# Class 10 RD Sharma Solutions – Chapter 4 Triangles – Exercise 4.7 | Set 2

Last Updated : 19 Apr, 2021

### Question 15. Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal.

Solution:

Given,

In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles

Each side = 10 cm and one diagonal AC = 16 cm

âˆ´ AO = OC = 16/2 = 8 cm

Now in âˆ†AOB,

By using Pythagoras theorem

AB2 = AO2 + OB2

(10)2 = (8)2 + (BO)2

100 = 64 + BO2

BO2 = 100 â€“ 64 = 36

BO = 6

So, BD = 2BO = 2 x 6 = 12 cm

Hence, the length of the other diagonal is 12 cm

### Question 16. Calculate the height of an equilateral triangle each of whose sides measures 12 cm.

Solution:

Given,

Side of equilateral triangle=12 cm

To find: Calculate the height of an equilateral triangle

Let us draw the figure. Let us draw the altitude AD.

BD = DC = 6cm               [ Altitude is also median of the equilateral triangle]

By using Pythagoras theorem

AD2 = 144 âˆ’ 36 = 108

Hence, the height of the equilateral triangle is 10.39 cm

### (ii) b2 = a2 + c2 â€“ 2ax

Solution:

Given : In âˆ†ABC, âˆ B < 90Â°

AD = c, BC = a, CA = b AD = h, BD = x, DC = a â€“ x

By using Pythagoras theorem

b2 = h2 + (a â€“ x)2

So, b2 = h2 + a2 + x2 â€“ 2ax

By using Pythagoras theorem

c2 = h2 + x2       …..â€¦.(i)

b2 = h2 + a2 + x2 â€“ 2ax

= h2 + x2 + a2 â€“ 2ax

= c2 + a2 â€“ 2ax         [From eq(i)]

So, b2 = a2 + c2 â€“ 2ax

Hence proved.

### Question 18. In an equilateral âˆ†ABC, AD âŠ¥ BC, prove that AD2 = 3 BD2.

Solution:

In right-angled âˆ†ABD,

By using Pythagoras theorem

AB2 = AD2 + BD2     â€¦.(1)

We know that in an equilateral triangle every altitude is also median.

We have BD = DC

Since âˆ†ABC is an equilateral triangle, AB = BC = AC

So, we can write equation (1) as

BC2 = AD2 + BD2       ….(2)

But BC = 2BD

Therefore, equation (2) becomes,

On simplifying the equation we get,

Hence proved

### Question 19. âˆ†ABD is a right triangle right-angled at A and AC âŠ¥ BD. Show that

(i) AB2 = BC.BD

(ii) AC2 = BC.DC

(iv) AB2/AC2 = BD/DC

Solution:

âˆ DAB = âˆ ACB = 90Â°

âˆ ABD = âˆ CBA                       (common angle)

âˆ ADB = âˆ CAB                      (remaining angle)

So, by AAA

Hence,

AB/CB = BD/AB

â‡’ AB2 = CB Ã— BD

(ii) Let âˆ CAB = y

In Î”CBA

âˆ CBA = 180Â° âˆ’ 90Â° âˆ’ y

âˆ CBA = 90Â° âˆ’ y

âˆ CDA = 90Â° âˆ’ âˆ CAB

= 90Â° âˆ’ y

âˆ CDA = 180Â° âˆ’90Â° âˆ’ (90Â° âˆ’ y)

âˆ CDA = y

âˆ CAB = âˆ CDA

âˆ ACB = âˆ DCA = 90Â°

So, by AAA rule

So, AC /DC = BC/AC

â‡’ AC2 = DC Ã— BC

(iii) In Î”DCA & Î”DAB

âˆ DCA = âˆ DAB     (both are equal to 90Â°)

âˆ CDA = âˆ ADB     (common angle)

âˆ DAC = âˆ DBA     (remaining angle)

So, by AAA rule

Î”DCA ~ Î”DAB

so, DC/DA = DA/DB

â‡’ AD2 = BD Ã— CD

(iv) From part (i) AB2 = CB x BD

From part (ii) AC2 = DC Ã— BC

So, AB2/AC2= CB x BD/DC x BC

AB2/AC2 = BD/DC

Hence proved

### Question 20. 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,

AC = 18m be the height of pole.

BC = 24m is the length of a guy wire and is attached to stake B.

Now,

In â–³ABC

By using Pythagoras theorem

BC2 = AB2 + AC2

242 = AB2 + 182

AB2 = 576 âˆ’ 324

= 252

So, AB = 6âˆš7 m

Hence, the stake has to be 6âˆš7 m from base A.

