Question Bank
Similarity of Triangle
1. On a map drawn to a scale 1 : 25000, a rectangular plot of land
ABCD has the following measurements. AB = 12 cm, BC = 16 cm.
Angles A, B, C, D are all 90° each. Calculate
(i) t he diagonal distance of the plot in km.
(ii) the area of the plot in km2
[6 marks] (1998)
Solution. We have,
1
AB = 12 cm, BC = 16 cm and K =
25000
(i) Length of the diagonal AC on the map
AB2 + BC2 = 144 + 256 cm
∴
⇒
⇒
(ii)
∴
⇒
⇒
= 400 cm
= 20 cm
Length of the diagonal on the map = K × Actual length of the
diagonal AC.
1
20 cm =
× Actual length of the diagonal AC
25000
Actual length of the diagonal AC = (20 × 25000) cm
20 × 25000
=
km = 5 km.
100 × 1000
Area of the plot in the map = (12 × 16) cm2 = 192 cm2
Area of the map = K2 × Actual area
2
⎛ 1 ⎞
2
192 cm
=⎜
⎟ × Actual area
⎝ 25000 ⎠
Actual area = (192 × 25000 × 25000) cm2
⎛ 192 × 25000 × 25000 ⎞
2
=⎜
⎟ km
⎝ 100 × 100 × 1000 × 1000 ⎠
= 12 km2
Math Class X
1
Question Bank
2. The model of a ship is made to a scale of 1 : 250. Find
(i) the length of the ship, if the length of its model is 1.2 m.
(ii) the area of the deck of the ship, if the area of the deck of its
model is 1.6 m2
(iii) the volume of its model, when the volume of the ship is 1
cubic kilometer.
1
Solution. We have, K =
250
(i) We have, length of the model = K × Actual length
1
⇒ 1.2 m =
× Actual length
250
⇒ Actual length = (1.2 × 250) m = 300 m
(ii) We have, area of the model = K2 × Actual area
2
⎛ 1 ⎞
2
⇒ 1.6 m = ⎜
⎟ × Actual area
250
⎝
⎠
⇒ Actual area = (1.6 × 250 × 250) m2
= 100000 m2
(iii) We have, volume of the model = K3 × Actual volume
⎛ 1 ⎞
⇒ Volume of the model = ⎜
⎟ × 1 km3
3
⎜ ( 250 ) ⎟
⎝
⎠
1000 × 1000 × 1000 3
=
m
250 × 250 × 250
= 65 m3
3. In the adjoining figure, LM is parallel to BC. AB = 6 cm,
AL = 2 cm and AC = 9 cm. Calculate :
(i) the length of CM
(ii) the value of
Area of ΔALM
Area of trapezium LBCM
Solution. In ΔALM and ΔABC, we have,
Math Class X
2
Question Bank
∠ALM = ∠ABC
[Corresponding angles]
∠LAM = ∠BAC
[Common]
∴ ΔALM ~ ΔABC
[AA similarity]
(i) Since, sides of similar triangles are proportional, therefore.
AL
AM
⇒
=
AB
AC
2 AM
=
⇒
6
9
18
⇒ AM =
= 3 cm
6
∴ CM = AC – AM
= (9 – 3) cm = 6 cm.
(ii) ∴ ΔALM ~ ΔABC, therefore
Area of ΔALM
AL2
4
1
=
=
=
Area of ΔABC
AB2
36 9
2
Let area of ΔALM be x cm .
Then area of ΔABC = 9x cm2
Area of ΔALM
∴
Area of trapezium LBCM
Area of ΔALM
x
1
=
=
=
Area of ΔABC – Area of ΔALM 9 x – x 8
area of ΔALM
1
=
Hence,
area of trapezium LBCM 8
4. In figure, ABCD is a trapezium
1
with AB || DC and AB = DC.
2
Prove that O is the point of trisection
of diagonals AC and BD.
Solution. In ΔOAB and ΔOCD, we have
∠OAB = ∠OCD
[Alternate angles]
∠OBA = ∠ODC
[Alternate angles]
∴
ΔOAB ~ ΔOCD
[AA similarity]
Math Class X
3
Question Bank
Since sides of similar triangles are proportional, therefore,
OA
OB
AB 1 ⎡ AB 1
⎤
=
=
=
=
,
given
∵
⎥⎦
OC
OD CD
2 ⎢⎣ CD
2
OA 1
OB
1
= and
=
⇒
OC
2
OD
2
⇒ OC = 2OA and OD = 2OB
Hence, O is the point of trisection of diagonals AC and BD.
