NCERT solutions for class 10 Maths
Chapter 6
Triangles
Exercise 6.5
Ex 6.5 Question 1.
Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Solution:
(i)Sides of the triangle = 7 cm, 24 cm, and 25 cm.
Squaring these sides
= 49, 576, and 625.
49 + 576 = 625
(7)2 + (24)2 = (25)2
The above equation satisfies, Pythagoras theorem. Hence, these are sides of a right angled triangle.
∴ Its lenght = 25 cm
(ii) Sides of the triangle =3 cm, 8 cm, and 6 cm.
Squaring these sides
=9, 64, and 36.
9 + 36 ≠ 64
Or, 32 + 62 ≠ 82
The sides do not satisfies the Pythagoras theorem.Hence, these are not the sides of a right angled triangle.
(iii)Sides of triangle = 50 cm, 80 cm, and 100 cm.
Squaring these sides
=2500, 6400, and 10000.
2500 + 6400 ≠ 10000
Or, 502 + 802 ≠ 1002
The sides do not satisfies the Pythagoras theorem.Hence, these are not the sides of a right angled triangle.
(iv) Sides of triangle = 13 cm, 12 cm, and 5 cm.
Squaring these sides
=169, 144, and 25.
144 +25 = 169
122 + 52 = 132
The above equation satisfies, Pythagoras theorem. Hence, these are sides of a right angled triangle.
∴ Its lenght = 13 cm
Ex 6.5 Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM × MR.
Solution:
To prove = PM2 = QM × MR
In ΔPQM,
Using pythagoras theorem
PQ2 = PM2 + QM2
Or, PM2 = PQ2 – QM2 ………..(i)
In ΔPMR, by Pythagoras theorem
PR2 = PM2 + MR2
Or, PM2 = PR2 – MR2 ……………..(ii)
Adding equation, (i) and (ii),
2PM2 = (PQ2 + PM2) – (QM2 + MR2)
= QR2 – QM2 – MR2 [∴ QR2 = PQ2 + PR2]
= (QM + MR)2 – QM2 – MR2
= 2QM × MR
∴ PM2 = QM × MR
Ex 6.5 Question 3.
In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB2 = BC × BD
(ii) AC2 = BC × DC
(iii) AD2 = BD × CD
Solution:
(i) In ΔADB and ΔCAB,
∠DAB = ∠ACB [Each 90°]
∠ABD = ∠CBA [Common]
∴ ΔADB ~ ΔCAB [AA similarity]
⇒ 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,
∠CBA = ∠CAD [Proved above]
∠CAB = ∠CDA [proved above]
∠ACB = ∠DCA [Each 90°]
∴ ΔCBA ~ ΔCAD [AAA similarity]
⇒ AC/DC = BC/AC
⇒ AC2 = DC × BC
(iii) In ΔDCA and ΔDAB,
∠DCA = ∠DAB [Each 90°]
∠CDA = ∠ADB [common]
∴ ΔDCA ~ ΔDAB [AA similarity]
⇒ DC/DA = DA/DA
⇒ AD2 = BD × CD
Ex 6.5 Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 .
Solution:
In ΔACB,
AC = BC [isosceles triangle property]
∠C = 90°
Using pythagoras theorem
AB2 = AC2 + BC2
= AC2 + AC2 [Because, AC = BC]
AB2 = 2AC2
Ex 6.5 Question 5.
ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ABC is a right triangle.
Solution:
Given, AB2 = 2AC2
AB2 = AC2 + AC2
= AC2 + BC2 [Because, AC = BC]
These sides satisfies Pythagoras theorem. Hence, the triangle is a right angled triangle.
Ex 6.5 Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
In ΔADB and ΔADC,
AB = AC
AD = AD
∠ADB = ∠ADC [Each 90°]
∴ ΔADB ≅ ΔADC [by RHS congruence rule]
Hence, BD = DC [by CPCT]
In ΔADB,
AB2 = AD2 + BD2
(2a)2 = AD2 + a2
⇒ AD2 = 4a2 – a2
⇒ AD2 = 3a2
⇒ AD = √3a
Ex 6.5 Question 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 that- ABCD is a rhombus. Diagonals AC and BD intersect at O.
TO prove – AB2 + BC2 + CD2 + AD2 = AC2 + BD2
The diagonals of a rhombus bisect each other at right angles.
∴AO = CO and BO = DO …….(i)
In ΔAOB,
∠AOB = 90°
AB2 = AO2 + BO2 ………….. (ii) [Using Pythagoras theorem]
Similarly,
AD2 = AO2 + DO2 ………….. (iii)
DC2 = DO2 + CO2 ………….. (iv)
BC2 = CO2 + BO2 ………….. (v)
Adding equations (ii) + (iii) + (iv) + (v),
AB2 + AD2 + DC2 + BC2 = 2(AO2 + BO2 + DO2 + CO2)
= 4AO2 + 4BO2 [from equation (i)]
= (2AO)2 + (2BO)2 = AC2 + BD2
AB2 + AD2 + DC2 + BC2 = AC2 + BD2
Ex 6.5 Question 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:
(i)In ΔAOF, By Pythagoras theorem
OA2 = OF2 + AF2………….. (i)
In ΔBOD, By Pythagoras theorem
OB2 = OD2 + BD2………….. (ii)
In ΔCOE,By Pythagoras theorem
OC2 = OE2 + EC2………….. (iii)
Adding equations, (i),(ii) and (iii)
OA2 + OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2
⇒ OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2.………….. (iv)
(ii) From equation (iv)
AF2 + BD2 + EC2 = (OA2 – OE2) + (OC2 – OD2) + (OB2 – OF2)
∴ ⇒ AF2 + BD2 + CE2 = AE2 + CD2 + BF2.
