Triangles

1. The greater of two supplementary angles exceeds the smaller by 18°. Find
measures of these two angles. (2024)

Answer. Let the measure of two angles be x and y (x > y)

Given x + y = 180 and x – y = 18
solving equations to get y = 81 and x = 99

2. If a line is drawn parallel to one side of a triangle to intersect the other two
sides in distinct points, then prove that other two sides are divided in the
same ratio. (2024)

Answer. (A)

Given: In  ABC, DE || BC

 BDE and  CDE are on the same base DE and between the same parallels DE
and BC.
 ar ( BDE) = ar ( CDE) ……………(iii)
From (i), (ii) and (iii)

3. Sides AB and BC and median AD of a AABC are respectively proportional to

sides PQ and PR and median PM of APQR. Show that (2024)
AABC - APQR.

Answer. (B)Produce AD to E and PM to N such that AD = DE, PM = MN.

 ADB   EDC  AB = CE, similarly PQ = RN.

  1 =  2, similarly  3 =  4
therefore  1 +  3 =  2 +  4 or BAC =  QPR

Therefore  ABC   PQR

4. In the given figure, ABCD is a quadrilateral.

Diagonal BD bisects ZB and D both. (2024)
Prove that:
(i) AABD ~ ACBD
(ii) AB = BC
Answer. (i) In ∆ABD & ∆CBD
∠3=∠4
∠1 = ∠ 2
∴ ∆ABD ∼ ∆CBD
(ii) ∆ABD ≅ ∆CBD
∴ AB = BC

5. If a line is drawn parallel to one side of a triangle to intersect the other two
sides in distinct points, then prove that the other two sides are divided in the
same ratio. (2024)
Answer. Correct figure, given, to prove and construction Correct proof

6. In the given figure PA, QB and RC are each perpendicular to AC. If (2024)
AP = x, BQ = y and CR = z, then

Answer.
6.2 Similar Figures
VSA (1 mark)
1. All concentric circles are __________ to each other. (2020)
2. Two polygons having same number of sides and corresponding sides
proportional are similar or not? (Board Term 1, 2016)
6.3 Similarity of Triangles
MCQ
3. In AABC, PQ||BC. If PB = 6 cm, AP = 4 cm, AQ = 8 cm,
find the length of AC.
(a) 12 cm
(b) 20 cm
(c) 6 cm
(d) 14 cm
4. In the given figure, PQ II AC. If BP = 4 cm, AP = 2.4 cm
and BQ = 5 cm, then length of BC is
5. In the figure given below, what value of x will make
PQ || AB?

(a) 2
(c) 4
(b) 3
(d) 5 (Term 1, 2021-22)
6.

(a) 2.0 cm
(b) 1.8 cm
(c) 4.0 cm
(d) 2.7 cm
VSA (1 mark)
7. In figure, GC||BD and GE||BF. If AC = 3 cm and

8. In AABC, X is middle point of AC. If XY||AB, then prove that Y is middle point
of BC. (Board Term I, 2017)
9. In AABC, D and E are point on side AB and AC respectively, such that DE ||
BC. If AE = 2 cm, AD = 3 cm and BD = 4.5 cm, then find CE. (Board Term I,
2017)
10. In AABC, DE || BC, then find the value of x.

11. In given figure, DE ||BC

SAI (2 marks)
12. In the given figure, DE || AC and DF || AE.

13. In figure, if PQ || BC and PR || CD, prove that

SA II (3 marks)
14.

15. In the figure, P is any point on side BC of AABC.

PQ || BA and PR || CA are drawn. RQ is extended to
meet BC produced at S. Prove that SP2 = SB × SC.

