CLASS: _______________________ SUBJECT: MATHEMATICS

DATE: ________________________ TOPIC: Triangle

Question 1:

Fill in the blanks using correct word given in the brackets:−

(i) All circles are __________. (congruent, similar)

(ii) All squares are __________. (similar, congruent)

(iii) All __________ triangles are similar. (isosceles, equilateral)

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are
__________ and (b) their corresponding sides are __________. (equal, proportional)
ANSWER:

(i) Similar

(ii) Similar

(iii) Equilateral

(iv) (a) Equal

(b) Proportional
Question 2:

Give two different examples of pair of

(i) Similar figures

(ii)Non-similar figures
ANSWER:

(i) Two equilateral triangles with sides 1 cm and 2 cm

Two squares with sides 1 cm and 2 cm

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(ii) Trapezium and square

Triangle and parallelogram

Question 1:

In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i)

(ii)

ANSWER:

(i)

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Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

(ii)

Let AD = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain

Question 2:

E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases,
state whether EF || QR.
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(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

ANSWER:

(i)

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

(ii)

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

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(iii)

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

Question 3:

In the following figure, if LM || CB and LN || CD, prove that

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ANSWER:

In the given figure, LM || CB

By using basic proportionality theorem, we obtain

Question 4:

In the following figure, DE || AC and DF || AE. Prove that

ANSWER:

In ΔABC, DE || AC

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Question 5:

In the following figure, DE || OQ and DF || OR, show that EF || QR.

ANSWER:

In Δ POQ, DE || OQ

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Question 6:

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

ANSWER:

In Δ POQ, AB || PQ

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Question 7:

Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a
triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
ANSWER:

Consider the given figure in which l is a line drawn through the mid-point P of line segment AB meeting
AC at Q, such that .

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Or, Q is the mid-point of AC.
Question 8:

Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two
sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
ANSWER:

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and
AC respectively.

i.e., AP = PB and AQ = QC

It can be observed that

Hence, by using basic proportionality theorem, we obtain

Question 9:

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show

that

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ANSWER:

Draw a line EF through point O, such that

In ΔADC,

By using basic proportionality theorem, we obtain

In ΔABD,

So, by using basic proportionality theorem, we obtain

From equations (1) and (2), we obtain

Question 10:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that Show
that ABCD is a trapezium.
ANSWER:

Let us consider the following figure for the given question.

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Draw a line OE || AB

In ΔABD, OE || AB

By using basic proportionality theorem, we obtain

However, it is given that

⇒ EO || DC [By the converse of basic proportionality theorem]

⇒ AB || OE || DC

⇒ AB || CD

∴ ABCD is a trapezium.

Question 1:

State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you
for answering the question and also write the pairs of similar triangles in the symbolic form:

(i)

(ii)

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(iii)

(iv)

(v)

(vi)

ANSWER:

(i) ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, ΔABC ∼ ΔPQR [By AAA similarity criterion]

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(ii)

(iii)The given triangles are not similar as the corresponding sides are not proportional.

(iv) In ∆ MNL and ∆ QPR, we observe that,

MNQP = MLQR = 12∠M = ∠Q = 70°∴ΔMNL ~ ΔQPR [By SAS similarity criterion]MNQP = MLQR
= 12∠M = ∠Q = 70°∴∆MNL ~ ∆QPR By SAS similarity criterion

(v)The given triangles are not similar as the corresponding sides are not proportional.

(vi) In ΔDEF,

∠D +∠E +∠F = 180º

(Sum of the measures of the angles of a triangle is 180º.)

70º + 80º +∠F = 180º

∠F = 30º

Similarly, in ΔPQR,

∠P +∠Q +∠R = 180º

(Sum of the measures of the angles of a triangle is 180º.)

∠P + 80º +30º = 180º

∠P = 70º

In ΔDEF and ΔPQR,

∠D = ∠P (Each 70°)

∠E = ∠Q (Each 80°)

∠F = ∠R (Each 30°)

∴ ΔDEF ∼ ΔPQR [By AAA similarity criterion]

Question 3:

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a

similarity criterion for two triangles, show that

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ANSWER:

In ΔDOC and ΔBOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠DCO = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ΔDOC ∼ ΔBOA [AAA similarity criterion]

Question 4:

In the following figure, Show that

ANSWER:

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR (i)

Given,
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Question 5:

S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ∼ ΔRTS.
ANSWER:

In ΔRPQ and ΔRST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ∼ ΔRTS (By AA similarity criterion)

Question 6:

In the following figure, if ΔABE ≅ ΔACD, show that ΔADE ∼ ΔABC.

