## Similarity

### Class-10-Mathematics-2-Chapter-1-Maharashtra Board

### Notes

**Practice set 1.1 (Textbook pages 5 and 6)**

**Question 1.1. **

**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 the base, height and area of the first triangle be b_{1}, h_{1} and A_{1} respectively. Let the base, height and area of the second triangle be b_{2}, h_{2} and A_{2} respectively.

b_{1 }= 9, h_{1 }= 5, b_{2 }= 10 and h_{2 }= 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.**

**Question 1.2. **

**In figure ****BC ⊥** **AB, AD ⊥** **AB, BC = 4, AD = 8, ****then find \(\frac{A(ΔABC)}{A(ΔADB)}\)**

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 Δ****ABC and Δ****ADB ****is 1 : 2.**

**Question 1.3. **

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

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**

**Question 1.4.**

**In the given figure, ****AP ⊥** **BC, AD ****|| ****BC, ****then find ****A(Δ****ABC) : A(Δ****BCD).**

**Δ**ABC and **Δ**BCD have same base BC.

The vertices of **Δ**ABC and **Δ**BCD 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(Δ****ABC) : A(Δ****BCD)** **= 1:1.**

**Question 1.5.**

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

**(i)\(\frac{A(ΔPQB)}{A(ΔPBC)}\)** **(ii) \(\frac{A(ΔPBC)}{A(ΔABC)}\)** **(iii) \(\frac{A(ΔABC)}{A(ΔADC)}\) ****(iv) \(\frac{A(ΔADC)}{A(ΔPQC)}\)**

**(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.

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

**Practice set ****1.2(Text Book Page 13)**

**Question 2.1**

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

**(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.**

**Question 2.2**

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

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.**

**Question 2.3**

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

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**

**Question 2.4**

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

** **

∠ 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)

**Question 2.5**

**In trapezium ****ABCD, ****side ****AB ****|| ****side ****PQ ****|| ****side ****DC, AP = 15, PD = 12, QC = 14, ****find ****BQ.**

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**

**Question 2.6**

**Find ****QP ****using given information in the figure****.**

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.**

**Question 2.7**

**In figure, if ****AB ****|| ****CD ****|| ****FE ****then find ***x ***and ****AE.**

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.**

**Question 2.8**

**In Δ****LMN, ****ray ****MT ****bisects ∠** **LMN. ****If ****LM = 6, MN = 10, TN = 8, ****then find ****LT.**

In **Δ**LMN,

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.**

**Question 2.9**

**In Δ****ABC, ****seg ****BD ****bisects ∠** **ABC.**

**If ****AB = ***x***, BC = ***x ***+ 5, AD = ***x ***– 2, DC = ***x ***+ 2, ****then find the value of ***x.*

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)*

*∴* *x ^{2 }+ 2x = x^{2 }+ 5x *

*-*

*2x*

*-*

*10*

*∴** 2x = 3x **-** 10*

*∴* *2x **-** 3x = **-**10*

*∴ **-** x = **-**10 *

*∴** x = 10*

**Answer is The value of ***x*** is 10.**

**Question 2.10**

**In the 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.**

**Proof : **In ΔXDE, PQ || DE ..........[…..]

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

In ΔXEF, QR || EF .......... […..]

*∴* \(\frac{[...]}{[....]}=\frac{[....]}{[...]}\)..........(II) […..]

