Chapter 6, Triangles, Exercise 6.5 begins on page number 150 of the NCERT textbook. The exercise covers the portions in which theorems related to right angles are present. In other words, the exercise-wise NCERT Class 10 Solutions contains questions that could be solved using the theorems that deal with right angles.

The subject experts at BYJUâ€™S make sure that the questions present in the NCERT textbooks are solved in the easiest way possible. The NCERT solutions for Class 10 Maths can help you not only with your board examination preparation but can also be used to check if the answers you gave to the questions are correct or not.

## NCERT Solutions for Class 10 Maths Chapter 6- Triangles Exercise 6.5

### 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 to Maths NCERT Class 10 Chapter 6 – Triangles Exercise 6.5

**1. Â Sides of 4 triangles are given below. Determine which of them are right triangles. In the 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, the 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}Â + (24)^{2}Â = (25)^{2}

Therefore, the above equation satisfies Pythagoras theorem. Hence, it is a right-angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, the 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 â‰ 64

Or, 3^{2}Â + 6^{2}Â â‰ 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 satisfy Pythagoras theorem.

(iii) Given, the sides of triangleâ€™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 â‰ 10000

Or, 50^{2}Â + 80^{2}Â â‰ 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 satisfy Pythagoras theorem.

Hence, it is not a right triangle.

(iv) Given, the 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}Â + 5^{2}Â = 13^{2}

The sides of the given triangle satisfy Pythagoras theorem.

Therefore, it is a right triangle.

Hence, the 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 âŠ¥ QR. Show that PM ^{2}Â = QMÂ Ã—Â MR.**

**Solution: **

Given, Î”PQR is right angled at P is a point on QR such that PMÂ âŠ¥QR

We have to prove, PM^{2}Â = QMÂ Ã—Â MR

In Î”PQM, by Pythagoras theorem

PQ^{2}Â = PM^{2}Â + QM^{2}

Or, PM^{2}Â = PQ^{2}Â – QM^{2}Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..**(i)**

In Î”PMR, by Pythagoras theorem

PR^{2}Â = PM^{2}Â + MR^{2}

Or, PM^{2}Â = PR^{2}Â – MR^{2}Â ………………………………………..**(ii)**

Adding equations **(i)**Â andÂ **(ii)**, we get,

2PM^{2}Â =Â (PQ^{2}Â + PM^{2}) – (QM^{2}Â + MR^{2})

= QR^{2}Â – QM^{2}Â – MR^{2 Â Â Â Â }Â [âˆ´ QR^{2}Â = PQ^{2}Â + PR^{2}]

= (QMÂ + MR)^{2}Â – QM^{2}Â – MR^{2}

= 2QM Ã— MR

âˆ´ PM^{2}Â = QMÂ Ã—Â MR

**3. In Figure, ABD is a triangle right angled at AÂ and AC âŠ¥ BD. Show that
(i) AB ^{2}Â = BC Ã—Â BD
(ii) AC^{2}Â = BC Ã—Â DC
(iii) AD^{2}Â = BDÂ Ã—Â CD**

**Solution:**

(i) In Î”ADB and Î”CAB,

âˆ DAB = âˆ ACB (Each 90Â°)

âˆ ABD = âˆ CBA (Common angles)

âˆ´ Î”ADB ~ Î”CAB [AA similarity criterion]

â‡’ AB/CB = BD/AB

â‡’ AB^{2}Â = CB Ã—Â BD

(ii) Let âˆ CAB = x

In Î”CBA,

âˆ CBA = 180Â° – 90Â° – x

âˆ CBA = 90Â° – x

Similarly, in Î”CAD

âˆ CAD = 90Â° – âˆ CBA

= 90Â° –*Â *x

âˆ CDA = 180Â° – 90Â° – (90Â° –Â x)

âˆ CDA =Â x

In Î”CBA and Î”CAD, we have

âˆ CBA = âˆ CAD

âˆ CAB = âˆ CDA

âˆ ACB = âˆ DCA (Each 90Â°)