5. In the figure alongside, PQRS
is a parallelogram. PQ = 16 cm,
QR = 10 cm. L is a point on PR
such that RL : LP = 2 : 3. QL
produced meets RS at M. and PS produced at N.
(i) Prove that ΔRLQ ~ ΔPLN. Hence, find PN.
(ii) Name a triangle similar to triangle RLM. Evaluate RM as a
fraction.
RL
2
=
Solution. We have, PQ = 16 cm, QR = 10 cm and
LP
3
(i) In ΔRLQ and ΔPLN, we have
∠RQL = ∠PNL
[Alternate angles]
∠LRQ = ∠NPL
[Alternate angles]
∴ ΔRLQ ~ ΔPLN
[AA similarity]
Since sides of similar triangles are proportional.
RQ
RL
10
2 ⎡ RL
2
⎤
∴
∵
=
=
=
(given)
⇒
⎥⎦
PN
PL
PN
3 ⎢⎣ LP
3
30
= 15
⇒ PN =
2
Hence, PN = 15 cm
(ii) In ΔRLM and ΔPLQ, we have
∠MRL = ∠LPQ
[Alternate angles]
∠RML = ∠PQL
[Alternate angles]
∴ ΔRLM ~ ΔPLQ
[AA similarity]
Math Class X
4
Question Bank
Since sides of similar triangles are proportional, therefore
RL RM
2 RM
32
2
=
⇒
=
⇒ RM =
= 10 cm
PL
PQ
3
16
3
3
2
Hence, RM = 10 cm
3
6. In ΔPQR, LM || QR and PM : MR = 3 : 4.
PL
LM
and
Find (i)
PQ
QR
Area (ΔLMN)
(ii)
Area (ΔMNR)
Area (ΔLQM)
(iii)
Area (ΔLQN)
Solution.
(i) In ΔPLM and ΔPQR, we have
∠PLM = ∠PQR
[Corresponding angles]
∠PML = ∠PRQ
[Corresponding angles]
∴
ΔPLM ~ ΔPQR
[AA - similarity]
PL
PM
LM
⇒
….(1)
=
=
PQ
PR
QR
[Corresponding sides of similar triangles are proportional]
PM
3
=
[Given]
Now, we have
MR
4
MR
4
⇒
=
PM
3
MR + PM
4+3
=
⇒
PM
3
PR
7
PM
3
=
=
⇒
⇒
….(2)
PM
3
PR
7
From (1) and (2), we get
Math Class X
5
Question Bank
PL
LM
3
=
=
PQ
QR
7
From M draw MD ⊥ LR
1
× LN × MD
Area (ΔLMN)
LN
2
…(3)
=
=
1
Area (ΔMNR)
NR
× NR × MD
2
In ΔLMN and ΔRQN, we have
∠LMN = ∠RQN
[Alternate angles]
∠MLN = ∠QRN
[Alternate angles]
∴
∠LMN ~ ∠RQN
[AA – similarity]
LM
LN
=
[Corresponding sides of similar triangles are
⇒
QR
NR
proportional]
LN
3
=
⇒
….(4)
NR
4
Area (ΔLMN)
3
=
From (3) and (4),
Area (ΔMNR) 7
(iii) From L, draw LE ⊥ QM.
1
× QM × LE
Area (ΔLQM)
QM
2
….(5)
=
=
1
Area (ΔLQN )
QN
× QN × LE
2
From part (ii), ΔLMN ~ ΔRQN
MN
LM
3
⇒
=
=
QN
QR
7
MN + QN 3 + 7
=
⇒
QN
7
QM 10
….(6)
⇒
=
QN
7
(ii)
Math Class X
6
Question Bank
From (5) and (6),
7.