*∴ * \(\frac{[...]}{[....]}=\frac{[....]}{[...]}\) .......... from (I) and (II)

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

**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)

**Question 2.11**

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

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.**

**Practice set ****1.3(Text Book Page 21)**

** ****Question 3.1**

**In 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.**

In ΔDCE and ΔBCA,

∠ CDE ≅ ∠ CBA (Each measures 75°)

∠ DCE ≅ ∠ BCA (Common angle)

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

**Question 3.2**

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

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

**∴** Δ**PQR ∼ Δ****LMN (SSS test of similarity)**

**Question 3.3**

**As shown in 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 ?**

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.**

**Question 3.4**

**In Δ****ABC, AP ⊥** **BC, BQ ⊥** **AC, B****-****P****-****C, A****-****Q****-****C ****then prove that, Δ****CPA ∼ ****Δ****CQB. ****If ****AP = 7, BQ = 8, BC = 12 ****then find ****AC.**

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**

**Question 3.5**

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

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**

**Question 3.6**

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

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.**

**Question 3.7**

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

**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.**

**Question 3.8**

**In the figure, seg ****AC ****and seg ****BD ****intersect each other in point ****P ****and ****Prove that****, Δ****ABP ∼**** Δ****CDP**

**Proof :**

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

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

ΔABP ∼ ΔCDP (SAS test of similarity)

**Question 3.9**

**In the figure, in Δ****ABC, ****point ****D ****on side ****BC ****is such that****, ∠** **BAC = ∠** **ADC. ****Prove that****, CA ^{2} = CB ×**

**CD**

**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)

**∴ CA ^{2} = CB × CD.**

**Practice set ****1.4(Text Book Page 25)**

**Question 4.1**

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

Let the corresponding sides of two similar triangles be x_{1} and x_{2}. Let their respective areas be A_{1} and A_{2}.

x_{1} : x_{2} = 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)]

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

A_{1 }: A_{2} = 9 : 25

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

**Question 4.2**

**If ****Δ****ABC ∼**** ****Δ****PQR ****and ****AB: PQ = 2:3, ****then fill in the blanks****.**

** \(\frac{A(ΔABC)}{A(ΔPQR)}=\frac{AB^2}{[...]}=\frac{2^2}{3^2}=\frac{[...]}{[...]}\)**

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

**Question 4.3**

**If ****Δ****ABC ∼**** ****Δ****PQR, A (****Δ****ABC) = 80, A (****Δ****PQR) = 125, ****then fill in the blanks.**

** \(\frac{A(ΔABC)}{A(Δ....)}=\frac{80}{125}\) **

**∴ \(\frac{AB}{PQ}=\frac{[...]}{[...]}\)**

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

∴ \(\frac{AB}{PQ}=\frac{[4]}{[5]}\)

**Explanation :**

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

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

∴ \(\frac{AB}{PQ}=\frac{[4]}{[5]}\)

**Question 4.4**

**Δ****LMN ∼**** ****Δ****PQR, 9 x** **A(****Δ****PQR ) = 16**** x** **A(****Δ****LMN). ****If ****QR = 20 ****then find ****MN.**

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**

**Question 4.5**

**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.**

Let the areas of two similar triangles be A_{1} and A_{2}. Let their respective corresponding sides be s_{1} and s_{2}.

A_{1} = 225 sq cm, A_{2} = 81 sq cm and s_{2} = 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)

∴ s_{1} = 20 cm

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

**Question 4.6**

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

ΔABC and **Δ**DEF 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)]

∴ DE^{2} = 4^{2} x 2

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

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

**Question 4.7**

**In figure seg ****PQ ****|| ****seg ****DE, A(****Δ****PQF) = 20 ****units****, PF = 2 DP, ****then find ****A([]**** ****DPQE) ****by completing the following activity****.**

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

**DF = DP + [...] = [...] + [...] = 3***x*

**In ****Δ** **FDE ****and ****Δ** **FPQ,**

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

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

**∴** **Δ****FDE ∼**** ****Δ****FPQ .......... ****AA test**

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

**A(****Δ****FDE) = ** ** \(frac{9}{4}\) A(****Δ****FPQ ) = \(frac{9}{4}\)**** ****x [...] = [ ..... ]**

**A([]**** D****PQE) = A(****Δ****FDE) − A(****Δ****FPQ)**

** = [ ... ] – [... ]**

** = [ .... ]**

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

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

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 ]

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