The Chapter 6, Triangles, Exercise 6.5 begins from page number 150 of the NCERT textbook. The exercise covers the portions in which theorems related to right angles are present. In other words, we can say that, the exercise contains questions that could be solved using the theorems that deals with right angles.
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Access other exercise solutions of class 10 Maths Chapter 6- Triangles
Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)
Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)
Exercise 6.3 Solutions 16 Questions (1 main question with 6 sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)
Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)
Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)
Access Answers of Maths NCERT class 10 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)^{2}
Therefore, 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, 3^{2} + 6^{2} ≠ 8^{2}
Therefore, 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, 50^{2} + 80^{2} ≠ 100^{2}
As 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, 12^{2} + 5^{2} = 13^{2}
The 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 PM^{2} = QM × MR.
Solution:
Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR
We have to prove, PM^{2} = QM × MR
In ΔPQM, by Pythagoras theorem
PQ^{2} = PM^{2} + QM^{2}
Or, PM^{2} = PQ^{2} – QM^{2} ……………………………..(i)
In ΔPMR, by Pythagoras theorem
PR^{2} = PM^{2} + MR^{2}
Or, PM^{2} = PR^{2} – MR^{2} ………………………………………..(ii)
Adding equation, (i) and (ii), we get,
2PM^{2} = (PQ^{2} + PM^{2}) – (QM^{2} + MR^{2})
= QR^{2} – QM^{2} – MR^{2 } [∴ QR^{2} = PQ^{2} + PR^{2}]
= (QM + MR)^{2} – QM^{2} – MR^{2}
= 2QM × MR
∴ PM^{2} = QM × MR
3. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) AB^{2} = BC × BD
(ii) AC^{2} = BC × DC
(iii) AD^{2} = BD × CD
Solution:
(i) In ΔADB and ΔCAB,
∠DAB = ∠ACB (Each 90°)
∠ABD = ∠CBA (Common angles)
∴ ΔADB ~ ΔCAB [AA similarity criterion]
⇒ AB/CB = BD/AB
⇒ AB^{2} = 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
⇒ AC^{2} = DC × BC
(iii) In ΔDCA and ΔDAB,
∠DCA = ∠DAB (Each 90°)
∠CDA = ∠ADB (common angles)
∴ ΔDCA ~ ΔDAB [AA similarity criterion]
⇒ DC/DA = DA/DA
⇒ AD^{2} = BD × CD
4. ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2} .
Solution:
Given, ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (By isosceles triangle property)
AB^{2} = AC^{2} + BC^{2} [By Pythagoras theorem]
= AC^{2} + AC^{2} [Since, AC = BC]
AB^{2} = 2AC^{2}
5. ABC is an isosceles triangle with AC = BC. If AB^{2} = 2AC^{2}, prove that ABC is a right triangle.
Solution:
Given, ΔABC is an isosceles triangle having AC = BC and AB^{2} = 2AC^{2}
In ΔACB,
AC = BC
AB^{2} = 2AC^{2}
AB^{2} = AC^{2 }+ AC^{2}
= AC^{2} + BC^{2 }[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,
AB^{2} = AD^{2 }+ BD^{2}
(2a)^{2} = AD^{2 }+ a^{2 }
⇒ AD^{2 =} 4a^{2} – a^{2}
⇒ AD^{2 =} 3a^{2}
⇒ 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,
AB^{2 }+ BC^{2 }+ CD^{2} + AD^{2 }= AC^{2 }+ BD^{2}
Since, the diagonals of a rhombus bisect each other at right angles.
Therefore, AO = CO and BO = DO
In ΔAOB,
∠AOB = 90°
AB^{2} = AO^{2 }+ BO^{2 }…………………….. (i) [By Pythagoras theorem]
Similarly,
AD^{2} = AO^{2 }+ DO^{2 }…………………….. (ii)
DC^{2} = DO^{2 }+ CO^{2 }…………………….. (iii)
BC^{2} = CO^{2 }+ BO^{2 }…………………….. (iv)
Adding equations (i) + (ii) + (iii) + (iv), we get,
AB^{2 }+ AD^{2 }+^{ }DC^{2 }+^{ }BC^{2} = 2(AO^{2 }+ BO^{2 }+ DO^{2 }+ CO^{2})
= 4AO^{2 }+ 4BO^{2 }[Since, AO = CO and BO =DO]
= (2AO)^{2 }+ (2BO)^{2} = AC^{2 }+ BD^{2}
AB^{2 }+ AD^{2 }+^{ }DC^{2 }+^{ }BC^{2} = AC^{2 }+ BD^{2}
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) OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2} ,
(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.
