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TRIANGLES
IMPORTANT CONCEPTS
Similar figures :Two figures having same shape (size may or may not same) are called similar
figures
Eg: (1) All Circles are similar
(2) Equilateral triangles are similar
Similar triangles: Two triangles are said to be similar if
(a) Corresponding angles of both the triangles are equal
(b) Corresponding sides of both the triangles are in proportion .
Basic Proportionality Theorem(Thales theorem) : If a line is drawn parallel to one side of a
triangle to intersect the other two sides in distinct points, then the other two sides are divided in the
same ratio.
Criterion of Similarity
● In two triangles, if the corresponding angles are equal, then the corresponding sides are in
proportion, then the triangles are similar (AAA similarity criterion)
● If the corresponding sides of any two triangles are proportional, then the corresponding
angles are equal and the two triangles are similar (SSS similarity criterion)
● If one angle of a triangle is equal to one angle of the another triangle and the corresponding
sides including these angle are proportional, then the triangles are similar (SAS similarity
criterion)

SECTION :A MCQ( 1 Mark each)

Q 1. If ΔABC is similar to ΔDEF such that 2 AB = DE and BC = 8 cm then EF is equal to.
a. 12 cm
b. 4 cm
c. 16 cm
d. 8 cm
Q 2. D and E are the midpoints of side AB and AC of a triangle ABC, respectively and BC = 6
cm. If DE || BC, then the length (in cm) of DE is:
(a) 2.5
(b) 3
(c) 5
(d) 6
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Q3. If ABC and EDF are two triangles and AB/DE=BC/FD, then the two triangles are similar
if
(a) ∠A=∠F
(b) ∠B=∠D
(c) ∠A=∠D
(d) ∠B=∠E
Q4. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE. Then, the two triangles
are
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Q5. In ΔLMN, ∠L = 50° and ∠N = 60°, If ΔLMN ~ ΔPQR, then find ∠Q
a. 50°
b. 70°
c. 60°
d. 40°
SECTION :B(2 Marks each)
Q6. If ∆ABC ~ ∆RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10,
then find QR.
Q7. In ∆ABC, DE || BC, find the value of x.

Q8. . X and Y are points on the sides AB and AC respectively of a triangle ABC such
that 𝐀𝐗/𝐀𝐁=1/4, AY = 2 cm and YC = 6 cm. Find whether XY || BC or not.
Q9. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
Show that ΔABE is similar to ΔCFB.
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Q10. In the given figure, QA ⊥ AB and PB ⊥ AB. If AO = 20 cm, BO = 12 cm, PB = 18 cm,

find AQ.

SECTION :C( 3 Marks each)

Q11. In the given figure, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and
EC = 2 cm.

Q12. If a line segment intersects sides AB and AC of a ∆ABC at D and E

𝐴𝐷 𝐴𝐸
respectively and is parallel to BC, prove that = 𝐴𝐶
𝐴𝐵
𝐴𝐷 3 𝐵𝐶
Q13. In a ∆ABC, DE || BC with D on AB and E on AC. If𝐷𝐵 =4 , find 𝐷𝐸

Q14. A vertical pole of length 8 m casts a shadow 6 cm long on the ground and at the same time a
tower casts a shadow 30 m long. Find the height of tower

Q15. In the figure, if DE || OB and EF || BC, then prove that DF || OC.

SECTION :D (5 Marks each)
Q16. If sides AB, BC and median AD of AABC are proportional to the
corresponding sides PQ, QR and median PM of PQR, show that ∆ABC ~
∆PQR
Q17. (a)“If a line is drawn parallel to one side of a triangle to intersect the other two sides in
distinct points, the other two sides are divided in the same ratio”. Prove
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(b). In ∆DEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then

find the value of DB.
𝐴𝑂
Q18. (a) The diagonals of a quadrilateral ABCD intersect each other at the point O such that 𝐵𝑂 =
𝐶𝑂
Show that ABCD is a trapezium.
𝐷𝑂

(b). In the given figure, if AB || DC, find the value of x.

Q19 (a). In the given Fig, 𝑃𝑆/𝑆𝑄 = 𝑃𝑇/𝑇𝑅 and ∠PST = ∠PRQ. Prove that PQR is an isosceles
triangle.

(b) In figure, DE|| BC and BD = CE. Prove that ∆ABC is an isosceles

triangle.