LA (4/5/6 marks)
16. If a line is drawn parallel to one side of a triangle to intersect the other two
sides at distinct points, prove that the other two sides are divided in the same
ratio. (2020, 2015)
OR
State and prove Basic Proportionality Theorem (Thales Theorem).(Board
Term 1, 2015)
17. ABCD is a trapezium with AB||CD. E and F are points on non parallel sides
AD and BC respectively, such

6.4 Criteria for Similarity of Triangles

MCQ
18. In the given figure, AABC ~ AQPR. If AC = 6 cm,
BC= 5 cm, QR = 3 cm and PR = x, then the value of x is

(a) 3.6 cm
(c) 10 cm
(b) 2.5 cm
(d) 3.2 cm (2023)
19. If AABC and APQR are similar triangles such that
ZA = 31° and ZR = 69°, then ZQ is
(a) 70°
(b) 100°
(c) 90°
(d) 80° (Term I, 2021-22)
20. A vertical pole of length 19 m casts a shadow 57 m long on the ground and
at the same time a tower casts a shadow 51 m long. The height of the tower is
(a) 171m
(b) 13 m
(c)17 m
(d) 117 m (Term I, 2021-22)
21. In the given figure, ZABC and ZACB are complementary to each other and
ADI BC. Then,
(a) BD.CD = BC2
(c) BD-CD = AD2
(b) AB-BC= BC2
(d) AB-AC = AD2 (Term I, 2021-22)
22. In the given figure, x expressed in terms of a, b, c, is

SAI (2 marks)
23. In the given figure, XZ is parallel to BC. AZ = 3 cm, ZC = 2 cm, BM = 3 cm
and MC = 5 cm. Find the length of XY.23. In the given figure, XZ is parallel to
BC. AZ = 3 cm, ZC = 2 cm, BM = 3 cm and MC = 5 cm. Find the length of XY.

25. State which of the two triangles given in the figure are similar. Also statthe
similarity criterion used.
26. Sides AB, BC and median AD of a ∆ABC are respectively proportional to
sides PQ, QR and median PM of ∆PQR. Show that ∆ABC ~∆PQR. (Board Term 1,
2015)
SA II (3 marks)
27. PA, QB and RC are each perpendicular to AC. If
AP = x, QB = z, RC = y, AB = a and BC = b, then prove

28. In the given figure, CD and RS are respectively the medians of ∆ABC and
∆PQR. If ∆ABC ~ ∆PQR then prove that:
(i) ∆ADC ~ ∆PSR
(ii) AD × PR= ACX PS

29. In the figure, if ∆BEA = ∆CDA, then prove that

∆DEA – ∆BCA.

30. In ∆ABC, ∠ADE = ∠B then prove that ∆ADE ~ ∆ABC also if AD = 7.6 cm, BD
= 4.2 cm and BC = 8.4 cm, then
find DE.

31. A girl of height 100 cm is walking away from the base of a lamp post at a
speed of 1.9 m/s. If the lamp is 5 m above the ground, find the length of her
shadow after 4 seconds. (Board Term 1, 2016)
32. AABC and AAMP are two right angled triangles right angled at B and M
respectively. Prove that CA × MP
= PAX BC.

LA (4/5/6 marks)
33. (A) In a APQR, N is a point on PR, such that QN ⊥ PR. If
PN × NR = QN², prove that /PQR = 90°. (2023)
34. In the given figure, AABC and ADBC are on the same base BC. If AD
intersects BC at O, prove that

35. In the given figure, E is a point on CB produced of an isosceles ∆ABC, with

side AB = AC. If AD ⊥ BC and EF ⊥ LAC, prove that ∆ABD ~ ∆AECF.

36. In the given figure, ABC is a triangle and GHED is a rectangle. BC = 12 cm,
HE = 6 cm, FC = BF and altitude AF = 24 cm. Find the area of the rectangle.

37. Two poles of height 'p' and 'q' metres are standing vertically on a level
ground, 'a' metres apart. Prove that the height of the point of intersection of
the lines joining the top of each pole to the foot of the opposite
38. In ∆ABC, from A and B altitudes AD and BE are drawn. Prove that
∆ADC ~ ∆BEC. Is ∆ADB ~ ∆AEB and
∆ADB ~ ∆ADC? (Board Term 1, 2016)
Pythagoras Theorem
MCQ
39. Assertion (A): The perimeter of AABC is a rational number.
Reason (R): The sum of the squares of two rational numbers is always
rational.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct
explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the
correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. (2023)
VSA (1 marks)
40. Aman goes 5 metres due west and then 12 metres due North. How far is he
from the starting point? (2021 C)
SA II (3 marks)
41. In ∆ABC, ZB = 90° and D is the mid point of BC. Prove
that AC2 = AD2+3CD2
42. Prove that the sum of squares of the sides of a rhombus is equal to the sum
of squares of its diagonals. (2019)
LA (4/5/6 marks)
43. In given figure BN and CM are medians of a right angled at A. Prove that 4
(BN² + CM²) = 5BC2

44. The perpendicular from A on the side BC of a ∆ABC intersects BC at D, such

that DB = 3CD. Prove that
2AB2 = 2AC² + BC².

