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MATHEMATICS–X TRIANGLES 87

CHAPTER 6TRIANGLES

Points to Remember :1. Two figures having the same shape but not necessarily the same size are called similar figures.2. All the congruent figures are similar but the converse is not true.3. Two polygons having the same number of sides are similar, if

(i) their corresponding angles are equal and(ii) their corresponding sides are proportional (i.e. in the same ratio)

4. Basic proportionality theorem (Thales theorem) : If a line is drawn parallel to one side of a triangle tointersect the other two sides in distinct points, then the others two sides are divided in the same ratio.

5. Converse of Thales’ theorem : If a line divides any two sides of a triangle in the same ratio, then the lineis parallel to the third side of the triangle.

6. The line drawn from the mid-point of one side of a triangle is parallel to another side bisects the thirdside.

7. The line joining the mid-points of two sides of a triangle is parallel to the third side.8. AAA similarity criterion : If in two triangles, corresponding angles are equal, then the triangles are

similar.9. AA similarity criterion : If in two triangles, two angles of one triangle are respectively equal to the two

angles of the other triangle, then the two triangles are similar.10. SSS similarity criterion : If in two triangles, corresponding sides are in the same ratio, then the two

triangles are similar.11. SAS similarity criterion : If one angle of a triangle is equal to one angle of another triangle and the

sides including these angles are in the same ratio, then the triangles are similar.12. The ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.13. If the areas of two similar triangles are equal, then the triangles are congruent.14. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then

the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other.15. Pythagoras Theorem : In a right triangle, the square of the hypotenuse is equal to the sum of the

squares of the other two sides.16. Converse of Pythagoras Theorem : If in a triangle, square of one side is equal to the sum of the squares

of the other two sides, then the angle opposite to first side is a right angle.

ILLUSTRATIVE EXAMPLES

Example 1. In the given figure, DE || OQ and DF || OR. Prove that EF || QR. [NCERT]

Solution. In POQ, DE || OR, by Thales’ theorem,

We have PE PD=EQ DO ...(1)

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88 TRIANGLES MATHEMATICS–X

Again, In POR, DF || OR, by Thales’ theorem,

We have, PF PD=FR DO ...(2)

from (1) and (2), we get

P

R

ED

O

F

QPE PF=EQ FR

EF || QR ( converse of Thales’s theorem)Example 2. Prove that the diagonals of a trapezium divide each other proportionally.

ORABCD is a trapezium such that AB || DC. The diagonals AC and BD intersect at O. Prove that

=AO COBO DO

. [CBSE 2004, NCERT]

Solution. Through O, draw OE || CD (or AB). In ABD, EO || AB, by Thales’ theorem,

We have AE BO=ED DO ...(1)

Again, In ADC, EO || DC, by Thales’ theorem, O

A

E

CD

B

We have AE AO=ED OC ...(2)


BO AO AO CO= =DO OC BO DO

Hence proved.

Example 3. If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium. [CBSE 2005]

OR

The diagonals of a quadrilateral ABCD intersect each other at the point O such that = .AO COBO DO

Show that ABCD is a trapezium. [NCERT]Solution. Through O, draw OE parallel to AB intersecting BC at E.

Now, in ABC, OE || AB ( by construction)

AO BE=OC EC ...(1) ( Thales’ theorem)

CD

O

BA

E

Now, AO CO=BO DO

AO BO=OC DO ...(2) ( given)


BE BOEC DO

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Now, In ABC, we have

BO BE=DO CE (proved above)

OE || DC ( Converse of Thales’ theorem)Now, In quadrilateral ABCD, OE || AB and OE || DC AB || DC. Quadrilateral ABCD is a trapezium.

Example 4. In the given figure, P is the mid-point of BC and Q is the mid point of AP. If BQ when produced

meets AC at R, prove that 1RA = CA3

. [CBSE 2006 C]

Solution. Draw PS || BR, meeting AC at S.In BCR, P is the mid-point of BC and PS || BR. S is the mid-point of CR. CS = SR ...(1)In APS, Q is the mid-point of AP and QR || PS.

A

S

R

C

Q

PB R is the mid-point of AS. AR = RS ...(2)from (1) and (2), we get

AR = RS = SCNow, AC = AR + RS + SC= 3AR

1AR= CA3 Hence proved.

Example 5. In the given figure, altitudes AD and CE of ABC intersect each other at the point P. Show that:(i) AEP ~ CDP (ii) ABD ~ CBE(iii) AEP ~ ADB (iv) PDC ~ BEC [NCERT]

Solution. (i) In AEP and CDP, we haveAEP = CDP (each = 90°)APE = CPD (vertically opposite angles)

AEP ~ CDP (AA similarity criterion)(ii) In ABD and CBE, we have

ADB = CEB (each = 90°)ABD = CBE (common)

ABD ~ CBE (AA similarity criterion)

C

D

B

P

EA(iii) In AEP and ADB, we have

A = A (common)AEP = ADB (each = 90°)

AEP ~ ADB (AA similarity criterion)(iv) In PDC and BEC, we have

PCD = BCE (common)CDP = CEB (each = 90°)

PDC ~ BEC (AA similarity criterion)

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Example 6. CD and GH (D and H lie on AB and FE) are respectively bisectors of ACB and EGF and

ABC ~ FEG. Prove that : (i) CD AC=GH FG

(ii) DCB ~ HGE (iii) DCA ~ HGF. [NCERT]

Solution. (i) Given ABC ~ FEG A = Fand C = G ...(1)

1 1C G2 2

C

BDA

12

G

EHF

43

1= 3and 2 = 4 ...(2)

[ CD and GH are bisector of C and G respectively]Now, In ACD and FGH, we have

A= F [from (1)]2 = 4 [from (2)]