### Question 21. Determine whether the triangle having sides (a â€“ 1) cm, 2 âˆša cm and (a + 1) cm is a right-angled triangle.

Solution:

Given,

The sides of triangle are (a – 1) cm, 2âˆša and (a + 1) cm.

Let us considered ABC be the triangle in which with the sides are

AB = (a – 1)cm, BC = (2âˆš a) cm, CA = (a + 1) cm

ABÂ² = (a – 1)

By using (a – b)2 = a2 + b2 – 2ab

= a2 + 12 – 2 Ã— a Ã— 1

AB2 = a2 + 1 -2a

BC2 = (2âˆša)2

âˆ´ BC = 4a

CA2 = (a + 1)2

By using (a + b)2 = a2 + b2 + 2ab

= a2 + 12 + 2 Ã— a Ã— 1

CA2 = a2+ 1 + 2a

By using Pythagoras Theorem

AC2 = AB2 + BC2

On putting value of AC2, AB2 and BC2 in the above equation,

a2 + 1 + 2a = a2 + 1 – 2a + 4a

a2 + 1 + 2a = a2 + 1 + 2a

AC2 = AB2 + BC2

âˆ†ABC is right-angled âˆ† at B.

Hence proved

### Question 22. In an acute-angled triangle, express a median in terms of its sides.

Solution:

Given,

In Î” ABC AD is median.

Construction: AE âŠ¥ BC

Now,

âˆ´ BD = CD = 1/2 BC ….(1)         [AD is the median]

In Î” AED,

By using Pythagoras Theorem

â‡’ AE2 = AD2 – DE2 …..(2)

In Î” AEB,

AB2 = AE2 + BE2

â‡’ AD2 – DE2 + BE2                       [From eq(2)]

= (BD + DE)2 + AD2 – DE2                [âˆ´ BE = BD + DE]

BD2 + DE2 + 2BD x DE + AD2 – DE2

= BD2 + AD2 + 2BD x DE

= (1/2BC)2 + AD2 + (2 Ã— 1/2BC Ã— DE)   [From eq(1)]

= (1/4BC)2 + AD2 + BC x DE ….(3)

In Î” AED,

By using Pythagoras Theorem

AC2 = AE2 + EC2

= AD2 – DE2 + EC2

= AD2 – DE2 + (DC – DE)2

= AD2 – DE2 + DC2 + DE2 – 2DC x DE

AD2 + DC2 – 2DC x DE

= AD2 + (1/2BC)2 – (2 Ã— 1/2BC x DE)

= AD2 + (1/4BC)2 – BC x DE ….(4)

On adding eq(3) and (4), we get

AB2 + AC2 = 1/4BC2 + AD2 + BC x DE + AD2 + 1/4BC2 – BC x DE

2(AB2 + AC2) = BC2 + 4AD2

2AB2 + 2AC2 = BC2 + 4AD2

Hence Proved

### Question 23. In right-angled triangle ABC in which âˆ C = 90Â°, if D is the mid point of BC, prove that AB2 = 4AD2 â€“ 3AC2.

Solution:

Given :

âˆ C = 90Â° and D is the mid-point of BC.

To prove : AB2 = 4AD2 â€“ 3AC2

In âˆ† ACD,

By using Pythagoras Theorem

CD2 = AD2 – AC2  â€¦â€¦â€¦.(1)

In âˆ†ACB,

By using Pythagoras Theorem

AB2 = AC2 + BC2

AB2 = AC2 + (2CD)2   [D is the mid-point of BC]

AB2 = AC2 + 4CD2

âˆ´ AB2 = AC2 + 4(AD2 – AC2)    [From eq(1)]

AB2 = AC2 + 4AD2 – 4AC2

AB2 = 4AD2 – 4AC2 + AC2

âˆ´ AB2 = 4AD2 – 3AC2

Hence Proved

### Question 24. In the figure, D is the mid-point of side BC and AE âŠ¥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that

(i) b2 = p2 + a2/4 + ax

(ii) c2 = p2– ax + a2/4

(iii) b2 + c2 = 2p2 + a2/2

Solution:

Given,

D is the midpoint of BC

(i) In âˆ† AEC

AC2 = AE2 + EC2

b2 = AE2 + (ED + DC)2

b2 = AD2 + DC2 + 2 x ED x DC            [Given BC = 2CD]

b2 = p2 + (a/2)2 + 2(a/2)x

b2 = p2 + a2/4 + ax      ………..(i)

