champstreet
www.champstreet.com
Triangles - NCERT Questions
Q 1.

Fill in the blanks using the correct word given in 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).

SOLUTION:

Soln. : (i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar.
(A) Their corresponding angles are equal and
(B) Their corresponding sides are proportional.

Q 2.

Give two different examples of pair of
(i) similar figures
(ii) non-similar figures

SOLUTION:

Soln. : (i) (a) Any two circles are similar figures.
(b) Any two squares are similar figures.
(ii) (a) A circle and a triangle are non-similar figures.
(b) An isosceles triangle and a scalene triangle are non-similar figures.

Q 3.

State whether the following quadrilaterals are similar or not :
ncert

SOLUTION:

Soln.: On observing the given figures, we find that
Their corresponding sides are proportional but their corresponding angles are not equal.
∴ The given figures are not similar.

Q 4.

In figures (i) and (ii), DEBC. Find EC in (i) and AD in (ii).
ncert

SOLUTION:

Soln. : (i) Since DEBC [Given]
∴ Using the Basic proportionality theorem,
We have ncert
Since, AD = 1.5 cm, DB = 3 cm and AE = 1 cm,
ncert
By cross-multiplication, we have EC × 1.5 = 1 × 3
ncert
EC = 2 cm
(ii) In ∆ABC, DEBC
∴ Using the Basic proportionality theorem, we have ncert
ncert
AD × 5.4 = 1.8 × 7.2
ncert
AD = 2.4 cm.

Q 5.

E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EFQR;
(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 = 0.36 cm

SOLUTION:

Soln. : (i) We have, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
ncert
ncert
And ncert
ncert
ncert
EF is not parallel to QR.
(ii) We have, PE = 4 cm, QE = 4.5 cm
PF = 8 cm and RF = 9 cm
ncert
ncert
And ncert
Since, ncert
EF is parallel to QR.
(iii) We have, PE = 0.18 cm, PQ = 1.28 cm
PF = 0.36 cm and PR = 2.56 cm
ncert
ncert
And ncert
Since, ncert
EF is parallel to QR.

Q 6.

In the figure, if LMCB and LNCD, prove that ncert
ncert

SOLUTION:

Soln.: In ∆ABC, LMCB [given]
∴ Using the Basic proportionality theorem, we have
ncert .....(1)
ncert
ncert
ncert
Similarly, in ∆ACDLNCD
∴ Using the Basic proportionality theorem, we have
ncert .....(2)
From (1) and (2),
ncert
ncert (Proved)

Q 7.

In the figure, DEAC and DFAE. Prove that ncert
ncert

SOLUTION:

Soln.: In ∆ABC,
DEAC [given]
ncert .....(1)
[By the basic proportionality theorem]
In ∆ABE,
DFAE [given]
∴ Using the basic proportionality theorem, we have
ncert .....(2)
From (1) and (2)
ncert
ncert

Q 8.

In the figure, DEOQ and DFOR. Show that EFQR.
ncert

SOLUTION:

Soln.: In ∆PQO,
DEOQ [given]
∴ Using the Basic proportionality theorem, we have
ncert .....(1)
Similarly, in ∆POR, (in which DFOR) [given]
∴ Using the Basic proportionality theorem, we have
ncert .....(2)
From (1) and (2),
ncert
ncert
Now, in ∆PQR,
E and F are two distinct points on PQ and PR respectively and , ie., E and F dividing the two sides PQ and PR in the same ratio in ∆PQR.
∴ By converse of Basic proportionality theorem, EFQR.

Q 9.

In the figure A, B and C are points on OP, OQ and OR respectively such that ABPQ and
ACPR. Show that BCQR.
ncert

SOLUTION:

Soln.: In ∆PQR, O is a point and OP, OQ and OR are joined. We have points A, B, and C on OP, OQ and OR respectively such that ABPQ and ACPR.
Now, in ∆OPQ,
ABPQ [Given]
ncert .....(1)
[By the Basic proportionality theorem]
Again, in ∆OPR,
ACPR [Given]
∴ Using the Basic proportionality theorem, we have
ncert .....(2)
From (1) and (2),
ncert
ncert
Now, in ∆OQR,
B is a point on OQ, C is a point on OR
ncert [Proved above]
i.e., B and C divide the sides OQ and OR in the same ratio in ∆OQR.
BCQR.

Q 10.

Using Basic proportionality theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

SOLUTION:

Soln. : We have ∆ABC, in which D is the midpoint of AB and E is a point on AC such that DEBC.
ncert
DEBC [given]
∴ Using the Basic proportionality theorem, we get
ncert .....(1)
But D is the mid-point of AB
AD = DB
ncert .....(2)
From (1) and (2),
ncert
EC = AE
E is the mid point of AC. Hence, it is proved that a line through the midpoint of one side of a triangle parallel to another side bisects the third side.