ACD ~ FGH (AA similarity criterion) DCA ~ HGF(ii) We have, ACD ~ FGH

AC CD=FG GH

(iii) In DCB and HGE, we have1 = 3 [from (2)]B = E ( ABC ~ FEG)

DCB ~ HGE (AA Similarity criterion)

Example 7. In the given figure,OA ODOC OB

. Prove that A = C and B = D

Solution. In AOD and COB, we have

OA OD=OC OB ( given)

OA OC=OD OB

and, 1 = 2 ( vertically opposite angles)

AOD ~ COB ( SAS similarity criterion)

A = C and B = D

Example 8. Two poles of height a metres and b metres are p metres apart. Prove that the height of the pointof intersection of the lines joining the top of each pole to the foot of the opposite pole is given by

aba b

metres.AMIT BAJA

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Solution. Let AB and CD be two poles of height a metres and b metres respectively such that poles are pmetres apart. Let the lines AD and BC meet at O such that OE = h metres.

D

BO

hC A

ab

x E y

Let CE = x and EA = y, so that x + y = p.Now, In ABC and EOC, we have

CAB = CEO (each = 90°)C = C (common)

CAB ~ CEO (AA similarity criterion)

CA ABCE EO

p a phxx h a

...(1)

In AEO and ACD, we haveAEO = ACD (each = 90°)A = A (common)

AEO ~ ACD (AA similarity criterion)

AE OEAC CD

y h phyp b b

...(2)

Now, x + y = p

ph ph pa b [using (1) and (2)]

1 1 1 1 1ph pa b h a b

1 a b abhh ab a b

Hence, the height of the intersection of the lines joining the top of each pole to the foot of the

opposite pole is ab

a b metres.

Example 9. Prove that area of an equilateral triangle described on one side of a square is equal to half thearea of the equilateral triangle described on one of its diagonal. [NCERT]

Solution. Let ABCD be the given square. Equilateral triangles BCE and ACF are described on side BCand diagonal AC respectively.

F C

E

D

BA

Since BCE and ACF are equilateral they are equiangular and hence BCE ~ ACF(AAA similarity criterion)

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2

2

( BCE) BC( ACF) AC

arar

2

2

BC( 2BC)

[ ABCD is a square diagonal = 2 (side) AC = 2 . BC]

2

2

BC 122.BC

( BCE) 1 .( ACF) 2

arar

Hence proved.

Example 10. In the given figure, XY || AC and XY divides triangular region ABC into two parts equal in area.

Determine AX .AB

Solution. We have, XY || AC and ar (BXY) = ar (quad. XYCA) ar (ABC) = 2.ar (BXY) ...(1)Now, XY || AC and BA is a transversal. BXY= BAC ...(2)Thus, In BAC and BXY, we have

XBY = ABC (common)BXY = BAC [from (2)]

BAC ~ BXY (AA similarity criterion)

A

C

X

YB

2

2

( BAC) BA( BXY) BX

arar

2

2

BA2BX

[using (1)]

BA 2.BX

BA 2(BA–AX)

( 2 1)BA 2AX

AX 2 – 1=AB 2

Ans.

Example 11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the sametime, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per

hour. How far apart will be the two planes after 112

hours? [NCERT]

Solution. Distance travelled by an aeroplane due north in 112

hours 31000 km = 1500 km2

OA = 1500 km

Distance travelled by an aeroplane due west in 112

hours = 1200 × 32

km = 1800 km

OB = 1800 km

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Now, In right angled AOB, AB2 = OA2 + OB2 (using pythagoras theorem)

= (1500)2 + (1800)2

= 2250000 + 3240000 = 5490000

A

B

North

West

1500 km

1800 km O AB 5490000 km 3 3 61 100 100 km

= 300 61 km Ans.Example 12. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of

its diagonals. [NCERT]OR

ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2. [CBSE 2005]Solution. Let the diagonals AC and BD of rhombus ABCD intersect at O. Since the diagonals of a rhombus

bisect each other at right angle. AOB = BOC = COD = DOA = 90° and AO = CO, BO = DO.Since AOB is a right triangle right-angled at O. AB2 = OA2 + OB2 (using Pythagoras theorem)

D C

BAO

2 21 1AC BD

2 2

[ OA = OC and OB = OD]

4AB2 = AC2 + BD2 ...(1)Similarly, we have

4BC2 = AC2 + BD2 ...(2)4CD2 = AC2 + BD2 ...(3)

and 4AD2 = AC2 + BD2 ...(4)adding all these results, we get

4 (AB2 + BC2 + CD2 + AD2) = 4 (AC2 + BD2) AB2 + BC2 + CD2 + AD2 = AC2 + BD2

Hence proved.

Example 13. In an equilateral triangle ABC, D is a point on side BC such that 1BD = BC.3

Prove that

9AD2 = 7AB2. [NCERT]Solution. Draw AP BC.

Since AP BC 1 1BP BC and BD BC2 3

1 1 1DP BC BC2 3 6

A

CPB D

Now, In right angled APB,

AB2 = BP2 + AP2

AB2 = BP2 + AD2 – DP2 ( AP2 = AD2 – DP2)AMIT B

AJAJ


2 2

2 21 1AB BC AD BC2 6

2 2 2 21 1AB AB AD AB

4 36 ( AB = BC = AC)

2 2 2

2 9AB +36AD –ABAB =36

36AB2 = 8AB2 + 36AD2

28AB2 = 36AD2

7AB2 = 9AD2

Hence proved.Example 14. In the given figure, O is a point in the interior of a triangle ABC, OD BC, OE AC and

OF AB. Show that(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2

(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2 [NCERT]Solution. (i) In right OFA, ODB and OEC,

using Pythagoras theorem, we haveOA2 = AF2 + OF2 ...(1)OB2 = BD2 + OD2 ...(2)