(ii) In âˆ† AEB

AB2 = AE2 + BE2

c2 = AD2 – ED2 + (BD – ED)2

c2 = p2 – ED2 + BD2 + ED2 – 2BD x ED

c2 = P2 + (a/2)2 – 2(a/2)2x

c2 = p2 – ax + a2/4 â€¦â€¦â€¦â€¦â€¦â€¦.(ii)

(iii) Adding eqn (i) and (ii) we get,

b2 + c2 = 2p2 + a2/2

Hence Proved

### Question 25. In âˆ†ABC, âˆ A is obtuse, PB x AC and QC x AB. Prove that:

(i) AB x AQ = AC x AP

(ii) BC2 = (AC x CP + AB x BQ)

Solution:

(i) Given :

âˆ A is obtuse.

PB is perpendicular to AC.

QC is perpendicular to AB.

To Prove :

AB Ã— AQ = AC Ã— AP.

Proof:

In Î”ACQ and Î”ABP,

â‡’ âˆ CAQ = âˆ BAP          [Vertically opposite âˆ ]

â‡’ âˆ Q = âˆ P                    [âˆ Q = âˆ P = 90 Âº]

So, by AA rule

Î”ACQ ~ Î”ABP

By Property of Similar Triangles,

â‡’ CQ/BP = AC/AB = AQ/AP

â‡’ AC/AB = AQ/AP

â‡’ AB Ã— AQ = AC Ã— AP  ……..(i)

Hence Proved.

(ii) To Prove :

BCÂ² = AB Ã— BQ + AC Ã— CP

Proof:

By using Pythagoras Theorem

â‡’ BC2 = CQ2 + QB2

â‡’ BC2 = CQ2 + (QA + AB)2

â‡’ BC2 = CQ2 + QA2 + AB2 + 2 QA Ã— AB

â‡’ BC2 = CQ2 + QA2 + AB2 + QA Ã— AB + QA Ã— AB

â‡’ BC2 = AC2 + AB2 + QA Ã— AB + QA Ã— AB           [In Î”ACQ, CQ2 + QA2 = AC2]

â‡’ BC2 = AC2 + AB2 + QA Ã— AB + AC Ã— CP            [By Eq (i)]

â‡’ BC2 = AC2 + AC Ã— CP + ABÂ² + QA Ã— AB

â‡’ BC2 = AC Ã— (AC + CP) + AB Ã— (QA + AB)

â‡’ BC2 = AC Ã— CP + AB Ã— BQ                                  [CP = AC + CP, BQ = AQ + AB]

â‡’ BC2 = AB Ã— BQ + AC Ã— CP.

Hence, Proved.

### Question 26. In a right âˆ†ABC right-angled at C, if D is the mid-point of BC, prove that BC2 = 4 (AD2 â€“ AC2).

Solution:

Given:

âˆ C = 90Â°

By using Pythagoras Theorem

AD2 = AC2 + (1/2BC)2    [DC = 1/2BC]

Taking minus common

Hence, proved

### Question 27. In a quadrilateral ABCD, âˆ B = 90Â°, AD2 = AB2 + BC2 + CD2, prove that âˆ ACD = 90Â°.

Solution:

Given: ABCD is a quadrilateral, âˆ B = 90Â° and AD2 = AB2 + BC2 + CD2

To prove: âˆ ACD = 90Â°

Proof: In right âˆ†ABC,

By using Pythagoras Theorem

AC2 = AB2 + BC2 â€¦ (1)

Given, AD2 = AB2 + BC2 + CD2

â‡’ AD2 = AC2 + CD2  [Using eq(1)]

In âˆ†ACD,

So, âˆ ACD = 90Â° [By converse of Pythagoras theorem]

### Question 28. An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/ hr. How far apart will be the two planes after 1and1/2 hours?

Solution:

Given that speed of first aeroplane = 1000 km/hr

Distance travelled by first aeroplane (due north) in  hours = 1000 x 3/2 km = 1500 km

Speed of second aeroplane = 1200 km/hr

Distance travelled by first aeroplane (due west) in  hours = 1200 Ã— 3/2 km = 1800 km

Now in Î”AOB,

Using Pythagoras Theorem,

AB2 = AO2 + OB2

â‡’ AB2 = (1500)2 + (1800)2

â‡’ AB = âˆš(2250000 + 3240000)

= âˆš5490000

â‡’ AB = 300âˆš61 km

Hence, in  hours the distance between two plane = 300âˆš61 km.

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