CBSE Sample Questions

6.3 Similarity of Triangles
VSA (1 mark)
1. In the ∆ABC, D and E are points on side AB and AC respectively such that DE
||BC. If AE = 2 cm, AD = 3 cm and BD = 4.5 cm, then find CE. (2020-21)
LA (4/5/6 marks)
2. Prove that if a line is drawn parallel to one side of a triangle intersecting the
other two sides in distinct points, then the other two sides are divided in the
same ratio.
Using the above theorem prove that a line through the point of intersection of
the diagonals and parallel to the base of the trapezium divides the non parallel
sides in the same ratio. (2022-23)
6.4 Criteria for Similarity of Triangles
MCQ
3. ∆ABC-∆PQR. If AM and PN are altitudes of ∆ABC and ∆PQR respectively and
∆AB2: PQ² = 4 : 9, then AM: PN=
(a) 3:2
(b) 16:81
(c) 4:9
(d) 2:3 (2022-23)
OR
∆ABC ~ ∆PQR. If AM and PN are altitudes of ∆ABC and ∆PQR respectively and
AB2: PQ2 = 4 : 9, then
AM: PN=
(a) 16:81
(b) 4:9
(c) 3:2
(d) 2:3 (Term I, 2021-22)
In the figure, if DE || BC, AD = 3 cm, BD = 4 cm and BC= 14 cm, then DE equals

(a) 7 cm
(b) 6 cm
(c) 4 cm
(d) 3 cm (Term 1, 2021-22)
5. ∆ABC is such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm. If
∆ABC ~∆DEF and EF = 4 cm, then perimeter of ∆DEF is
(a) 7.5 cm
(b) 15 cm
(c) 22.5 cm
(d) 30 cm (Term I, 2021-22)
6. In the given figure, <ZACB = <CDA, AC = 8 cm,
AD = 3 cm, then BD is

7. Sides AB and BE of a right triangle, right angled at B are of lengths 16 cm

and 8 cm respectively. The length of the side of largest square FDGB that can
be inscribed in the triangle ABE is

Case study-based questions are compulsory. Attempt any 4 sub parts. Each
question carries 1 mark.
8. SCALE FACTOR AND SIMILARITY
Scale Factor
A scale drawing of an object is the same shape as the object but a different
size. The scale of a drawing is a comparison of the length used on a drawing to
the length it represents. The scale is written as a ratio.
Similar Figures
The ratio of two corresponding sides in similar figures is called the scale
factor.
Hence, two shapes are Similar when one can become the other after a resize,
flip, slides or turn.
(i) A model of a boat is made on the scale of 1: 4. The model is 120 cm long.
The full size of the boat has a width of 60 cm. What is the width of the scale
model?

(a) 20 cm (b) 25 cm (c) 15 cm (d) 240 cm

(ii) What will effect the similarity of any two polygons?
(a) They are flipped horizontally
(b) They are dilated by a scale factor
(c) They are translated down
(d) They are not the mirror image of one another
(iii) If two similar triangles have a scale factor of a: b. Which statement
regarding the two triangles is true?
(a) The ratio of their perimeters is 3a: b
(b) Their altitudes have a ratio a: b

(d) Their angle bisectors have a ratio a²: b2

(iv) The shadow of a stick 5 m long is 2 m. At the same time the shadow of a
tree 12.5m high is

(a) 3m (b) 3.5 m (c) 4.5m (d) 5m

(v) Below you see a student's mathematical model of a farmhouse roof with
measurements. The attic floor, ABCD in the model, is a square. The beams that
support the roof are the edges of a rectangular prism, EFGHKLMN. E is the
middle of AT, F is the middle of BT, G is the middle of CT, and H is the middle of
DT. All the edges of the pyramid in the model have length of 12 m.