A

CDB

OF E

OC2 = CE2 + OE2 ...(3)Adding (1), (2) and (3), we get

OA2 + OB2 + OC2 = AF2 + BD2 + CE2 + OF2 + OD2 + OE2

OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2

which proves the (i) part.(ii) In right ODB and ODC, we have

OB2 = OD2 + BD2

and OC2 = OD2 + CD2

OB2 – OC2 = BD2 – CD2 ...(4)Similarly, we have

OC2 – OA2 = CE2 – AE2 ...(5)and OA2 – OB2 = AF2 – BF2 ...(6)Adding (4), (5) and (6), we get

(OB2 – OC2)+ (OC2 – OA2) + (OA2 – OB2)= (BD2 – CD2) + (CE2 – AE2) + (AF2 – BF2) 0 = (BD2 + CE2 + AF2) – (AE2 + CD2 + BF2) AF2 + BD2 + CE2 = AE2 + CD2 + BF2 , which proves the (ii) part.

Example 15. ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length ofperpendicular from C on AB, prove that :

(i) cp = ab (ii) 2 2 2

1 1 1= +p a b

[CBSE 2002]AMIT B

AJAJ


Solution. (i) ar (ABC) 1 AB CD2

12

cp ...(1)

also, ar (ABC) 1 BC AC2

A

b

C

pc

Ba

D

12

ab ...(2)


1 1 ,2 2

cp ab cp ab which proves (i) part.

Since ABC is a right triangle right-angled at C. AB2 = BC2 + AC2 c2 = a2 + b2

22 2ab a b

p

abcp ab cp

2 22 2

2

a b a bp

2 2

2 2 2

1 a bp a b

2 2

2 2 2 2 2

1 a bp a b a b

2 2 2

1 1 1p b a

2 2 2

1 1 1p a b

, which proves the (ii) part.

PRACTICE EXERCISE

Questions based on Basic Proportionality Theorem :

1. In the given figure, DE || BC and AD 3 .DB 5

If AC = 4.8 cm, find AE. [CBSE 2002, 2003]

A

D E

B CAMIT B

AJAJ


2. In the given figure, find the value of x if DE || AB.

A

D E

B

C4 x–3

3 –19xx–4

3. In the given figure, DE || BC. If :(i) AD = 4 cm, DB = 6 cm and AC = 12.5 cm, find AE.(ii) AD : DB = 3:5 and AE = 3.6 cm, find AC.(iii) AE : EC = 5 : 3 and AB = 15.6 cm, find DB.

A

D E

B C(iv) AE : AC= 5 : 9 and DB = 7.2 cm, find AB.(v) If AB = 2 AD and AE = 4.8 cm, find EC.

4. In a ABC, D and E are points on the sides AB and AC respectively such that DE || BC.(i) If AD = 4, AE = 8, DB = x – 4, and EC = 3x – 19, find x.(ii) If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x. [CBSE 2002](iii) If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find x.(iv) If BD = x – 3, AB = 2x, CE = x – 2, AC = 2x + 3, find x.

5. In a ABC, D and E are points on the sides AB and AC respectively. For each of the following casesshow that DE || BC :(i) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.(ii) AB = 11.2 cm, DB = 2.8 cm, AC = 14.4 cm and AE = 10.8 cm(iii) AB = 24 cm, AD = 15 cm, AE = 22.5 cm and AC = 36 cm(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm

6. In a ABC, D and E are points on AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm,DE = 2 cm and BC = 5 cm, find BD and CE. [CBSE 2001C]

7. M and N are points on the sides PQ and PR respectively of PQR. For each of the following cases, statewhether MN || QR.(i) PM = 12 cm, QM = 15 cm, PN = 10 cm and NR = 12.5 cm(ii) PQ = 1.8 cm, QM = 1.2 cm, PN = 2.7 cm and PR = 3.6 cm(iii) PQ = 1.25 cm, PR = 2.5, PM = 0.17 cm and PN = 0.34 cm(iv) PQ = 30 cm, PR = 36 cm, PM = 12 cm and NR = 21.6 cm.

8. In the given figure, EF || AB || DC. Prove that AE BF= .ED FC

A B

E F

D COR

Prove that a line-segment drawn parallel to the parallel sides of a trapezium divides the non-parallel sidesproportionally.

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9. Let X be any point on the side BC of ABC (see figure given below). If XM, XN are drawn parallel to BAand CA meeting CA, BA in M, N respectively; MN meets BC produced in T, prove that TX2 = TB × TC.

XB C

A

MN

T

10. In the given figure, DE || AC and DC || AP. Prove that BE BCEC CP

. [CBSE 2005]

CE P

D

A

B11. Let ABC be a triangle and D and E be two points on side AB such that AD = BE (see figure). If DP || BC

and EQ || AC, then prove that PQ || AB.A

C

D

E

Q

P

B

12. Any point X inside DEF is joined to its vertices. From a point P in DX, PQ is drawn parallel to DEmeeting XE at Q and QR is drawn parallel to EF meeting XF in R. Prove that PR || DF.

13. If three or more parallel lines are intersected by two transversals, prove that the intercepts made by themon the transversals are proprotional.

14. ABC is a triangle in which AB = AC. Points D and E are points on the sides AB and AC respectively suchthat AD = AE. Show that the points B, C, E and D are concyclic (lie on the same circle).

15. Use the given figure to find the value of x if :(i) OA = 4x – 2, OB = 4x + 1, OC = 2x + 2, OD = 3x – 1 O

D C

BA(ii) OA = 3x – 1, OB = 2x + 1, OC = 5x – 3, OD = 6x – 5

16. In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.Show that BC || QR.

P

A

O

B C

Q R17. ABCD is a quadrilateral. P, Q, R and S are the points of trisection of sides AB, BC, CD and DA respec-

tively and are adjacent to A and C. Prove that PQRS is a parallelogram.

