NCERT Solutions Class 10 Maths Chapter 6 Triangles Ex 6.5

NCERT Solutions Class 10 Maths
Chapter – 6 (Triangles)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 6 Triangles Exercise 6.5 has been provided here to help the students in solving the questions from this exercise.

Chapter : 6 Triangles

Exercise – 6.5

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) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of the sides of the, we will get 49, 576, and 625.
49 + 576 = 625
(7)² + (24)² = (25)²Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.
Length of Hypotenuse = 25 cm

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will get 9, 64, and 36.
Clearly, 9 + 36 ≠ 64
Or, 3² + 6² ≠ 8²Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.
Hence, the given triangle does not satisfies Pythagoras theorem.

(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will get 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 50² + 80² ≠ 100²As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle does not satisfies Pythagoras theorem.
Hence, it is not a right triangle.

(iv) Given, sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will get 169, 144, and 25.
Thus, 144 +25 = 169
Or, 12² + 5² = 13²The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
Hence, length of the hypotenuse of this triangle is 13 cm.

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 – Given : ΔPQR is right angled at P is a point on QR such that PM ⊥QR
NCERT Class 10 Maths Solution
We have to prove, PM² = QM × MR
In ΔPQM, by Pythagoras theorem
PQ² = PM² + QM²Or, PM² = PQ² – QM² ——————- (i)
In ΔPMR, by Pythagoras theorem
PR² = PM² + MR²Or,
PM² = PR² – MR² ——————- (ii)
Adding equation, (i) and (ii), we get,
2PM² = (PQ² + PM²) – (QM² + MR²)
= QR² – QM² – MR² [∴ QR² = PQ² + PR²]
= (QM + MR)² – QM² – MR²= 2QM × MR
∴ PM² = QM × MR

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
NCERT Class 10 Maths Solution

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.
NCERT Class 10 Maths Solution
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 AC = BC and AB² = 2AC² NCERT Class 10 Maths Solution
In ΔACB,
AC = BC
AB² = 2AC²AB² = 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.
NCERT Class 10 Maths Solution
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⇒ AD^{2 =} 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.
NCERT Class 10 Maths Solution
We have to prove, as per the question,
AB²+ BC²+ CD² + AD²= AC²+ BD²Since, 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+ BD²AB²+ AD²+DC²+BC² = AC²+ BD²

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.
NCERT Class 10 Maths Solution
And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have
OA² = OF² + AF²Similarly, in ΔBOD
OB² = OD² + BD²Similarly, in ΔCOE
OC² = OE² + EC²Adding these equations,
OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE²+ EC²OA² + 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.
NCERT Class 10 Maths Solution
Let BA be the wall and AC be the ladder,
Therefore, by Pythagoras theorem,
AC² =AB² + BC²10² = 8² + BC²BC²= 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.
NCERT Class 10 Maths Solution
Let AB be the pole and AC be the wire.
By Pythagoras theorem,
AC² =AB² + BC²24² = 18² + BC²BC²= 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 $\mathbf{1\frac{1}{2}}$ hours?

Solution –
Given : Speed of first aeroplane = 1000 km/hr
Distance covered by first aeroplane flying due north in $1\frac{1}{2}$ hours (OA) = 1000 × 3/2 km = 1500 km
Speed of second aeroplane = 1200 km/hr
Distance covered by second aeroplane flying due west in $1\frac{1}{2}$ hours (OB) = 1200 × 3/2 km = 1800 km
NCERT Class 10 Maths Solution
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.
Distance between the feet of the poles is 12 m.
NCERT Class 10 Maths Solution
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)²AC² = (144+25) m² = 169 m²AC = 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.
NCERT Class 10 Maths Solution
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 + CE²In ΔABC, by Pythagoras theorem, we get
AB² =AC² + CB²Putting 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 = $\mathbf{\frac{1}{3}}$ BC. Prove that 9AD² = 7AB².

Solution –
Given : ABC is an equilateral triangle.
D is a point on side BC such that BD = $\frac{1}{3}$ BC
NCERT Class 10 Maths Solution
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 = $\frac{1}{3}$ BC
∴ BD = a/3
DE = BE – BD = a/2 – a/3 = a/6
In ΔADE, by Pythagoras theorem,
AD² = AE² + DE²AD² = $\left ( \frac{a\sqrt{3}}{2} \right )^2 + \left ( \frac{a}{6} \right )^2$

AD² = $\frac{3a^2}{4} +\frac{a^2}{36}$

AD² = $\frac{28a^2}{36}$ = $\frac{7}{9}AB^2$
⇒ 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,
NCERT Class 10 Maths Solution
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

a² = AE² + $\left ( \frac{a}{2} \right )^2$

AE² = a² – $\frac{a^2}{4}$

AE² = $\frac{3a^2}{4}$

4AE² = 3a²⇒ 4 × (Square of altitude) = 3 × (Square of one side)

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 : ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.
NCERT Class 10 Maths Solution
We can observe that,
AB² = 108
AC² = 144
And,
BC² = 36
AB² + BC² = AC²The 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).

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NCERT Solutions Class 10 Maths Chapter 6 Triangles Ex 6.5