Area (ΔLQM) 10
=
Area (ΔLQN)
7
In ΔABC, D, E and F are the
mid-points of BC, AC and AB
respectively. P, Q and R are the
mid-points of EF, FD and DE
respectively. Find :
Area (ΔABC)
Area (ΔPQR)
Area (ΔPQR)
(i)
(ii)
(iii)
Area (ΔBDF)
Area (ΔABC)
Area (ΔEDC)
Solution. We know that the line segment joining the mid-points of
any two sides of a triangle is parallel to the third side and equal to
half of it. [Mid-point Theorem]
1
…..(1)
∴
DE || AB and DE = AB
2
1
…..(2)
EF || BC and EF = BC
2
1
FD || AC and FD = AC
…..(3)
2
1
…..(4)
PQ || DE and PQ = DE
2
1
…..(5)
QR || EF and QR = EF
2
1
PR || FD and PR = FD
…..(6)
2
(i) In ΔABC and ΔBDF
∠ABC = ∠DBF
[Common]
∠ACB = ∠FDB
[Corresponding angles]
∴
ΔABC ~ ΔFBD
[AA similarity]
Math Class X
7
Question Bank
Area (ΔABC) AB2
AB2
=
=
2
2
Area (ΔFBD)
FB
⎛1
⎞
⎜ AB ⎟
⎝2
⎠
[∴ F is the mid-point of AB]
(ii) From (1) and (4)
1
AB || PQ || ED and PQ = AB
4
Similarly from (2) and (5)
1
BC || QR || FE and QR = BC and AC || PR || FD and PR ||
4
1
FD and PR = AC
4
∴ ∠PQR ~ ∠ABC [Sides are proportional]
2
Area (ΔPQR)
PQ 2
1
⎛1⎞
=
=
=
∴
⎜
⎟
Area (ΔABC) AB2 ⎝ 4 ⎠
16
(iii) In ΔABC and ΔEDC
∠ACB = ∠ECD
[Common]
∠ABC = ∠EDC
[Corresponding angles]
∴ ΔABC ~ ΔEDC
[AA similarity]
From part (ii), ΔABC ~ ΔPQR
ΔPQR ~ ΔEDC
1
ED 2
2
1
Area (ΔPQR)
PQ
=
= 2 2 =
2
4
Area (ΔEDC) ED
ED
Math Class X
8
Question Bank
8. In the figure, AD and BE are two medians of ΔABC intersecting at
G. DF || BE. Find :
Area (ΔCDF)
(i)
Area (ΔCBE)
Area (ΔAGE)
(ii)
Area (trap. GDFE)
Area (ΔCDF)
(iii)
Area (ΔABC)
Area (ΔAGE)
(iv)
Area (ΔABC)
Solution.
(i) In ΔCDF and ΔCBE,
∠DCF = ∠BCE
[Common]
∠FDC = ∠EBC
[Corresponding angles]
∴ ΔCDF ~ ΔCBE
[AA similarity]
2
Area (ΔCDF)
CF
∴
…(1)
=
Area (ΔCBE) CE 2
In ΔBCE, D is the mid-point of BC and DF || BE.
∴ F is the mid-point of CE. i.e., 2CF = CE
Area (ΔCDF)
CF2
1
∴
=
=
Area (ΔCBE) 4CF2
4
(ii) In ΔADF and ΔAGE,
∠ADF = ∠AGE
[Corresponding angles]
∠AFD = ∠AEG
[Corresponding angles]
∴
ΔADF ~ ΔAGE
[AA similarity]
2
Area (ΔADF)
AF
…(2)
=
∴
Area (ΔAGE) AE 2
Math Class X
9
Question Bank
In ΔADC, E is the mid-point of AC and F is the mid-point of EC.
1
1
i.e.,
AE = AC and CF = EF = AE
2
2
1
⇒
ΕF = AC
4
1
3
AF = AC – FC = AC – EF = AC – AC = AC.
4
4
2
⎛3
⎞
AC ⎟
⎜
Area (ΔADF) ⎝ 4
⎠ = 9 × 4 = 9
∴
=
2
Area (ΔAGE) ⎛ 1
16 1
4
⎞
AC
⎜
⎟
⎝2
⎠
Area (ΔADF) – Area (ΔAGE)
9–4
⇒
=
Area (ΔAGE)
4
Area (trap. GDFE)
5
⇒
=
Area (ΔAGE)
4
Area(ΔAGE)
4
⇒
=
Area (trap. GDFE) 5
(iii) In ΔABC, E is the mid-point of AC
1
Area (ΔABC) = Area (ΔBEC)
…(3)
∴
2
Area (ΔCDF)
Area (ΔCBE)
∴
=
[From part (i)]
Area (ΔABC) 4 Area (ΔABC)
Area(ΔABC)
1
=
=
2 × 4(Area ΔABC) 8
(iv) AD is the median of ΔABC
1
∴ Area (ΔADC) = Area (ΔABC)
2
1
⇒ Area (ΔADF) + Area (ΔCDF) = Area (ΔABC)
2
Math Class X
10
Question Bank
⇒
⇒
⇒
9
1
Area (ΔAGE) + Area (ΔABC)
4
8
1
= Area (ΔABC)
2
9
3
Area (ΔAGE) = Area (ΔABC)
4
8
Area(ΔAGE)
3 4 1
= × =
(Area ΔABC)
8 9 6
Math Class X
11
[From part (ii) and (iii)]
Question Bank