Q 11.

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.

SOLUTION:

Soln. : We have ∆ABC, in which D and E are the mid-points of sides AB and AC respectively.
AD = DB and AE = EC
From (1) and (2), we have
ncert
ncert
ncert
DEBC (By converse of Basic proportionality theorem).

Q 12.

ABCD is a trapezium in which ABDC and its diagonals intersect each other at the point O. Show that ncert

SOLUTION:

Soln. : We have, a trapezium ABCD such that
ABDC. The diagonals AC and BD intersect each other at O.
ncert
Let us draw OE parallel to either AB or DC.
In ∆ADC
OEDC [By construction]
∴ Using the Basic proportionality theorem, we get
ncert .....(1)
In ∆ABD
OEAB [By construction]
∴ Using the Basic proportionality theorem, we get
ncert
ncert .....(2)
From (1) and (2),
ncert
ncert

Q 13.

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

SOLUTION:

Soln. : It is given that ncert
From ncert
(Through O, draw OEBA)
In ∆ADB
OEAB [By construction]
ncert
∴ Using the Basic proportionality theorem, we get
ncert
ncert .....(1)
Also, ncert .....(2)
From (1) and (2), we have
ncert
i.e., the points O and E on the sides AC and AD (of ∆ADC) respectively are in the same ratio.
∴ Using the Basic proportionality theorem, we get
OEDC and OEAB
ABDC
ABCD is a trapezium.

Q 14.

State which pairs of triangles in the given figures 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) ncert
(ii) ncert
(iii) ncert
(iv) ncert
(v) ncert
(vi) ncert

SOLUTION:

Soln. : (i) In ∆ABC and ∆PQR,
We have : ∠A = ∠P = 60°
B = ∠Q = 80°
C = ∠R = 40°
∴ The corresponding angles are equal.
∴ Using the AAA similarity rule,
ABC ~ ∆PQR
(ii) In ∆ABC and ∆QRP,
ncert
ncert
ncert
i.e., , ncert
i.e., Using the SSS similarity, ∆ABC ~ ∆QRP
(iii) In ∆LMP and ∆DEF,
ncert
ncert
⇒ ∆s are not similar.
(iv) In ∆MNL and ∆QPR,
ncert and ∠NML = ∠PQR
∴ Using SAS criteria of similarity, we have
MNL ~ ∆QPR.
(v) In ∆ABC and ∆FDE,
A = ∠F = 80°
ncert are unknown.
∴ The ∆s cannot said to be similar.
(vi) In ∆DEF and ∆PQR,
D = ∠P = 70°
[∵ ∠P = 180° - (80° + 30°) = 180° - 110° = 70°]
E = ∠Q = 80°
F = ∠R = 30° [( ∠F = 180°]
∴ Using the AAA similarity rule, ∆DFE ~ ∆PRQ.

Q 15.

2. In the figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
ncert

SOLUTION:

Soln.: We have, ∠BOC = 125° and ∠CDO = 70°
since, ∠DOC + ∠BOC = 180° [Linear pair]
⇒ ∠DOC = 180° - 125° = 55° .....(1)
In ∆DOC,
Using the angle sum property for DODC,
we get
DOC + ∠ODC + ∠DCO = 180°
⇒ 55° + 70° + ∠DCO = 180°
⇒ ∠DCO = 180° - 55° - 70° = 55°
Again,
DOC = ∠BOA .....(2)
[vertically opposite angles]
and ∠OCD = ∠OAB = 55° .....(3)
(corresponding angles of similar triangles)
Thus, from (1), (2) and (3)
DOC = 55°, ∠DCO = 55° and ∠OAB = 55°.

Q 16.

Diagonals AC and BD of a trapezium ABCD with ABDC intersect each other at the point O. Using a similarity criterion for two triangles, show that ncert
ncert

SOLUTION:

Soln.: We have a trapezium ABCD in which ABDC. The diagonals AC and BD intersect at O.
In ∆OAB and ∆OCD,
ABDC and AC and BD intersect each other at point O.
∴ ∠OBA = ∠ODC (Alternate angles)
and ∠OAB = ∠OCD (Alternate angles)
∴ Using AA similarity rule, ∆OAB ~ ∆OCD
So, ncert (Ratios of corresponding sides of the similar triangles)
ncert

Q 17.