âˆ´ Î”CBA ~ Î”CAD [AAA similarity criterion]

â‡’ AC/DC = BC/AC

â‡’ AC^{2}Â = Â DC Ã— BC

(iii) In Î”DCA and Î”DAB,

âˆ DCA = âˆ DAB (Each 90Â°)

âˆ CDA = âˆ ADB (common angles)

âˆ´ Î”DCA ~ Î”DAB [AA similarity criterion]

â‡’ DC/DA = DA/DA

â‡’ AD^{2}Â = BDÂ Ã—Â CD

**4. ABC is an isosceles triangle right angled at C. Prove that AB ^{2}Â = 2AC^{2}.**

**Solution:**

Given, Î”ABC is an isosceles triangle right angled at C.

In Î”ACB, âˆ C = 90Â°

AC = BC (By isosceles triangle property)

AB^{2}Â = AC^{2}Â + BC^{2}Â [By Pythagoras theorem]

= AC^{2}Â +Â AC^{2}Â [Since, AC = BC]

AB^{2}Â = 2AC^{2}

**5.Â ABC is an isosceles triangle with AC = BC. IfÂ AB ^{2}Â = 2AC^{2}, prove that ABC is a right triangle.**

**Solution:**

Given, Î”ABC is an isosceles triangle having AC = BC and AB^{2}Â = 2AC^{2}

In Î”ACB,

AC = BC

AB^{2}Â = 2AC^{2}

AB^{2}Â = AC^{2Â }+ AC^{2}

= AC^{2}Â + BC^{2Â }[Since, AC = BC]

Hence, by Pythagoras theorem, Î”ABC is a right angle triangle.

**6.Â ABC is an equilateral triangle of side 2a. Find each of its altitudes**.

**Solution: **

Given, ABC is an equilateral triangle of side 2a.

Draw, ADÂ âŠ¥Â BC

In Î”ADB and Î”ADC,

AB = AC

AD = AD

âˆ ADB = âˆ ADC [Both are 90Â°]

Therefore, Î”ADBÂ â‰…Â Î”ADC by RHS congruence.

Hence, BD = DC [by CPCT]

In the right angled Î”ADB,

AB^{2}Â = AD^{2Â }+ BD^{2}

(2*a*)^{2}Â = AD^{2Â }+Â *a*^{2Â }

â‡’Â AD^{2 =} 4*a*^{2}Â –Â *a*^{2}

â‡’Â AD^{2 =} 3*a*^{2}

â‡’Â AD^{ =}Â âˆš3a

**7. Prove that the sum of the squares of the sides of the 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^{2Â }+ BC^{2Â }+ CD^{2}Â + AD^{2Â }= AC^{2Â }+ BD^{2}

Since the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

In Î”AOB,

âˆ AOB =Â 90Â°

AB^{2}Â = AO^{2Â }+ BO^{2Â }â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â **(i)**Â [By Pythagoras theorem]

Similarly,

AD^{2}Â = AO^{2Â }+ DO^{2Â }â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â **(ii)**

DC^{2}Â = DO^{2Â }+ CO^{2Â }â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â **(iii)**

BC^{2}Â = CO^{2Â }+ BO^{2Â }â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..Â **(iv)**

Adding equationsÂ **(i) + (ii)Â + (iii)Â + (iv)**, we get,

AB^{2Â }+ AD^{2Â }+^{Â }DC^{2Â }+^{Â }BC^{2}Â = 2(AO^{2Â }+ BO^{2Â }+ DO^{2Â }+ CO^{2})

= 4AO^{2Â }+ 4BO^{2Â }[Since, AO = CO and BO =DO]

= (2AO)^{2Â }+ (2BO)^{2}Â = AC^{2Â }+ BD^{2}

AB^{2Â }+ AD^{2Â }+^{Â }DC^{2Â }+^{Â }BC^{2}Â = AC^{2Â }+ BD^{2}

Hence, proved.