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18. In the given figure, D is a point on AB and E is a point on AC of ABC such that DE || BC. If AD : DB =1:2, AB = 18 cm and AC = 30 cm, find the values of AE, EC, AD and DB.

A

E

CB

D

19. In the given figure, DE || AB and FE || DB.Prove that DC2 = CF . AC [CBSE 2007]

C

F

ED

BA20. In the given figure, DE || BC and BD = CE. Prove that ABC is an isosceles triangle.

C

ED

B

A

21. In the given figure, A = B and AD = BE. Show that AB || DE.

B

ED

A

C

22. Using basic proportionality theorem, prove that line segment joining the mid-points of any two sides ofa triangle is parallel to the third side.

23. Determine a point on a line segment AB = 8 cm, which divides AB internally in the ratio 2 : 3. Prove yourassertion.

24. In the given figure, ABC is a triangle; and P, Q and R are points on BC, CA and AB respectively such thatPQ || AB and QR || BC. Prove that RP || CA, if R is the mid-point of AB.

C

QR

B

A

P25. ABCD is a parallelogram. P is a point on side BC, and DP when produced meets AB produced at L. Prove

that :(a) DP : PL = DC : BL (b) DL : DP = AL : DC

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D C

P

LBA

Questions based on Criteria for similarity of Triangles :

26. In the given figure, PQ || RS. Find PT and QT.

3 cm 2 cm3.8 cm

3.2 cm

RP

Q S

T

27. In the given figure, ABC = 90° and BD AC. If BD = 8cm and AD = 4 cm, find CD.

C

A

B

D4 cm

8 cm

28. In the given figure, AB || QR. Find PB.

R

BA

Q

P

9 cm

6 cm3 cm

29. In the given figure, DE || BC, AD = 2 cm, BD = 2.5 cm, AE = 3.2 cm and DE = 4 cm. Determine AC and BC.

C

ED

B

A

2.5 cm4 cm

2 cm 3.2 cm

30. In the given figure, find x in terms of a, b and c.

C

E

B

A

a

53° 53°x

b cD31. The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, find

AB.

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32. A vertical stick 15 m long casts its shadow 10 m long on the ground. At the same time, the flag pole castsa shadow 60 m long. Find the height of the pole.

33. In the given figure, AQ and PB are perpendiculars to AB. If AO = 10 cm, BO = 6cm and PB = 9 cm, find AQ.

P

BOA

Q34. P and Q are points on the sides AB and AC respectively of a triangle ABC. If AP = 3 cm, PB = 6 cm,

AQ = 4.5 cm and QC = 9 cm; prove that 1PQ BC.3

35. In the given figure, ABC is a triangle, right angled at B. If FG || DE || CB and AG = GE = EB, find DE + FG.

C

BA

F

G

D

E

12 cm

36. In the given figure, EDC ~ EBA, BEC = 115° and EDC = 70°. Find :(i) DEC(ii) DCE(iii) EAB

D C

BA

E

70°

115°(iv) AEB(v) EBA

37. In the given figure, AO BO 1=OC OD 2

and AB = 5 cm. Find DC.

O

A B

CD

5 cm

38. In ABC, DE is parallel to base BC, with D on AB and E on AC. If AD 2 BC, find .DB 3 DE

[CBSE 2002C]

C

ED

B

AAMIT BAJA

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39. In the given figure, AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm. Find xand y.

B

F

EA C

D6 cm

4 cm10 cm

x cmy cm

40. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is3.6 m above the ground, find the length of her shadow after 4 seconds.

41. D is a point on the side BC of ABC such that ADC = BAC. Prove that CA CB=CD CA or, CA2 = CB × CD.

[CBSE 2004]42. Two right triangles ABC and DBC are drawn on the same hypotenuse BC on the same side of BC. If AC

and DB intersect at P, prove that AP × PC = BP × PD. [CBSE 2000]43. In a ABC, AB = AC and D is a point on side AC, such that BC2 = AC × CD. Prove that BD = BC.

44. In the given figure, BC AB, DE AC and DA AB. Prove that : DE × BC = AE × AB.C

D

A BE

45. In the given figure, PA, QB and RC are each perpendicular to AC and AP = x, QB = z, RC = y, AB = a and

BC = b. Prove that 1 1 1 .x y z

P

QR

BA C

yzx

a b46. In ABC, if AD BC and AD2 = BD × CD, prove that BAC = 90°. [CBSE 2004, 2006]

A

B CD

47. In a right angled triangle ABC, A = 90° and AD BC. Prove that AD2 = BD × CD. [CBSE 2004 (C)]48. In the given figure, AB || DE and BD || EF. Prove that DC2 = CF × AC. [CBSE 2004 (C)]

F

B

C

D E

A

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49. In an isosceles ABC, the base AB is produced both ways to P and Q such that AP × BQ = AC2. Provethat ACP ~ BCQ.

B

C

AP Q50. ABCD is a trapezium with AB || DC. If AED ~ BEC, prove that AD = BC.

E

D C

BA

51. In a quadrilateral ABCD, AB || CD and E and F are the mid-points of AB and CD respectively. If thediagonal BD intersects EF in G, prove that DF.BG = EB.GD.

52. In a given figure, the diagonal BD of a parallelogram ABCD intersects the segment AE at F, where E is anypoint on the side BC. Prove that DF.EF = BF.AF.

A D

CB E

F

53. In the given figure, QT QRPR QS

and 1 = 2. Prove that PQS ~ TQR.

P

R

T

Q S1 2

54. CD and GH are respectively the medians of ABC and EFG. If ABC ~ FEG, prove that

(i) ADC ~ FHG (ii) CD AB=GH FE (iii) CDB ~ GHE

55. Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting ACin L and AD produced in E. Prove that EL = 2.BL.

