Exercise 6.1
1. Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
Answer: Similar
(ii) All squares are __________. (similar, congruent)
Answer: Similar
(iii) All __________ triangles are similar. (isosceles, equilateral)
Answer: 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: (a) Equal
(b) Proportional
2. Give two different examples of pair of
(i) Similar figures
(ii) Non-similar figures
Solution:
Exercise 6.2
1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution:
(i) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒1.5/3 = 1/EC
⇒EC = 3/1.5
EC = 3×10/15 = 2 cm
Hence, EC = 2 cm.
(ii) Given, in △ ABC, DE∥BC
∴ AD/DB = AE/EC [Using Basic proportionality theorem]
⇒ AD/7.2 = 1.8 / 5.4
⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10
⇒ AD = 2.4
Hence, AD = 2.4 cm.
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.
(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
Solution:
Given, in ΔPQR, E and F are two points on side PQ and PR respectively. See the figure below;
(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm
Therefore, by using Basic proportionality theorem, we get,
PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3
And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5
So, we get, PE/EQ ≠ PF/FR
Hence, EF is not parallel to QR.
(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm
Therefore, by using Basic proportionality theorem, we get,
PE/QE = 4/4.5 = 40/45 = 8/9
And, PF/RF = 8/9
So, we get here,
PE/QE = PF/RF
Hence, EF is parallel to QR.
(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
From the figure,
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm
So, PE/EQ = 0.18/1.10 = 18/110 = 9/55…………. (i)
And, PE/FR = 0.36/2.20 = 36/220 = 9/55………… (ii)
So, we get here,
PE/EQ = PF/FR
Hence, EF is parallel to QR.
3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD
Solution:
In the given figure, we can see, LM || CB,
By using basic proportionality theorem, we get,
AM/AB = AL/AC……………………..(i)
Similarly, given, LN || CD and using basic proportionality theorem,
∴AN/AD = AL/AC……………………………(ii)
From equation (i) and (ii), we get,
AM/AB = AN/AD
Hence, proved.
4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC
Solution:
In ΔABC, given as, DE || AC
Thus, by using Basic Proportionality Theorem, we get,
∴BD/DA = BE/EC ………………………………………………(i)
In ΔABC, given as, DF || AE
Thus, by using Basic Proportionality Theorem, we get,
∴BD/DA = BF/FE ………………………………………………(ii)
From equation (i) and (ii), we get
BE/EC = BF/FE
Hence, proved.
5. In the figure, DE||OQ and DF||OR, show that EF||QR.
Solution:
Given,
In ΔPQO, DE || OQ
So by using Basic Proportionality Theorem,
PD/DO = PE/EQ……………… ..(i)
Again given, in ΔPOR, DF || OR,
So by using Basic Proportionality Theorem,
PD/DO = PF/FR………………… (ii)
From equation (i) and (ii), we get,
PE/EQ = PF/FR
Therefore, by converse of Basic Proportionality Theorem,
EF || QR, in ΔPQR.
6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Solution:
Given here,
In ΔOPQ, AB || PQ
By using Basic Proportionality Theorem,
OA/AP = OB/BQ…………….(i)
Also given,
In ΔOPR, AC || PR
By using Basic Proportionality Theorem
∴ OA/AP = OC/CR……………(ii)
From equation (i) and (ii), we get,
OB/BQ = OC/CR
Therefore, by converse of Basic Proportionality Theorem,
In ΔOQR, BC || QR.
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).
Solution:
Given, in ΔABC, D is the midpoint of AB such that AD=DB.
A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.
We have to prove that E is the mid point of AC.
Since, D is the mid-point of AB.
∴ AD=DB
⇒AD/DB = 1 …………………………. (i)
In ΔABC, DE || BC,
By using Basic Proportionality Theorem,
Therefore, AD/DB = AE/EC
From equation (i), we can write,
⇒ 1 = AE/EC
∴ AE = EC
Hence, proved, E is the midpoint of AC.
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).
Solution:
Given, in ΔABC, D and E are the mid points of AB and AC respectively, such that,
AD=BD and AE=EC.
We have to prove that: DE || BC.
Since, D is the midpoint of AB
∴ AD=DB
⇒AD/BD = 1……………………………….. (i)
Also given, E is the mid-point of AC.