**8. In Fig. 6.54, O is a point in the interior of a triangle.**

**ABC, OD âŠ¥ BC, OEÂ âŠ¥Â AC and OF âŠ¥ AB. Show that:
(i) OA ^{2}Â + OB^{2}Â + OC^{2}Â â€“ OD^{2}Â â€“ OE^{2}Â â€“ OF^{2}Â = AF^{2}Â + BD^{2}Â + CE^{2}Â ,
(ii) AF^{2}Â + BD^{2}Â + CE^{2}Â = AE^{2}Â + CD^{2}Â + BF^{2}.**

**Solution: **

Given, in Î”ABC, O is a point in the interior of a triangle.

And OD âŠ¥ BC, OEÂ âŠ¥Â AC and OF âŠ¥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in Î”AOF, we have

OA^{2}Â = OF^{2}Â +Â AF^{2}

Similarly, inÂ Î”BOD

OB^{2}Â = OD^{2}Â + BD^{2}

Similarly,Â inÂ Î”COE

OC^{2}Â = OE^{2}Â + EC^{2}

Adding these equations,

OA^{2}Â + OB^{2}Â + OC^{2}Â = OF^{2}Â + AF^{2}Â + OD^{2}Â + BD^{2}Â + OE^{2Â }+ EC^{2}

OA^{2}Â + OB^{2}Â + OC^{2}Â â€“ OD^{2}Â â€“ OE^{2}Â â€“ OF^{2}Â = AF^{2}Â + BD^{2}Â + CE^{2}.

(ii) AF^{2}Â + BD^{2}Â + EC^{2}Â = (OA^{2}Â – OE^{2})Â + (OC^{2}Â – OD^{2})Â + (OB^{2}Â – OF^{2})

âˆ´ AF^{2}Â + BD^{2}Â + CE^{2}Â = AE^{2}Â + CD^{2}Â + BF^{2}.

**9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance between the foot of the ladder from the 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}Â =^{Â }AB^{2}Â + BC^{2}

10^{2}Â = 8^{2}Â + BC^{2}

BC^{2Â }= 100 – 64

BC^{2Â }= 36

BC^{Â }= 6m

Therefore, the distance between 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}Â =^{Â }AB^{2}Â + BC^{2}

24^{2}Â = 18^{2}Â + BC^{2}

BC^{2Â }= 576 – 324

BC^{2Â }= 252

BC^{Â }= 6âˆš7m

Therefore, the distance from the base is 6âˆš7m.

**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 **

** Â hours?**

**Solution: **

Given,

Speed of first aeroplane = 1000 km/hr

Distance covered by the first aeroplane flying due north in

**Â **Â hours (OA) = 1000 Ã— 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance covered by the second aeroplane flying due west in

**Â **Â hours (OB) = 1200 Ã— 3/2 km = 1800 km

In right angle Î”AOB, by Pythagoras Theorem,

AB^{2}Â =^{Â }AO^{2}Â + OB^{2}

â‡’ AB^{2}Â =^{Â }(1500)^{2}Â + (1800)^{2}

â‡’ AB = âˆš(2250000Â + 3240000)

= âˆš5490000

â‡’ AB = 300âˆš61Â km

Hence, the distance between two aeroplanes will be 300âˆš61Â 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 the 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 Î”APC, we get,

AP^{2}Â =^{Â }PC^{2}Â + AC^{2}

(12m)^{2}Â + (5m)^{2}Â = (AC)^{2}

AC^{2}Â = (144+25) m^{2}Â = 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}Â + BD^{2}Â = AB^{2}Â + 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 Î”ACE, we get

AC^{2}Â +^{Â }CE^{2}Â = AE^{2}Â ………………………………………….**(i)**

In Î”BCD, by Pythagoras theorem, we get

BC^{2}Â +^{Â }CD^{2}Â = BD^{2}Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..**(ii)**