A D

CB

E

L MAMIT BAJA

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56. Prove that the line segments joining the mid-points of the sides of a triangle form four triangle, each ofwhich is similar to the original triangle.

57. In the given figure, FEC GDB and 1 = 2. Prove that ADE ~ ABC.A

ED

F B GC

21

58. If the angles of one triangles are respectively equal to the angles of another triangle, prove that the ratioof their corresponding sides is the same as the ratio of their corresponding :(i) medians (ii) angle bisectors (iii) altitudes

59. In the given figure, ABC is a right triangle right angled at B and D is the foot of the perpendicular drawnfrom B on AC. If DM BC and DN AB, prove that :(i) DM2 = DN × MC (ii) DN2 = DM × AN

A

C

N

B M

D

60. In the given figure, DEFG is a square and BAC = 90°. Prove that :(i) AGF ~ DBG (ii) AGF ~ EFC(iii) DBG ~ EFC (iv) DE2 = BD × EC

E

A

DB C

G F

Questions based on Areas of similar Triangles :

61. D and E are points on the sides AB and AC respectively of ABC such that DE is parallel to BC andAD : DB = 4:5. CD and BE intersect each other at F. Find the ratio of the areas of DEF and BCF.

[CBSE 2003]62. The areas of two similar triangles are 100 cm2 and 49 cm2 respectively. If the altitude of the bigger triangle

is 5 cm, find the corresponding altitude of the other. [CBSE 2002]63. The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is

12.1 cm, find the corresponding median of the other. [CBSE 2001]64. In ABC, P divides the side AB such that AP : PB = 1 : 3. Q is a point on AC such that PQ || BC. Find the

ratio of the areas of APQ to trapezium BPQC.

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65. ABC is an isosceles triangle right angled at B. Two equilateral triangles are constructed with side BC and

AC as shown. Prove that : 1( BCD) ( ACE)2

ar ar . [CBSE 2001]

C

BA

D

E

66. In the given figure, DE || BC and AD : DB = 5 : 4. Find ( DFE)( CFB)

arar

. [CBSE 2000]

A

CB

DE

F

67. In the given figure, DE || BC and DE : BC = 4 : 5. Find ratio of the areas of ADE and trapezium BCED.

A

D E

B C

68. If the area of two similar triangles are equal, prove that they are congruent.69. Prove that area of an equilateral triangle described on the side of a square is half the area of the

equilateral triangle described on its diagonal.70. Equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the

hypotenuse is equal to the sum of the areas of triangles on the other two sides. [CBSE 2002]71. ABCD is a trapezium in which AB || DC and AB = 2DC. Determine the ratio of the areas of AOB and

COD.72. In the trapezium ABCD, AB || CD and AB = 2CD. If the area of AOB = 84 cm2, find the area of COD.

[CBSE 2005]73. D and E are points on the sides AB and AC respectivley of a ABC such that DE || BC and divides ABC

into two parts, equal in area, find BD .AB

[CBSE 2000]

74. In the given figure, ABC and DBC are on the same base BC. If AD and BC intersect at O, prove that

( ABC) AO( DBC) DO

arar

. [CBSE 2002, 2005]

A C

DBOAMIT B

AJAJ


75. In ABC, D is a point on AB, E is a point on AC and DE || BC. If AD : DB = 5 : 4, find :

(i) DE, if BC = 27 cm

(ii) ar (ADE) : ar (ABC)

A

D E

B C(iii) ar (ADE) : ar (trapezium DBCE)

Questions based on Pythagoras Theorem and its converse76. The sides of a some triangles are given below. Determine which of them are right triangles.

(i) 15 cm, 20 cm, 25 cm (ii) 5 cm, 12 cm, 13 cm(iii) 7 cm, 24 cm, 25 cm (iv) 5 cm, 8 cm, 12 cm(v) 8 cm, 15 cm, 17 cm (vi) 9 cm, 40 cm, 41 cm

77. Show that the ABC is right triangle, if :(i) AB = 24 cm, BC = 25 cm and AC = 7 cm (ii) AB = 12 cm, BC = 16 cm and AC = 20 cm

78. In PQR, Q = 90°, find :(i) PQ, if PR = 34 cm and QR = 30 cm (ii) PR, if PQ = 24 cm and QR = 7 cm

79. In PQR, QPR = 90° and QR = 26 cm. If PS SR, PS = 6 cm and SR = 8 cm, find the area of PQR.P

RQ

6 cm

8 cmS

26 cm80. A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the

foot of the ladder from the building.81. A ladder 25 m long reaches a window which is 24 m above the ground on side of the street. Keeping the

foot at the same point, the ladder is turned to the other side of the street to reach a window 7 m high. Findthe width of the street.

82. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance betwen their feet is 12 m, findthe distance between their tops. [CBSE 2002 C]

83. In an equilateral ABC, AD BC, prove that AD2 = 3BD2.84. The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.85. In an equilateral triangle with side a, prove that :

(i) Altitude 3

2a [CBSE 2001 C]

(ii) Area 234

a [CBSE 2002]

86. In the given figure, ABC is a right triangle right-angled at B. AD and CE are the two medians drawn from

A and C respectively. If AC = 5 cm and AD 3 5 cm.

2 Find the length of CE.

A

C

E

B D

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87. ABC is an isosceles triangle right angled at C. prove that AB2 = 2AC2. [CBSE 2005]88. ABC is an isosceles triangle with AB = BC. If AB2 = 2AC2, prove that ABC is a right triangle.

[CBSE 2004 (C)]89. In a quadrilateral ABCD, CA = CD, B = 90° and AD2 = AB2 + BC2 + CA2. Prove that ACD = 90°.