In the figure, ncert and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.
ncert

SOLUTION:

Soln.: In ∆PQR
∵ ∠1 = ∠2 [Given]
PR = QP .....(1)
[( In a ∆, sides opposite to equal angles are equal]
ncert [Given] .....(2)
From (1) and (2),
ncert .....(3)
(By taking reciprocals)
Now, in ∆PQS and ∆TQR,
ncert [From (3)]
and ∠SQP = ∠RQT = ∠1
Now, using SAS similarity rule,
PQS ~ ∆TQR.

Q 18.

S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that
RPQ ~ ∆RTS.

SOLUTION:

Soln. : In ∆PQR,
T is a point on QR and S is a point on PR such that ∠RTS = ∠P.
ncert
Now in ∆RPQ and ∆RTS,
RPQ = ∠RTS [Given]
PRQ = ∠TRS [Common]
∴ Using AA similarity, we have ∆RPQ ~ ∆RTS.

Q 19.

In the figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
ncert

SOLUTION:

Soln. : We have, ∆ABE ≅ ∆ACD
∴ Their corresponding parts are equal, i.e.,
AB = AC, AE = AD
ncert
ncert
ncert .....(1) [∵ AE = AD]
Now in ∆ADE and ∆ABC, ncert [from (1)]
and ∠DAE = ∠BAC (each = ∠A)
∴ Using the SAS similarity, we have
ADE ~ ∆ABC.

Q 20.

In the 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:

Soln.: We have a ∆ABC in which altitude AD and CE intersect each other at P.
⇒ ∠D = ∠E = 90° .....(1)
(i) In ∆AEP and ∆CDP,
AEP = ∠CDP [From (1)]
EPA = ∠DPC [Vertically opp. angles]
∴ Using AA similarity, we get ∆AEP ~ ∆CDP
(ii) In ∆ABD and ∆CBE,
ADB = ∠CEB [From (1)]
Also, ∠ABD = ∠CBE [Common]
∴ Using AA similarity, we have
ABD ~ ∆CBE
(iii) In ∆AEP and ∆ADB,
∵ ∠AEP = ∠ADB [From (1)]
Also, ∠EAP = ∠DAB [Common]
∴ Using AA similarity, we have
AEP ~ ∆ADB
(iv) In ∆PDC and ∆BEC,
∵ ∠PDC = ∠BEC [From (1)]
Also, ∠DCP = ∠ECB [Common]
∴ Using AA similarity, we have
PDC ~ ∆BEC

Q 21.

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
ncert

SOLUTION:

Soln.: We have a parallelogram ABCD in which AD is produced to E and BE is joined such that BE intersect CD at F.
Now, in ∆ABE and ∆CFB
BAE = ∠FCB [Opp. angles of a ∥ gm are always equal]
AEB = ∠CBF [∵ Parallel sides are intersected by the transversal BE]
Now, using AA similarity, we have ∆ABE ~ ∆CFB.

Q 22.

In the figure, ABC and AMP are two right triangles, right angled at B and M respectively.
ncert
Prove that : (i) ∆ABC ~ ∆AMP
(ii) ncert

SOLUTION:

Soln. : We have ∆ABC, right angled at B and ∆AMP, right angled at M.
∴ ∠B = ∠M = 90° .....(1)
(i) In ∆ABC and ∆AMP,
∵ ∠ABC = ∠AMP [From (1)]
and ∠BAC = ∠MAP [Common]
∴ Using AA similarity, we get ∆ABC ~ ∆AMP
(ii) ( ∆ABC ~ ∆AMP [As proved above]
∴ Their corresponding sides are proportional.
ncert

Q 23.

CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that :
(i) ncert
(ii) ∆DCB ~ ∆HGE
(iii) ∆DCA ~ ∆HGF
ncert

SOLUTION:

Soln.: We have, two similar ∆ABC and ∆FEG such that CD and GH are the bisectors of ∠ACB and ∠FGE respectively.
(i) In ∆ACD and ∆FGH,
A = ∠F [∵ ∆ABC ~ ∆FEG] .....(1)
since ∆ABC ~ ∆FEG
∴ ∠C = ∠G ncert
⇒ ∠ACD = ∠FGH .....(2)
From (1) and (2),
ACD ~ ∆FGH [AA similarity]
∴ Their corresponding sides are proportional,
ncert
(ii) In ∆DCB and ∆HGE,
B = ∠E [∵ ∆ABC ~ ∆FEG] .....(1)
Again, ∆ABC ~ ∆FEG ⇒ ∠ACB = ∠FGE
ncert
⇒ ∠DCB = ∠HGE .....(2)
From (1) and (2),
DCB ~ ∆HGE [AA similarity]
(iii) In ∆DCA and ∆HGF,
∵ ∆ABC ~ ∆FEG ⇒ ∠CAB = ∠GFE
⇒ ∠CAD = ∠GFH ⇒ ∠DAC = ∠HFG .....(1)
Also, ∆ABC ~ ∆FEG ⇒ ∠ACB = ∠FGE
ncert
⇒ ∠DCA = ∠HGF .....(2)
From (1) and (2), we have
DCA ~ ∆HGF [AA similarity]

Q 24.