From equationsÂ **(i)**Â andÂ **(ii)**, we get

AC^{2}Â +^{Â }CE^{2}Â + BC^{2}Â +^{Â }CD^{2}Â = AE^{2}Â + BD^{2}Â â€¦â€¦â€¦â€¦..**(iii)**

In Î”CDE, by Pythagoras theorem, we get

DE^{2}Â =^{Â }CD^{2}Â + CE^{2}

In Î”ABC, by Pythagoras theorem, we get

AB^{2}Â =^{Â }AC^{2}Â + CB^{2}

Putting the above two values in equationÂ **(iii)**, we get

DE^{2}Â + AB^{2}Â = AE^{2}Â + BD^{2}.

**14. The perpendicular from A on side BC of a Î” ABC intersects BC at D such that DB = 3CD (see figure). Prove that 2AB ^{2}Â = 2AC^{2}Â + BC^{2}.**

**Solution: **

Given, the perpendicular from A on side BC of aÂ Î” ABC intersects BC at D such that;

DB = 3CD.

In Î” ABC,

AD âŠ¥BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

AB^{2}Â =^{Â }AD^{2}Â + BD^{2}Â â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦….**(i)**

AC^{2}Â =^{Â }AD^{2}Â + DC^{2}Â ……………………………..**(ii)**

Subtracting equationÂ **(ii)**Â from equationÂ **(i)**, we get

AB^{2}Â – AC^{2}Â = BD^{2}Â – DC^{2}

= 9CD^{2}Â – CD^{2}Â [Since, BD = 3CD]

= 8CD^{2}

= 8(BC/4)^{2Â }[Since, BC = DBÂ + CD = 3CDÂ + CD = 4CD]

Therefore, AB^{2}Â – AC^{2}Â = BC^{2}/2

â‡’ 2(AB^{2}Â – AC^{2}) = BC^{2}

â‡’ 2AB^{2}Â – 2AC^{2}Â = BC^{2}

âˆ´ 2AB^{2}Â = 2AC^{2}Â + BC^{2}.

**15. Â In an equilateral triangle, ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD ^{2}Â = 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 beÂ *a*, and AE be the altitude of Î”ABC.

âˆ´ BE = EC = BC/2 = a/2

And, AE =Â aâˆš3/2

Given, BD = 1/3BC

âˆ´ BD =Â a/3

DE = BE – BD =Â a/2 –Â a/3 =Â a/6

In Î”ADE, by Pythagoras theorem,

AD^{2}Â = AE^{2}Â + DE^{2Â }

â‡’ 9 AD^{2}Â = 7 AB^{2
}

^{
}**16.Â 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 Î”ABC.

âˆ´ BE = EC = BC/2 =Â a/2

In Î”ABE, by Pythagoras Theorem, we get

AB^{2}Â = AE^{2}Â + BE^{2}

4AE^{2}Â = 3a^{2}

â‡’ 4 Ã— (Square of altitude) = 3 Ã— (Square of one side)

Hence, proved.^{
}

**17. Tick the correct answer and justify: In Î”ABC, AB = 6âˆš3Â cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120Â°**

**(B) 60Â°
(C) 90Â°Â **

**(D) 45Â°**

**Solution: **

Given, in Î”ABC, AB = 6âˆš3Â cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB^{2}Â = 108

AC^{2}Â = 144

And, BC^{2}Â = 36

AB^{2}Â + BC^{2}Â = AC^{2}

The given triangle, Î”ABC, satisfies Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

âˆ´ âˆ B = 90Â°

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. 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 3 theorems that are covered in this exercise. The three theorems that form the base of 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 the 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 yourself with all the concepts of Maths present in Class 10, practising with the solutions for the questions given in the NCERT textbook is mandatory. These NCERT Solutions will help the students in scoring excellent marks in the board examination.

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