What is the length of EF, where EF is one of the horizontal edges of the block?
(a) 24 m
(c) 6m
(b) 3m
(d) 10 m (2020-21)
SAI (2 marks)
9.

SA II (3 marks)
10. The perimeters of two similar triangles are 25 cm and 15 cm respectively.
If one side of the first triangle is 9 cm, find the length of the corresponding
side of the second triangle. (2020-21)

SOLUTIONS
Previous Years' CBSE Board Questions
1. All concentric circles are similar to each other.
2. Two polygons having same number of sides and corresponding sides
proportional are not similar.
3. (b): Since, PQ||BC
4. (a): Since, PQ || AC.

5. (a): Suppose PQ || AB
:- By Basic Proportionality theorem, we have

6. (b): In AABC, DE || BC

7. Here in the given figure,

GC || BD and GE || BF
AC = 3 cm and CD = 7 cm
By Basic Proportionality theorem,
8. Given, X is middle point of AC and XY || AB. We have to prove Y is middle
point of BC. In AABC, XY || AB

9.

10.

11.
12.

13.

14.

:- DE || BC
[By Converse of Basic Proportionality Theorem]
Also, <D= <B and ZE = /C [Corresponding angles] ...(ii)
From (i) and (ii), we get <B = <C ⇒ AB = AC
[Sides opposite to equal angles are equal]
:- AABC is an isosceles triangle.
15.

16. Consider AABC in which DE || BC, DE intersects AB at D and AC at E.

17.
First join AC to intersect EF at G.
Given AB || DC and EF || AB
⇒ EF || DC ...(i)
[:: Lines parallel to same line are parallel to each other.]
Now in ∆ADC, we have
EG||DC (:- EF||DC)

18.

19.

Given, <ZA = 31°, <R = 69° and ∆ABC ~ ∆PQR

:- <ZP = <A = 31°
:- <ZQ=180° - (31° + 69°) [By angle sum property]
= <ZQ=80°
20. (c): Let AB be the pole and PQ be the tower.
Let height of tower be h m.
Now, ∆ABC ~ ∆PQR

21. (c): We have,

<ABC+<ACB = 90° [Given] . (i)
In ∆BDA, <ABD+<BAD = 90° ... (ii) [By angle sum property]
From (i) and (ii), we get, <ACB = <BAD
In ∆ADC and ∆BDA
<ADC=<BDA = 90°
<ACD=<BAD [Proved above]
:- ∆ADC~ ∆BDA [By AA similarity]
22. (b): In ∆RST and ARPQ,
<R = <R [Common]
<RTS = <RQP = 47° [Given]
∆RST ~ ∆RPQ [By AA similarity]

23. (a) Given, AZ = 3 cm, ZC = 2 cm, BM = 3 cm and MC = 5 cm

In ∆ABC, XZ || BC

Now, AC = AZ + ZC=3+2 = 5 cm
BC= BM+MC=3+5=8 cm and
In ∆AXY and ∆ABM

24. Given, PQ||BC

PQ = 3 cm, BC = 9 cm and AC = 7.5 cm
Since, PQ || BC
:- <APQ = <ABC (Corresponding angles are equal)
Now, in ΔAPQ and ΔABC
<APQ = <ABC (Corresponding angles)
<A = <A (Common)
ΔΑΡΟ - ΔΑΒΕ (AA similarity)

25. In AABC and ADEF

26. We have ΔABC and ΔPQR in which AD and PM are medians corresponding
to sides BC and QR respectively
27. Given: PA, QB and RC are each perpendicular to AC.

Now in ΔACR and ΔABQ, :: BQ||CR

<ARC=<AQB (Corresponding angles)
and <RAC = <QAB (Common)
:- ΔACR-ΔABQ

Hence proved.
28. (i) Given: Two similar triangles are ABC and PQR. CD and RS are medians
of ΔABC and ΔPQR.

