac{TE}{DE}\)    ….(Corresponding sides of similar CE DE triangles are in proportion)

∴ DE × BE = CE × TE.

Proof :

\(\frac{AP}{CP}=\frac{BP}{DP}\)   (Given)

∠ APB ≅ ∠ CPD   …..(Vertically opposite angles)

ΔABP ∼ ΔCDP  (SAS test of similarity)

Proof; In ΔBAC and ΔADC,

∠ BAC ≅ ∠ ADC     ….(Given)

∠ ACB ≅ ∠ DCA      …..(Common angle)

ΔBAC ∼ ΔADC     ….(AA test of similarity)

∴  \(\frac{CA}{CD}=\frac{CB}{CA}\)     …….(Corresponding sides of similar triangles are in proportion)

∴ CA² = CB × CD.

Let the corresponding sides of two similar triangles be x₁ and x₂. Let their respective areas be A₁ and A₂.

x₁ : x₂ = 3 : 5   ….(Given)

\(\frac{x_1}{x_2}=\frac{3}{5}\)   …(1)

The two triangles are similar.

∴ \(\frac{A_1}{A_2}=\frac{{x_1}^2}{{x_2}^2}\)   ……(Theorem of areas of similar triangles)

∴  \(\frac{A_1}{A_2}=(\frac{x_1}{x_2})^2 =(\frac{3}{5})^2\) …[From (1)]

\(\frac{A_1}{A_2}=\frac{9}{25}\)

A₁ : A₂ = 9 : 25

Answer is - The ratio of the areas of two similar triangles is 9 : 25.

\(\frac{A(ΔABC)}{A(ΔPQR)}=\frac{AB^2}{[PQ^2]}=\frac{2^2}{3^2}=\frac{[4]}{[9]}\)

\(\frac{A(ΔABC)}{A(ΔPQR)}=\frac{80}{125}\)

∴ \(\frac{AB}{PQ}=\frac{[4]}{[5]}\)

Explanation :

ΔABC ∼ ΔPQR. By the theorem of areas of similar triangles

\(\frac{A(ΔABC)}{A(ΔPQR)}=\frac{AB^2}{PQ^2}=\frac{80}{125}=\frac{16}{25}\)

∴ \(\frac{AB}{PQ}=\frac{[4]}{[5]}\)

ΔLMN ∼ ΔPQR  ....(Given)

9 x A(ΔPQR) = 16 x A(ΔLMN)

∴  \(\frac{A(ΔLMN)}{A(ΔPQR)}=\frac{9}{16}\)  … (1)

Now

  \(\frac{A(ΔLMN)}{A(ΔPQR)}=\frac{MN^2}{QR^2}\) ….(Theorem of areas of similar triangles)

∴ \( \frac{9}{16}=\frac{MN^2}{QR^2}\)…..from (1)

∴ \(\frac{MN}{QR}=\frac{3}{4}\) 

∴  \(MN=\frac{20×3}{4}=15\)  ….(…QR=20 given)

Answer is –MN=15

Let the areas of two similar triangles be A₁ and A₂. Let their respective corresponding sides be s₁ and s₂.

A₁ = 225 sq cm, A₂ = 81 sq cm and s₂ = 12 cm.  …(Given)

The two triangles are similar

∴ by the theorem of areas of similar triangles,

\(\frac{A_1}{A_2}=\frac{{s_1}^2}{{s_2}^2}\)

∴ \(\frac{225}{81}=\frac{{s_1}^2}{12^2}\)

∴ \({s_1}^2=\frac{225×12^2}{81}\)

∴ \(s_1=\frac{15×12}{9}\)     …(Taking square roots on both the sides)

∴ s₁ = 20 cm

Answer is-The corresponding side of the bigger triangle is 20 cm.

ΔABC and ΔDEF are equilateral triangles.

 ∠ A = ∠ B = ∠ C = ∠ D = ∠ E = ∠ F = 60°   …(Angles of equilateral triangle)

In ΔABC and ΔDEF,

∠ A ≅ ∠ D and ∠ B ≅ ∠ E  ….(Each measures 60°)

∴ ΔABC ∼ ΔDEF  ….(AA test of similarity)

By theorem of areas of similar triangles,

\(\frac{A(ΔABC)}{A(ΔDEF)}=\frac{AB^2}{DE^2}\) .....(1)

A(ΔABC) : A(ΔDEF) = 1 : 2 and AB = 4   ….(Given) .....(2)

∴ \(\frac{1}{2}=\frac{4^2}{DE^2}\)   ….[From (1) and (2)]

∴  DE² = 4² x 2

∴ DE = 4\(\sqrt{2}\)    ….(Taking square roots of both the sides)

Answer is DE = 4\(\sqrt{2}\)

A(ΔPQF) = 20 sq units, PF = 2 DP, Let us assume DP = x. ∴ PF = 2x

DF = DP + [PF  ] = [ x ] + [ 2x ] = 3x

In Δ FDE and Δ FPQ,

∠ FDE  ≅ ∠ [ FPQ ] .......... corresponding angles

∠ FED ≅ ∠ [FQP  ] .......... corresponding angles

∴ ΔFDE ~ ΔFPQ .......... AA test

∴\(\frac{A(ΔFDE)}{A(ΔFPQ)}= \frac{[DF^2]}{[PF^2]}=\frac{(2x)^2}{(2x)^2}=\frac{9}{4}\)

A(ΔFDE) = \(frac{9}{4}\) A(ΔFPQ ) = \(frac{9}{4}\) x [20] = [ 45 sq units ]

A([] DPQE) = A(ΔFDE) − A(ΔFPQ)

                  = [ 45 ] – [ 20 ]

                  = [ 25 sq units ]