Ex 6.5 Question 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:
Let
BA be the wall = 8 m
and AC be the ladder = 10 m
Using Pythagoras theorem,
AC2 = AB2 + BC2
102 = 82 + BC2
BC2 = 100 – 64
BC2 = 36
BC = 6m
Hence, the distance of the foot of the ladder from the base of the wall is 6 m
Ex 6.5 Question 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:
Let
AB be the pole = 18 m
and AC be the wire = 24 m
Udsing Pythagoras theorem,
AC2 = AB2 + BC2
242 = 182 + BC2
BC2 = 576 – 324
BC2 = 252
BC = 6√7m
Hence, the distance from the base is 6√7m.
Ex 6.5 Question 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
1½ hours?
Solution:
Speed of first aeroplane = 1000 km/hr
Distance travelled by first aeroplane (due north) in 1½ hours = 1000 x 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance travelled by first aeroplane (due west) in 1½ hours = 1200 × 3/2 km = 1800 km
Now in ΔAOB,
Using Pythagoras Theorem,
AB2 = AO2 + OB2
⇒ AB2 = (1500)2 + (1800)2
⇒ AB = √(2250000 + 3240000)
= √5490000
⇒ AB = 300√61 km
Hence, in 1½ hours the distance between two plane = 300√61 km.
Ex 6.5 Question 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:
Let AB and CD be the poles of height 6m and 11 m.
Therefore, CP = 11 – 6 = 5 m and AP = 12 m
In ΔAPC,
Using Pythagoras theorem
AP2 = PC2 + AC2
(12m)2 + (5m)2 = (AC)2
AC2 = (144+25) m2 = 169 m2
AC = 13m
Therefore, the distance between top of the two pole = 13 m
Ex 6.5 Question 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:
In ΔACE,
Using Pythagoras theorem
AC2 + CE2 = AE2 …………….(i)
In ΔBCD,
Using Pythagoras theorem,
BC2 + CD2 = BD2 ……………..(ii)
From equations (i) and (ii),
AC2 + CE2 + BC2 + CD2 = AE2 + BD2 …………..(iii)
In ΔCDE,
Using Pythagoras theorem,
DE2 = CD2 + CE2…………..(iv)
In ΔABC,
Using Pythagoras theorem,
AB2 = AC2 + CB2…………..(v)
Putting the values of equation (iv) & (v) in equation (iii),
DE2 + AB2 = AE2 + BD2.
Ex 6.5 Question 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:
In Δ ADB
By Pythagoras theorem,
AB2 = AD2 + BD2 …………….(i)
In ΔADC
By Pythagoras theorem,
AC2 = AD2 + DC2 ………..(ii)
Subtracting equation (ii) from equation (i),
AB2 – AC2 = BD2 – DC2
= 9CD2 – CD2 [Given, DB = 3CD]
= 8CD2
= 8(BC/4)2 [BC = DB + CD = 3CD + CD = 4CD]
∴ AB2 – AC2 = BC2/2
⇒ 2(AB2 – AC2) = BC2
⇒ 2AB2 – 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
Ex 6.5 Question 15.
In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD2 = 7AB2.
Solution:
In ΔABC
Let the side of the equilateral triangle = a,
and the altitude = AE
∴ BE = EC = BC/2 = a/2
AE = a√3/2
BD = 1/3BC [Given]
∴ BD = a/3
DE = BE – BD = a/2 – a/3 = a/6
In ΔADE,
By Pythagoras theorem,
AD2 = AE2 + DE2
AD2 = (a√3/2)2 + (a/6)2
= (3a2/4) + (a2/36)
= 28a2/36
= 7/9AB2
⇒ 9 AD2 = 7 AB2
Ex 6.5 Question 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:
In ΔABC
Let the sides of the equilateral triangle be of length= a,
and the altitude = AE
∴ BE = EC = BC/2 = a/2
In ΔABE,
Using Pythagoras Theorem
AB2 = AE2 + BE2
a2= AE2 + (a/2)2
AE2 = a2 – a2/4
AE2 = 3a2/4
4AE2 = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
Ex 6.5 Question 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: AB = 6√3 cm, AC = 12 cm and BC = 6 cm
Squaring these sides
AB2 = 108
AC2 = 144
And, BC2 = 36
AB2 + BC2 = AC2
The above ΔABC satisfies, Pythagoras theorem. Hence, the triangle is a right angled triangle.
∴ ∠B = 90°
Hence,C is the correct answer .