∴ AE=EC
⇒ AE/EC = 1
From equation (i) and (ii), we get,
AD/BD = AE/EC
By converse of Basic Proportionality Theorem,
DE || BC
Hence, proved.
9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Solution:
Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.
We have to prove, AO/BO = CO/DO
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔADC, we have OE || DC
Therefore, By using Basic Proportionality Theorem
AE/ED = AO/CO ……………..(i)
Now, In ΔABD, OE || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/BO…………….(ii)
From equation (i) and (ii), we get,
AO/CO = BO/DO
⇒AO/BO = CO/DO
Hence, proved.
10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.
Solution:
Given, Quadrilateral ABCD where AC and BD intersects each other at O such that,
AO/BO = CO/DO.
We have to prove here, ABCD is a trapezium
From the point O, draw a line EO touching AD at E, in such a way that,
EO || DC || AB
In ΔDAB, EO || AB
Therefore, By using Basic Proportionality Theorem
DE/EA = DO/OB ……………………(i)
Also, given,
AO/BO = CO/DO
⇒ AO/CO = BO/DO
⇒ CO/AO = DO/BO
⇒DO/OB = CO/AO …………………………..(ii)
From equation (i) and (ii), we get
DE/EA = CO/AO
Therefore, By using converse of Basic Proportionality Theorem,
EO || DC also EO || AB
⇒ AB || DC.
Hence, quadrilateral ABCD is a trapezium with AB || CD
Exercise 6.3
Question 1.
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 :
Solution:
Exercise 6.5
We have to prove, PM2 = QM × MR
Solution:
(i) In ΔADB and ΔCAB,
∠DAB = ∠ACB (Each 90°)
∠ABD = ∠CBA (Common angles)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB2 = CB × BD
(ii) Let ∠CAB = x
In ΔCBA,
∠CBA = 180° – 90° – x
∠CBA = 90° – x
Similarly, in ΔCAD
∠CAD = 90° – ∠CBA
= 90° – x
∠CDA = 180° – 90° – (90° – x)
∠CDA = x
In ΔCBA and ΔCAD, we have
∠CBA = ∠CAD
∠CAB = ∠CDA
∠ACB = ∠DCA (Each 90°)
∴ ΔCBA ~ ΔCAD [AAA similarity criterion]
⇒ AC/DC = BC/AC
⇒ AC2 = DC × BC
(iii) In ΔDCA and ΔDAB,
∠DCA = ∠DAB (Each 90°)
∠CDA = ∠ADB (common angles)
∴ ΔDCA ~ ΔDAB [AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD2 = BD × CD
4. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 .
Solution:
Given, ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (By isosceles triangle property)
AB2 = AC2 + BC2 [By Pythagoras theorem]
= AC2 + AC2 [Since, AC = BC]
AB2 = 2AC2
5. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Solution:
Given, ΔABC is an isosceles triangle having = BC and AB2 = 2AC2
In ΔACB,
AC = BC
AB2 = 2AC2
AB2 = AC2 + AC2
= AC2 + BC2 [Since, AC = BC]
Hence, by Pythagoras theorem ΔABC is right angle triangle.
6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Given, ABC is an equilateral triangle of side 2a.
Draw, AD ⊥ BC
In ΔADB and ΔADC,
AB = AC
AD = AD
∠ADB = ∠ADC [Both are 90°]
Therefore, ΔADB ≅ ΔADC by RHS congruence.
Hence, BD = DC [by CPCT]
In right angled ΔADB,
AB2 = AD2 + BD2
(2a)2 = AD2 + a2
⇒ AD2 = 4a2 – a2
⇒ AD2 = 3a2
⇒ AD = √3a
7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
Solution:
Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.
We have to prove, as per the question,
AB2 + BC2 + CD2 + AD2 = AC2 + BD2
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB2 = AO2 + BO2 …………………….. (i) [By Pythagoras theorem]
Similarly,
AD2 = AO2 + DO2 …………………….. (ii)
DC2 = DO2 + CO2 …………………….. (iii)
BC2 = CO2 + BO2 …………………….. (iv)
Adding equations (i) + (ii) + (iii) + (iv), we get,
AB2 + AD2 + DC2 + BC2 = 2(AO2 + BO2 + DO2 + CO2)
= 4AO2 + 4BO2 [Since, AO = CO and BO =DO]
= (2AO)2 + (2BO)2 = AC2 + BD2
AB2 + AD2 + DC2 + BC2 = AC2 + BD2
Hence, proved.