29. We have, ΔBEA = ΔCDA

:- AB = AC and AE = AD [By C.P.C.T.]
30. In AADE and AABC
<ADE = <ABC [Given]
and <DAE = <BAC [Common]
:- ΔADE ~ ΔABC [By AA similarity criterion]

31. Let AB be the lamp post and CD be the girl's height and DE = x be the
length of the shadow of the girl.

We know that, Distance = Speed × Time

Distance of the girl from the lamp post after 4
seconds, BD = 1.9x4 = 7.6m
In ΔABE and ΔCDE
We know that, Distance = Speed × Time
Distance of the girl from the lamp post after 4
seconds, BD = 1.9x4 = 7.6m
In ΔABE and ΔCDE

:- Length of shadow of girl after 4 seconds is

1.9 metres.
32. In ΔABC and ΔAMP
<ABC = <AMP [Each 90°]
<BAC = <PAM [Common]
:- ΔΑΒΕ - ΔΑΜΡ [By AA similarity criterion]

33. Given, in ΔPQR, N is a point on PR, such that QN PR

and PN. NR = QN²
To prove: <PQR = 90°
Proof: We have, PN. NR = QN²
⇒ PN . NR = QN . QN
and <QNP = <RNQ [each equal to 90° ]
:- ΔANP~ΔRNQ [by SAS similarity criterion]
Then, ΔQNP and ΔRNQ are equiangulars.
i.e., <PQN= <QRN ...(i)
and <RQN = <QPN ...(ii)
Adding (i) and (ii), we get
<PQN + <RQN = <QRN+ <QPN
= <PQR = <QRN + <QPN ...(iii)
We know that, sum of angles of a triangle is 180°
In ΔPQR, <PQR + <QPR + <QRP = 180°
= <PQR + <QPN+ <QRN = 180°
[:: <QPR = <QPN and<ZQRP = <QRN]
<PQR+ZPQR = 180° [using (iii)]
2<PQR = 180°

34. We have, ΔABC and ΔDBC are on the same base BC.
Also, BC and AD intersects at O.
Let us draw AE ⊥ BC and DF ⊥ BC.
In ΔAOE and ΔDOF,
<AEO = <DFO = 90° ...(i)
Also, <AOE = <DOF ...(ii)

:- From (i) and (ii), we get

ΔAOE ~ ΔDOF [By AA similarity]
:- Their corresponding sides are proportional.

35. Given, ΔABC is an isosceles triangle with AB = AC

:- <ABC= <ACB = <ABD = <ECF ...(i)
Now, in ΔABD and ΔECF
<ADB = <EFC [Each 90°]
<ABD = <ECF [Using (i)]
:- By AA similarity criterion, ΔABD ~ ΔECF.
36. In ΔABC, BC = 12 cm,
EH = DG = 6 cm,
BC = 12 cm

and AF = 24 cm, DE = GH
Now, in ΔAFC and <EHC
<AFC = <EHC [Each 90°]
<ACF = <ECH [Common]
:- By AA similarity criterion,
ΔAFC ~ ΔEHC

Now, FHFC-HC = (6-1.5) cm = 4.5 cm

GH = 2 × FH = 2x4.5 = 9 cm
Area of rectangle DEHG=HEX GH = 6 x 9 = 54 cm²
37. Let AB and CD be two poles
of height p and q metres
respectively and poles are 'a'
metres apart i.e., AC = a metres.
Let AD and BC meet at 'O' such
that OL = h metres
Let CL=x and LA = y
:- x+y=a
In ΔABC and ΔLOC, we have
<CAB = <CLO [Each 90°]
<C=<C [Common]
:- AABC - ALOC [By AA similarity criterion]
Hence, the height of the point of intersection of the lines joining the top of
each pole to the foot of the opposite pole

38. In ΔADC and ΔBEC

<DC = <BEC [Each 90°]
<ACD = <BCE [Common]
:- ΔADC ~ ΔBEC [By AA similarity criterion]
In ΔADB and ΔAEB
<ADB = <AEB [Each 90°]
<BAD+<EAB
:- ΔADB is not similar to ΔAEB.
Again, in ΔADB and ΔADC
<BAD + <DAC
<ADB = <ADC [Each 90°]
:- ΔADB is not similar to ΔADC
39. (d): In AABC, AC² = AB² + BC2
⇒ AC² = 2²+32
⇒ AC²=4+9
⇒ AC=√13 cm
So, perimeter is (2+3+√13) cm
=(5+√13) cm which is irrational.
Hence, Assertion in false but Reason is true.
40. Let Aman starts from A point and continues 5 m towards west and reached
at B point, from which he goes 12 m towards North reached at C point finally.
In AABC, we have
AC² = AB² + BC2
AC² = 52 + 122
(By Pythagoras theorem)
AC² = 25+ 144 = 169
⇒ AC = 13 m
So, Aman is 13 m away from his starting point.
41.