A

C

B

D

90. In the given figure, BCA = 90°. Q is the mid-point of BC. Prove that : AB2 = 4AQ2 – 3AC2.[CBSE 2004 (C)]

A

CB Q91. In the given figure, ABCD is a square. F is the mid-point of AB, BE is one-third of BC. If the area of FBE

= 108 cm2, find the length of AC.D C

A

E

F B92. Prove that three times the square of any side of an equilateral triangle is equal to four times the square

of the altitude. [CBSE 2002]93. ABC is an obtuse triangle, obtuse angled at B. If AD CB, prove that AC2 = AB2 + BC2 + 2BC × BD.94. B of ABC is an acute angle and AD BC, prove that AC2 = AB2 + BC2 – 2BC × BD.95. P and Q are the mid-points of the sides CA and CB respectively of a ABC, right angled at C. Prove that:

(i) 4 AQ2 = 4AC2 + BC2

(ii) 4BP2 = 4BC2 +AC2 [CBSE 2001](iii) 4 (AQ2 + BP2) = 5AB2 [CBSE 2001, 2006 (C)]

96. ABC is a triangle in which AB = AC and D is any point in BC. Prove that AB2 – AD2 = BD . CD.[CBSE 2005]

97. The perpendicular AD on the base BC of a ABC intersects BC at D so that DB = 3CD. Prove that2AB2 = 2AC2 + BC2. [CBSE 2005]

98. ABC is a right triangle right-angled at B. Let D and E be any points on AB and BC respectively. Prove thatAE2 + CD2 = AC2 + DE2. [CBSE 2002 (C)]

99. In a ABC, AD BC and AD2 = BC × CD. Prove that ABC is a right triangle. [CBSE 2006 (C), 2007]100. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of squares of

its sides.101. A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that

OB2 + OD2 = OC2 + OA2. [CBSE 2006 (C)]

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102. In the given figure, D and E trisect BC. Prove that :8AE2 = 3AC2 + 5AD2 [CBSE 2006(C)]

A

B D E C103. ABCD is a rhombus. Prove that : AB2 + BC2 + CD2 + DA2 = AC2 + BD2. [CBSE 2005, 2006]104. In a ABC, AC > AB, D is the mid-point of BC and AE BC. Prove that :

(i) AC2 = AD2 + BC.DE + 14

BC2

(ii) AB2 = AD2 – BC.DE + 14

BC2

DE

A

B C(iii) AB2 + AC2 = 2AD2 +

12

BC2 [CBSE 2006(C)]105. P and Q are points on the side CA and CB respectively of ABC, right angled at C, prove that :

AQ2 + BP2 = AB2 + PQ2.

HINTS TO SELECTED QUESTIONS

9. In TXM, XM || BN TB TNTX TM

...(1)

In TMC, XN || CM TX TNTC TM

...(2)

from (1) and (2), we will get desired result.17. In ADC,

DS 2SA DR 2RC2 and, 2SA SA RC RC

D

RCQ

BPA

S

DS DR SR || ACSA RC

.

Similarly, PQ || ACAlso, by joining BD, we can prove that PS || RQ.

25. (a) In ALD, BP || AD BL LP .BA PD

Now use the fact that AB = CD.

41. In ABC and DAC,ADC = BAC and C= C ABC ~ DAC ( AA similarity)

AB BC AC= = etc.DA AC DC

45. Clearly, ABQ ~ ACR z ay a b

...(1)

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also, CBQ ~ CAP z bx a b

...(2)

adding (1) and (2), we get

1 1 1 1 11z z a b zy x a b x y x y z

55. Clearly, BMC EMD ( AAS congruence rule) BC = DE (cpct)Then, AE = 2BCAlso, AEL ~ CBL ( AA Similarity)

EL AE 2BC 2 EL 2BLBL CB BC

59. Clearly, quadrilateral BMDN is a rectangle. Then, BM = NDhere, BMD ~ DMC ( AA similarity)Also, BND ~ DNA ( AA similarity)

74. Draw AL BC and DM BC.Clearly, ALO ~ DMO ( AA similarity)

AL AODM DO

...(1)

1 BC ALar ( ABC) AL AO21ar ( DBC) DM DOBC DM2

[ using (1)]

75. (i) In ABC, DE || BC ADE ~ ABC.

AD DE 5 DE DE 15cmAB BC 5 4 27

(ii) 22

2

ar ( ADE) AD 5 25 :81ar ( ABC) 9AB

(iii) Consider, ar (trap. DBCE) ar ( ABC) ar( ADE)ar ( ADE) ar ( ADE)

ar ( ABC) 81 561 1ar ( ADE) 25 25

[ using (ii)]

ar ( ADE) 25 : 56ar (trap. DBCE)

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93. In ADB, AB2 = AD2 + DB2 ...(1)

In ADC, AC2 = AD2 + DC2

= AD2 + (DB + BC)2

A

DB C

= (AD2 + DB2) + BC2 + 2BC.BD

= AB2 + BC2 + 2BC.BD [using (1)]

100. We know that if AD is a median of ABC, then AB2 + AC2 = 2AD2 21 BC .2

Since diagonals of a parallelogram bisect each other.

BO and DO are medians of ABC and ADC respectively.

AB2 + BC2 = 2 BO2 + 21 AC2

...(1)

D C

B

O

A

and, AD2 + CD2 = 2DO2 + 21 AC2

...(2)

add (1) and (2), and simplify now.

102. Here, BD = DE = CE = x (say). Then, BE = 2x and BC = 3x.

In right ABD, ABE and ABC,

AD2 = AB2 + x2 ...(1)

AE2 = AB2 + 4x2 ...(2)

and AC2 = AB2 + 9x2 ...(3)

Consider, 8 AE2 – 3 AC2 – 5AD2. Using (1), (2) and (3) in this expression, we get

8AE2 – 3AC2 – 5AD2 = 0 8AE2 = 3AC2 + 5AD2.