Q20. In ∆ABC, if ∠ADE = ∠B, then prove that ∆ADE ~ ∆ABC. Also, if AD = 7.6
cm, AE = 7.cm, BE = 4.2 cm and BC = 8.4 cm, then find DE.
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ANSWERS
SECTION :A
1) ( c ) 16 cm 2) (b) 3 3) (b) ˂B = ˂D 4) ( b) Similar not congruent 5) (b) 700
SECTION :B (2 Marks each)
Q6. ∆ABC ~ ∆RPQ …[Given

∴ QR = 12 cm
Q7. In ∆ABC, DE || BC …[Given

x(x + 5) = (x + 3)(x + 1)
x2 + 5x = x2 + 3x + x + 3
x2 + 5x – x2 – 3x – x = 3
∴ x = 3 cm
Q8. Given: 𝐴𝑋/𝐴𝐵=1/4

AX = 1K, AB = 4K
∴ BX = AB – AX
= 4K – 1K = 3K
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Q9.
In ∆ABE and ∆CFB, we have
∠AEB = ∠CBF (Alternate angles)
∠A = ∠C (Opposite angles of a parallelogram)
∴ ∆ABE ~ ∆CFB (By AA criterion of similarity)

Q10. In ∆OAQ and ∆OBP,

∠OAQ = ∠OBP … [Each 90°
∠AOQ = ∠BOP … [vertically opposite angles

SECTION :C
Q11. In ∆ABL, CD || LA
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Q12. Given. In ∆ABC, DE || BC

To prove. AD/AB=AE/AC
Proof.
In ∆ADE and ∆ABC
∠1 = ∠1 … Common
∠2 = ∠3 … [Corresponding angles
∆ADE ~ ∆ABC …[AA similarity
∴ 𝐀𝐃/𝐀𝐁=𝐀𝐄/𝐀𝐂
…[In ~∆s corresponding sides are proportional]
Q13. Given: In a ∆ABC, DE || BC with D on AB and E
on AC and 𝐴𝐷/𝐷𝐵=3/4
To find: BCDE
Proof. Let AD = 3k,

DB = 4k
∴ AB = 3k + 4k = 7k
In ∆ADE and ∆ABC,
∠1 = ∠1 …[Common
∠2 = ∠3 … [Corresponding angles
∴ ∆ADE ~ ∆ABC …[AA similarity]
AD/AB = DE/BC [In ~∆s corresponding sides are proportional]
3k/7k = DE/BC
Therefore, BC/DE = 7/3
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Q14.

Let BC be the pole and EF be the tower Shadow AB = 6 m and DE = 30 m.

In ∆ABC and ∆DEF,
∠2 = ∠4 … [Each 90°
∠1 = ∠3 … [Sun’s angle of elevation at the same time]
∆ABC ~ ∆DEF …[AA similarity]
𝐴𝐵/𝐷𝐸=𝐵𝐶/𝐸𝐹 … [In similar triangles corresponding sides are proportional]
⇒ 6/30=8/EF ∴ EF = 40 m
Q15.

Solution:
Given. In ∆ABC, DE || OB and EF || BC
To prove. DF || OC
Proof. In ∆AOB, DE || OB … [Given

Section :D (5 Marks each)

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Q 16.Solution:

Q17.(a) Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB
and AC at D and E respectively.

To prove:

Construction: Join BE and CD and draw DM ⊥ AC and EN ⊥ AB.

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Proof: area of Δ ADE

(Taking AD as base)

So, [The area of Δ ADE is denoted as ar (ADE)].

Similarly,

[Δ BDE and DEC are on the same base DE and between the same parallels BC and DE.]

Therefore, from (i), (ii) and (iii), we have:

(b))
Let BD = x cm
then BW = (24 – x) cm, AE = 12 – 4 = 8 cm
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In ∆DEW, AB || EW

Q18. Given: Quadrilateral ABCD in which

AC and BD intersect each other at 0.
Such that 𝐴𝑂/𝐵𝑂=𝐶𝑂/𝐷𝑂
To prove: ABCD is a trapezium
Const.: From O, draw OE || AB.

Solution:

AB || DC

In quad ABCD, AB || DC
⇒ ABCD is a trapezium.

(b)

Given AB||DC

∴∠ODC=∠OBA

(Alternate interior angles)

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and
∠OCD=∠OAB
(Alternate interior angles)
∴ΔDOC ~ΔBOA
(By AA similarity criterion)

∴OD/OB=OC/OA⇒x−2/x−1=x+3/x+5
⇒(x−2)(x+5)=(x+3)(x−1)
⇒x2+3x−10=x2+2x−3⇒x=7
Q19. (a)Given: 𝑃𝑆/𝑆𝑄 = 𝑃𝑇/𝑇𝑅 and ∠PST = ∠PRQ
To Prove: PQR is an isosceles triangle.
Proof: 𝑃𝑆/𝑆𝑄 = 𝑃𝑇/𝑇𝑅
By converse of BPT we get
ST || QR
∴ ∠PST = ∠PQR (Corresponding angles) ….(i)
But, ∠PST = ∠PRQ (Given) ….(ii)
From equation (i) and (ii)
∠PQR = ∠PRQ
⇒ PR = PQ
So, ∆PQR is an isosceles triangle.

(b)

Q20.
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