In ΔABC, ZB = 90°
AB² + BC² = AC² (By Pythagoras theoram)
⇒ AB2 = AC2 - BC2 ...(i)
Similarly by pythagoras theoram, In ΔABD,
AD² = AB² + BD2 ...(ii)
⇒ AB² = AD² - BD2 ...(iii)
From eqn (i) and (ii), we get AC² - BC² = AD2 - BD2
⇒ AC² = AD² - BD² + BC²
As D is the mid point of BC,

AC² = AD2 - CD2+ BC2

AC² = AD2 - CD² + (2CD)2 [::: BC= 2CD]
⇒ AC² = AD²+3CD2
Hence proved.
42. Let we have a rhombus ABCD.
:- Diagonal of a rhombus bisect each other at right
angles.
:- OA = OC and OB = OD
Also, <AOB = <BOC [Each = 90°]
and <COD = <DOA [Each = 90°]
In ΔAOB, we have
AB² = OA² + OB2 ...(i)
(By Pythagoras theorem)
Similarly in ABOC, we have
BC² = OB² + OC² ...(ii)
In ACOD,
CD² = OC²+OD2 ...(iii)
In right ΔAOD
DA² = OD² + OA² ...(iv)
On adding (i), (ii), (iii) and (iv), we get
AB² + BC² + CD2 + DA²
= [OA² + OB²] + [OB² + OC²] + [OC² + OD²] + [OD² + OA²]
= 2 [OA² + OB² + OC² + OD²]
:- OA² = OC² and OB² = OD²

Thus, sum of the squares of the sides of a rhombus is equal to the sum of the
squares of its diagonals.
43. In ΔABC, BN and CM are medians and <A = 90°
To prove : 4(BN² + CM²) = 5BC2
In ΔABC, ZA = 90°
:- BC² = AB² + AC² ...(i) (By Pythagoras theorem)
In ΔCAM, ZA = 90°
:- CM² = AC2+ AM²
44. We have, ΔABC such that AD ⊥BC. ΔABC intersect BC
at D such that BD = 3CD.
In right ΔADB, by Pythagoras theorem, we have
AB² = AD² + BD2 ...(i)
Similarly in ΔACD, we have AC² = AD2² + CD2 ...(ii)
Subtracting (ii) from (i), we get
AB2 - AC² = BD2-CD2 ...(iii)
Now, BC= DB + CD = 4 CD [::: BD = 3CD]

CBSE Sample Questions

1.

2. Consider AABC in which DE || BC, DE intersects AB at D and AC at E.

Since, triangles EDB and ECD are on the same base DE and between the same
parallel lines DE and BC. So, area (ΔEDB) = area (ΔECD) ...(iii) (1/2)
3.

4. (b): Since, DE || BC ⇒ AABC ~ AADE

(By AA similarity criterion)

5. (b): : AABC - ADEF

6. (c): In ΔACD and AABC, we have

<A = <A (Common)
<CDA = ZBCA (Given)
:- ΔACD ~ ΔABC (By AA similarity)
7. (b): AABE is a right triangle and FDGB is a square of side x cm (say).

8.

(ii) (d): They are not the mirror image of one another. (1)
(iii) (b): Their altitudes have a ratio a: b. (1)
(iv) (d): Since the two triangles are similar so the ratio of their corresponding
sides are equal.
(v) (c): Since E is the middle point of AT and F is the middle point of BT.

9.

10. Given one side of first triangle is 9 cm.

Let the length of the corresponding side of the second
triangle be x cm. (1)
Now, ratio of perimeter

[:- In similar triangles, the perimeter of the triangle will be in the ratio of their
corresponding sides.]