ANSWER:

It is given that ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] (1)

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And, AD = AE [By CPCT] (2)

In ΔADE and ΔABC,

[Dividing equation (2) by (1)]

∠A = ∠A [Common angle]

∴ ΔADE ∼ ΔABC [By SAS similarity criterion]

Question 7:

In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

(i) ΔAEP ∼ ΔCDP

(ii) ΔABD ∼ ΔCBE

(iii) ΔAEP ∼ ΔADB

(v) ΔPDC ∼ ΔBEC

ANSWER:

(i)

In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔCDP

(ii)

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In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion,

ΔABD ∼ ΔCBE

(iii)

In ΔAEP and ΔADB,

∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔADB

(iv)

In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

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ΔPDC ∼ ΔBEC
Question 8:

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that
ΔABE ∼ ΔCFB
ANSWER:

In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴ ΔABE ∼ ΔCFB (By AA similarity criterion)

Question 9:

In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove
that:

(i) ΔABC ∼ ΔAMP

(ii)
ANSWER:

In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ΔABC ∼ ΔAMP (By AA similarity criterion)

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Question 10:

CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE
of ΔABC and ΔEFG respectively. If ΔABC ∼ ΔFEG, Show that:

(i)

(ii) ΔDCB ∼ ΔHGE

(iii) ΔDCA ∼ ΔHGF

ANSWER:

It is given that ΔABC ∼ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴ ΔACD ∼ ΔFGH (By AA similarity criterion)

In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ ΔDCB ∼ ΔHGE (By AA similarity criterion)

In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Proved above)

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∠A = ∠F (Proved above)

∴ ΔDCA ∼ ΔHGF (By AA similarity criterion)

Question 11:

In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If
AD ⊥ BC and EF ⊥ AC, prove that ΔABD ∼ ΔECF

ANSWER:

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠ABD = ∠ECF (Proved above)

∴ ΔABD ∼ ΔECF (By using AA similarity criterion)

Question 12:

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR
and median PM of ΔPQR (see the given figure). Show that ΔABC ∼ ΔPQR.
ANSWER:

Median divides the opposite side.

Given that,
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In ΔABD and ΔPQM,

(Proved above)

∴ ΔABD ∼ ΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

Question 13:

D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that
ANSWER:

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ΔADC ∼ ΔBAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

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Question 14:

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR
and median PM of another triangle PQR. Show that
ANSWER:

Given that,

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join
B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides.

Therefore, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

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Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given that

∴ ΔABE ∼ ΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠BAE = ∠QPL … (1)

Similarly, it can be proved that ΔAEC ∼ ΔPLR and

∠CAE = ∠RPL … (2)

Adding equation (1) and (2), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒ ∠CAB = ∠RPQ … (3)

In ΔABC and ΔPQR,

(Given)

∠CAB = ∠RPQ [Using equation (3)]

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

Question 15:

A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower
casts a shadow 28 m long. Find the height of the tower.
ANSWER:

Let AB and CD be a tower and a pole respectively.

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Let the shadow of BE and DF be the shadow of AB and CD respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, ∠DCF = ∠BAE

And, ∠DFC = ∠BEA

∠CDF = ∠ABE (Tower and pole are vertical to the ground)

∴ ΔABE ∼ ΔCDF (AAA similarity criterion)

Therefore, the height of the tower will be 42 metres.

Question 16:

If AD and PM are medians of triangles ABC and PQR, respectively

where
ANSWER:

It is given that ΔABC ∼ ΔPQR

We know that the corresponding sides of similar triangles are in proportion.

∴ … (1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PM are medians, they will divide their opposite sides.

∴ … (3)

From equations (1) and (3), we obtain

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 … (4)

In ΔABD and ΔPQM,

∠B = ∠Q [Using equation (2)]

[Using equation (4)]

∴ ΔABD ∼ ΔPQM (By SAS similarity criterion)

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