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
OA^{2} = OF^{2} + AF^{2}
Similarly, in ΔBOD
OB^{2} = OD^{2} + BD^{2}
Similarly, in ΔCOE
OC^{2} = OE^{2} + EC^{2}
Adding these equations,
OA^{2} + OB^{2} + OC^{2} = OF^{2} + AF^{2} + OD^{2} + BD^{2} + OE^{2 }+ EC^{2}
OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2}.
(ii) AF^{2} + BD^{2} + EC^{2} = (OA^{2} – OE^{2}) + (OC^{2} – OD^{2}) + (OB^{2} – OF^{2})
∴ AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.
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,
AC^{2} =^{ }AB^{2} + BC^{2}
10^{2} = 8^{2} + BC^{2}
BC^{2 }= 100 – 64
BC^{2 }= 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,
AC^{2} =^{ }AB^{2} + BC^{2}
24^{2} = 18^{2} + BC^{2}
BC^{2 }= 576 – 324
BC^{2 }= 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,
AB^{2} =^{ }AO^{2} + OB^{2}
⇒ AB^{2} =^{ }(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,
AP^{2} =^{ }PC^{2} + AC^{2}
(12m)^{2} + (5m)^{2} = (AC)^{2}
AC^{2} = (144+25) m^{2} = 169 m^{2}
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 AE^{2} + BD^{2} = AB^{2} + DE^{2}.
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
AC^{2} +^{ }CE^{2} = AE^{2} ………………………………………….(i)
In ΔBCD, by Pythagoras theorem, we get
BC^{2} +^{ }CD^{2} = BD^{2} ………………………………..(ii)
From equations (i) and (ii), we get,
AC^{2} +^{ }CE^{2} + BC^{2} +^{ }CD^{2} = AE^{2} + BD^{2} …………..(iii)
In ΔCDE, by Pythagoras theorem, we get
DE^{2} =^{ }CD^{2} + CE^{2}
In ΔABC, by Pythagoras theorem, we get
AB^{2} =^{ }AC^{2} + CB^{2}
Putting the above two values in equation (iii), we get
DE^{2} + AB^{2} = AE^{2} + BD^{2}.
14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB^{2} = 2AC^{2} + BC^{2}.
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,
AB^{2} =^{ }AD^{2} + BD^{2} ……………………….(i)
AC^{2} =^{ }AD^{2} + DC^{2} ……………………………..(ii)
Subtracting equation (ii) from equation (i), we get
AB^{2} – AC^{2} = BD^{2} – DC^{2}
= 9CD^{2} – CD^{2} [Since, BD = 3CD]
= 8CD^{2}
= 8(BC/4)^{2 }[Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB^{2} – AC^{2} = BC^{2}/2
⇒ 2(AB^{2} – AC^{2}) = BC^{2}
⇒ 2AB^{2} – 2AC^{2} = BC^{2}
∴ 2AB^{2} = 2AC^{2} + BC^{2}.
15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD^{2} = 7AB^{2}.
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,
AD^{2} = AE^{2} + DE^{2 }
⇒ 9 AD^{2} = 7 AB^{2}
^{}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
AB^{2} = AE^{2} + BE^{2}
4AE^{2} = 3a^{2}
⇒ 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,
AB^{2} = 108
AC^{2} = 144
And, BC^{2} = 36
AB^{2} + BC^{2} = AC^{2}
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).
The Class 10 Maths Chapter 6, Triangles covers a wide range of topics with the six exercises present in the chapter. The Exercise 6.5 deals with the theorems related to Pythagoras Theorem. 17 Questions are present in this exercise, out of which, 15 questions are short answer type questions and 2 questions are long answer type questions.There are three theorems that comes are covered in this exercise.The three theorem that forms the base of the Exercise 6.5 is given below:
- Theorem 6.7 : If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
- Theorem 6.8 : In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. It is also known as pythagoras Theorem.
- Theorem 6.9 : In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
The best way to get acquainted with a mathematical concept is to solve as many problems as possible from that particular topic. To familiarise with all the concepts of maths present in class 10, practising with the solutions for the questions given in the NCERT textbook is more than enough. These NCERT Solutions will help the students in scoring excellent marks in the examination.