104. We have AED = 90°, ADE < 90° and ADC > 90°.

(i) In ADC, ADC is an obtuse angle.

AC2 = AD2 + DC2 + 2DC.DE

22 1 1AD BC 2. BC.DE.

2 2

Simplify now..

(ii) In ABD, ADE is an acute angle.

AB2 = AD2 + BD2 – 2BD . DE

22 1 1AD BC 2. BC.DE.

2 2

Simplify now..

(iii) Adding (i) and (ii), we will get the desired result.

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MULTIPLE CHOICE QUESTIONS

Mark the correct alternative in each of the following :1. D and E are respectively the points on the sides AB and AC of a ABC and DE || BC, such that AD = 3

cm, AB = 12 cm, AE = 4 cm, then the value of CE is :(a) 6 cm (b) 9 cm (c) 12 cm (d) 15 cm

2. In the given figure, PQ || DC || AB. The value of x is :A B

P Q

D C

x x–1

xx+2

(a) 1 (b) 2 (c) 3 (d) 4

3. In the given figure, DE || BC, BE || XC and AD 2 ,DB 1

then the value of AX is:XB

A

D E

CB

X(a) 2 : 1 (b) 3 : 1 (c) 3 : 2 (d) none of these

4. In the given figure, AB || CD. Then the value of x is :D C

A B

4

3 –19x

x–3

x+4

(a) 8 (b) 11 (c) 8 or 11 (d) none of these5. In the given figure, ACB ~ APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, the value of AC

is:B P

QCA

(a) 5.6 cm (b) 2.8 cm (c) 8 cm (d) none of these6. A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time a tower casts the

shadow 40 m long on the ground. The height of the tower is :(a) 50 m (b) 60 m (c) 80 m (d) none of these

7. The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 8 cm, thenthe value of AB is :(a) 12 cm (b) 15 cm (c) 18 cm (d) none of these

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8. In the given figure, BAC = ADB= 90°. The length of DC is :

D

A

B C

5 cm

3 cm

4 cm(a) 2 cm (b) 2.25 cm (c) 2.5 cm (d) 2.75 cm

9. The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, the length of QR is:(a) 5 cm (b) 5.5 cm (c) 6 cm (d) none of these

10. The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. If the altitude of the bigger triangleis 4.5 cm, the length of corresponding altitude of smaller triangle is :(a) 3 cm (b) 4 cm (c) 3.5 cm (d) 4.5 cm

11. Two isosceles triangles have equal vertex angles and their areas are in the ratio 9 : 16. The ratio of theircorresponding altitudes is :(a) 3 : 4 (b) 9 : 16 (c) 81 : 256 (d) none of these

12. In the given figure, DE || BC and AD : DB = 4 : 5. Then the value of ar ( DEF)ar ( BCF)

is :

C

A

DF

B

E

(a) 4 : 5 (b) 4 : 9 (c) 16 : 81 (d) none of these13. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas

of the triangles ABC and BDE is :(a) 2 : 1 (b) 1 : 2 (c) 4 : 1 (d) 1 : 4

14. D, E and F are the mid-points of the sides BC, CA and AB respectively of a ABC. The ratio of the areasof DEF and ABC is :(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

15. Two vertical poles are 24 m apart. If the heights of the poles be 17 m and 27 m; the distance between thetips of the poles is :(a) 20 m (b) 26 m (c) 28 m (d) none of these

16. Each side of a rhombus is 13 cm. If one of the diagonals is 24 cm, the length of the other diagonal is :(a) 5 cm (b) 8 cm (c) 10 cm (d) 12 cm

17. The area of a right-angled triangle is 60 cm2. If the larger side of the triangle exceeds the smaller side by7 cm, the length of hypotenuse is :(a) 10 cm (b) 12 cm (c) 15 cm (d) 17 cm

18. In the given figure, the value of AD is :

C

D

B A

20 cm

24 cm

7 cm

(a) 25 cm (b) 15 cm (c) 30 cm (d) none of these

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19. In an equilateral triangle ABC, if AD BC, then:(a) 2AB2 = 3AD2 (b) 4AB2 = 3AD2 (c) 3AB2 = 4AD2 (d) 3AB2 = 2AD2

20. The perpendicular AD on the base BC of a ABC intersects BC at D so that 1CD . DB.4

(a) 3BC2 = 5 (AC2 – AB2) (b) 3BC2 = 5 (AB2 – AC2)(c) 5BC2 = 3 (AC2 – AB2) (d) 5BC2 = 3 (AB2 – AC2)

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK QUESTIONS)

1. In figure, DE || BC. If AD 4 ,DB 3

then what is the value of AE ?AC

A

D E

B C2. In figure, XMN ~ XYZ, what is the measure of XZY?

X

NM

YZ

75°65°

3. In figure, DE || BC, what is the value of x?

A

DE

B

C

3 cm5 cm

5 cmx

4. If the ratio of the corresponding sides of two similar triangles is 3 : 5, then what is the ratio of theircorresponding heights?

5. If the ratio of the corresponding medians of two similar triangles is 7 : 5, then what is the ratio of theircorresponding sides?

6. State the basic theorem of proportionality.7. State the converse of the basic theorem of proportionality.8. In figure, state whether EF || QR ?

6 cm

3.5 cm

5 cm

4 cm

P

Q R

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9. In figure, AD is bisector of BAC, what is the value of DC.