In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If ADBC and EFAC, prove that ∆ABD ~ ∆ECF.
ncert

SOLUTION:

Soln.: We have an isosceles ∆ABC in which AB = AC.
In ∆ABD and ∆ECF,
AB = AC [Given]
⇒ Angles opposite to them are equal
∴ ∠ACB = ∠ABC
⇒ ∠ECF = ∠ABD .....(1)
Again, ADBC and EFAC
⇒ ∠ADB = ∠EFC = 90° .....(2)
From (1) and (2), we have
ABD ~ ∆ECF [AA similarity]

Q 25.

Side AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR (see figure). Show that ∆ABC ~ ∆PQR.
ncert

SOLUTION:

Soln.: We have ∆ABC and ∆PQR in which AD and PM are medians corresponding to sides BC and QR
respectively such that
ncert
ncert
∴ Using SSS similarity, we have
ABD ~PQM
∴ Their corresponding angles are equal.
⇒ ∠ABD =PQM ⇒ ∠ABC =PQR
Now, in ∆ABC and ∆PQR, ncert .....(1)
Also, ∠ABC =PQR .....(2)
∴ Using SAS similarity from (1) and (2), we get ∆ABC ~ ∆PQR.

Q 26.

D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CBCD.
ncert

SOLUTION:

Soln.: We have a ∆ABC and a point D on its side BC such that ∠ADC = ∠BAC.
In ∆BAC and ∆ADC,
∵ ∠BAC =ADC [Given]
and ∠BCA =DCA [Common]
∴ Using AA similarity, we have ∆BAC ~ ∆ADC.
∴ Their corresponding sides are proportional.
ncert
CA × CA = CB × CD
CA2 = CB × CD

Q 27.

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 ∆ABC ~ ∆PQR.

SOLUTION:

Soln. : Given : ∆ABC and ∆PQR in which AD and PM are medians.
Also, ncert .....(1)
To prove : ∆ABC ~ ∆PQR
Construction : Produce AD to E and PM to N such that AD = DE and PM = MN. Join BE, CE, QN and RN.
ncert
Proof : Quadrilaterals ABEC and PQNR are parallelograms, since their diagonals bisect each other at point D and M respectively.
BE = AC and QN = PR
ncert
[By (1)]
i.e., ncert .....(2)
From (1), ncert
ncert .....(3)
From (2) and (3), we have
ncert
⇒ ∆ABE ~ ∆PQN ⇒ ∠1 = ∠3 .....(4)
Similarly, we can prove
ACE ~ ∆PRN ⇒ ∠2 = ∠4 .....(5)
From (4) and (5)
⇒ ∠1 + ∠2 = ∠3 + ∠4
⇒ ∠A = ∠P .....(6)
Now, in ∆ABC and ∆PQR, we have
ncert [From (1)]
and ∠A = ∠P [From (6)]
∴ ∆ABC ~ ∆PQR (SAS similarity criterion)

Q 28.

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

SOLUTION:

Soln. : Let AB = 6m be the pole and BC = 4 m be its shadow (in right ∆ABC), whereas DE and EF denote the tower and its shadow respectively.
EF = Length of the shadow of the tower = 28 m
and DE = h = Height of the tower
ncert
In ∆ABC and ∆DEF, we have ∠B = ∠E = 90°
A = ∠D [∵ Angular elevation of the
sun at the same time is equal]
∴ Using AA criteria of similarity, we have
ABC ~ ∆DEF
∴ Their sides are proportional i.e., ncert
ncert
Thus, the required height of the tower is 42 m.

Q 29.

16. If AD and PM are medians of triangles ABC and PQR, respectively where, ∆ABC ~ ∆PQR, prove that ncert

SOLUTION:

Soln. : We have ∆ABC ~ ∆PQR such that AD and PM are the medians corresponding to the sides BC and QR respectively.
ncert
∵ ∆ABC ~ ∆PQR
And the corresponding sides of similar triangles are proportional.
ncert .....(1)
∵ Corresponding angles are also equal in two similar triangles.
∴ ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R .....(2)
Since, AD and PM are medians.
BC = 2BD and QR = 2QM
∴ From (1), ncert .....(3)
And ∠B = ∠Q ⇒ ∠ABD = ∠PQM .....(4)
∴ From (3) and (4), we have
ABD ~ ∆PQM
∴ Their corresponding sides are proportional.
ncert