8. In Fig. 6.54, O is a point in the interior of a triangle.
ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that:
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2 ,
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Solution:
Given, in ΔABC, O is a point in the interior of a triangle.
And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
Join OA, OB and OC
(i) By Pythagoras theorem in ΔAOF, we have
OA2 = OF2 + AF2
Similarly, in ΔBOD
OB2 = OD2 + BD2
Similarly, in ΔCOE
OC2 = OE2 + EC2
Adding these equations,
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2
OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.
(ii) AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)
∴ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Solution:
Given, a ladder 10 m long reaches a window 8 m above the ground.
Let BA be the wall and AC be the ladder,
Therefore, by Pythagoras theorem,
AC2 = AB2 + BC2
102 = 82 + BC2
BC2 = 100 – 64
BC2 = 36
BC = 6m
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.
10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:
Given, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.
Let AB be the pole and AC be the wire.
By Pythagoras theorem,
AC2 = AB2 + BC2
242 = 182 + BC2
BC2 = 576 – 324
BC2 = 252
BC = 6√7m
Therefore, the distance from the base is 6√7m.
11. An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after
hours?
Solution:
Given,
Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane flying due north in
hours (OA) = 100 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane flying due west in
hours (OB) = 1200 × 3/2 km = 1800 km.
In right angle ΔAOB, by Pythagoras Theorem,
AB2 = AO2 + OB2
⇒ AB2 = (1500)2 + (1800)2
⇒ AB = √(2250000 + 3240000)
= √5490000
⇒ AB = 300√61 km
Hence, the distance between two aeroplanes will be 300√61 km.
12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:
Given, Two poles of heights 6 m and 11 m stand on a plane ground.
And distance between the feet of the poles is 12 m.
Let AB and CD be the poles of height 6m and 11m.
Therefore, CP = 11 – 6 = 5m
From the figure, it can be observed that AP = 12m
By Pythagoras theorem for ΔAPC, we get,
AP2 = PC2 + AC2
(12m)2 + (5m)2 = (AC)2
AC2 = (144+25) m2 = 169 m2
AC = 13m
Therefore, the distance between their tops is 13 m.
13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.
Solution:
Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
By Pythagoras theorem in ΔACE, we get
AC2 + CE2 = AE2 ………………………………………….(i)
In ΔBCD, by Pythagoras theorem, we get
BC2 + CD2 = BD2 ………………………………..(ii)
From equations (i) and (ii), we get,
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii)
In ΔCDE, by Pythagoras theorem, we get
DE2 = CD2 + CE2
In ΔABC, by Pythagoras theorem, we get
AB2 = AC2 + CB2
Putting the above two values in equation (iii), we get
DE2 + AB2 = AE2 + BD2.
14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB2 = 2AC2 + BC2
Solution:
Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;
DB = 3CD.
In Δ ABC,
AD ⊥BC and BD = 3CD
In right angle triangle, ADB and ADC, by Pythagoras theorem,
AB2 = AD2 + BD2 ……………………….(i)
AC2 = AD2 + DC2 ……………………………..(ii)
Subtracting equation (ii) from equation (i), we get
AB2 – AC2 = BD2 – DC2
= 9CD2 – CD2 [Since, BD = 3CD]
= 8CD2
= 8(BC/4)2 [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.
Solution:
Given, ABC is an equilateral triangle.
And D is a point on side BC such that BD = 1/3BC
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
And, AE = a√3/2
Given, BD = 1/3BC
∴ BD = a/3
DE = BE – BD = a/2 – a/3 = a/6
In ΔADE, by Pythagoras theorem,
AD2 = AE2 + DE2
⇒ 9 AD2 = 7 AB2
16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:
Given, an equilateral triangle say ABC,
Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.
∴ BE = EC = BC/2 = a/2
In ΔABE, by Pythagoras Theorem, we get
AB2 = AE2 + BE2
4AE2 = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
Hence, proved.
17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Solution:
Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
We can observe that,
AB2 = 108
AC2 = 144
And, BC2 = 36
AB2 + BC2 = AC2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct answer is (C).