3 cm 5 cm

A

B CD10 cm

10. The perimeters of two similar triangles ABC and PQR are respectively 30 cm and 24 cm. If PQ = 8 cm, findAB.

11. In figure, express x in the terms of a, b and c.

A

B CDb c

Ea

36°36°x

12. The lengths of sides of a triangle are 12 cm, 16 cm and 21 cm. The bisector of the greatest angle dividesthe opposite side into two parts. Find the length of these two parts.

13. In figure, find AD.

D

BA 4 cm

2 cm3 cm

C

14. In figure, find x.

B CE

12 cm 9 cm

6 cmx cm

A

D

120°

30°

15. The area of the two similar triangles are in the ratio 81 : 121. What is the ratio of their correspondingsides?

16. If ABC ~ DEF such that BC = 5 cm, EF = 3 cm and ar (DEF) = 36 cm2, find the area of ABC.17. If ABC ~ PQR such that ar (ABC) = 81 cm2 and ar (PQR) = 121 cm2. If QR = 22 cm, find BC.18. If ABC ~ XYZ and the ratio of area of ABC to that of area of XYZ is 25 : 49, then what is the ratio

of their corresponding medians?19. If ABC ~ PQR and their corresponding altitudes are in the ratio 7 : 8, then what is the ratio of the area

of PQR to that of area of ABC?

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20. State Pythagoras theorem.

21. State converse of Pythagoras theorem.

22. Is the triangle with sides 9 cm, 41 cm, 40 cm a right angled triangle?

23. Two poles of height 6 m and 11 m are standing 12 m apart. What is the distance between their tops?

24. In figure, ABC and DEC are the right triangles with B = E = 90°, find BE.

A

D

C EB

17 cm

10 cm6 cm

8 cm

25. In figure, ADB = BAC = 90°. What is the value of x?

A

B C

6 cm

8 cmD

26 cm

x

PRACTICE TEST

M.M : 30 Time : 1 hourGeneral Instructions :

Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.

1. Determine, by using the pythagoras theorem, whether the triangles having sides (2a – 1) cm, 2 2 cmaand (2a + 1) cm is a right angled triangle.

2. Prove that the diagonals of a trapezium divide each other proportionally.

3. If ABC ~ DEF such that BC = 3 cm, EF = 4 cm and area of ABC= 54 cm2. Determine the area of DEF.

4. The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any pointon the side BC. Prove that DF × FE = BF × FA.

5. In the given figure, PQ = PR, PN QR and LM PR. Prove that PQ QNLR MR

.

PM

NRL Q

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6. ABC is a right angled triangle in which C = 90°. If p is the length of the perpendicular from C to AB and

AB = c, BC = a, CA = b, prove that 2 2 2

1 1 1p a b

.

7. In ABC, A = 90°, if AD BC, prove that AB2 + CD2 = BD2 + AC2

8. D is a point on the side BC of ABC such that ADC = BAC. Prove that CA CB= .CD CA

9. In an equilateral ABC, D is a point on BC such that 1BD = BC.3

Prove that 9AD2 = 7AB2.

10. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, theother two sides are divided in the same ratio. Prove it.Using above do the following : In the given figure, DE || BC and BD = CE. Prove that ABC is an isoscelestriangle.

A

D E

B C

ANSWERS OF PRACTICE EXERCISE

1. 1.8 cm 2. x = 8 or 11 3. (i) 5 cm (ii) 9.5 cm (iii) 5.85 cm (iv) 16.2 cm (v) 4.8 cm4. (i) x = 11 (ii) x = 1 (iii) x = 4 (iv) x = 9 6. BD = 3.6 cm, CE = 4.8 cm

7. (i) yes (ii) no (iii) yes (iv) yes 15. (i) x = 5 (ii) x = 2 or 12

x

18. AE = 10 cm, EC = 20 cm, AD = 6 cm and DB = 12 cm26. PT = 4.8 cm, QT = 5.7 cm 27. CD = 16 cm 28. PB = 2 cm

29. AC = 7.2 cm, BC = 9 cm 30. acx

b c

31. AB = 15 cm 32. 72 m

33. AQ = 15 cm 35. 12 cm 36. (i) 65 (ii) 45° (iii) 45° (iv) 65° (v) 70°

37. 10 cm 38. 52

39. x = 3.75 cm, y = 6.67 cm

40. 1.6 m 61. 16 : 81 62. 3.5 cm 63. 8.8 cm 64. 1 : 366. 25 : 81 67. 16 : 9 71. 4 : 1 72. 21 cm2

73.2 2

2

75. (i) 15 cm (ii) 25 : 81 (iii) 25 : 56 76. (i), (ii), (iii), (v), (vi)

78. (i) 16 cm (ii) 25 cm 79. 120 cm2 80. 15 m 81. 31 m

82. 13 m 84. 13 cm 86. 2 5 cm 91. 12 2 cm

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ANSWERS OF MULTIPLE CHOICE QUESTIONS

1. (c) 2. (b) 3. (b) 4. (c) 5. (a)6. (b) 7. (a) 8. (b) 9. (c) 10. (c)

11. (a) 12. (c) 13. (c) 14. (c) 15. (b)16. (c) 17. (d) 18. (b) 19. (c) 20. (b)

ANSWERS OF VERY SHORT ANSWER TYPE QUESTIONS

1.47 2. 40° 3. 25

3 cm 4. 3 : 5 5. 7 : 5

8. No 9. 6 cm 10. 10 cm 11. ac

b c 12. 9 cm, 12 cm

13. 1.5 cm 14. 8 cm 15. 9 : 11 16. 100 cm2 17. 18 cm

18. 5 : 7 19. 64 : 49 22. yes 23. 13 m 24. 23 cm

25. 24 cm

ANSWERS OF PRACTICE TEST

1. yes 3. 96 cm2

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6_tr1

Documents

aaa similarity

sas similarity

aa similarity

vertically

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parallelogram

aabc intersects

aa similarity