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

### 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

∴ AO = OC = 16/2 = 8 cm

Now in ∆AOB,

By using Pythagoras theoremAB

^{2}= AO^{2}+ OB^{2}(10)

^{2}= (8)^{2}+ (BO)^{2}100 = 64 + BO

^{2}BO

^{2}= 100 – 64 = 36BO = 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]

In ∆ADB,

By using Pythagoras theoremAB

^{2 }= AD^{2 }+ BD^{2 }144 = AD

^{2}+ 36AD

^{2}= 144 − 36 = 108AD = 10.39 cm

Hence, the height of the equilateral triangle is 10.39 cm

### Question 17. In the figure, ∠B < 90° and segment AD ⊥ BC. Show that:

### (i) b^{2} = h^{2} + a^{2} + x^{2} – 2ax

### (ii) b^{2} = a^{2} + c^{2} – 2ax

**Solution:**

Given : In ∆ABC, ∠B < 90°

AD ⊥ BC

AD = c, BC = a, CA = b AD = h, BD = x, DC = a – x

(i)In ∆ADC,

By using Pythagoras theoremAC

^{2}= AD^{2}+ DC^{2}b

^{2}= h^{2}+ (a – x)^{2}So, b

^{2}= h^{2}+ a^{2}+ x^{2}– 2ax

(ii)Similarly in right ∆ADB

By using Pythagoras theoremAB

^{2}= AD^{2}+ BD^{2}c

^{2}= h^{2}+ x^{2}…..….(i)b

^{2}= h^{2}+ a^{2}+ x^{2}– 2ax= h

^{2}+ x^{2}+ a^{2}– 2ax= c

^{2 }+ a^{2}– 2ax [From eq(i)]So, b

^{2}= a^{2}+ c^{2}– 2axHence proved.

### Question 18. In an equilateral ∆ABC, AD ⊥ BC, prove that AD^{2} = 3 BD^{2}.

**Solution:**

In right-angled ∆ABD,

By using Pythagoras theoremAB

^{2}= AD^{2}+ BD^{2 }….(1)We know that in an equilateral triangle every altitude is also median.

So, AD bisects BC.

We have BD = DC

Since ∆ABC is an equilateral triangle, AB = BC = AC

So, we can write equation (1) as

BC

^{2}= AD^{2}+ BD^{2 }….(2)But BC = 2BD

Therefore, equation (2) becomes,

(2BD)

^{2}= AD^{2}+ BD^{2}On simplifying the equation we get,

4BD

^{2}– BD^{2}= AD^{2}3BD

^{2}= AD^{2}So, AD

^{2 }= 3BD^{2}Hence proved

### Question 19. ∆ABD is a right 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**

**(iv) AB ^{2}/AC^{2} = BD/DC **

**Solution:**

(i)In ΔADB and ΔCAB∠DAB = ∠ACB = 90°

∠ABD = ∠CBA (common angle)

∠ADB = ∠CAB (remaining angle)

So, by AAA

ΔADB ~ ΔCAB

Hence,

AB/CB = BD/AB

⇒ AB2 = CB × BD

(ii)Let ∠CAB = yIn ΔCBA

∠CBA = 180° − 90° − y

∠CBA = 90° − y

Similarly, in ΔCAD

∠CAD = 90° − ∠CAD = 90° − y

∠CDA = 90° − ∠CAB

= 90° − y

∠CDA = 180° −90° − (90° − y)

∠CDA = y

Now in ΔCBA and ΔCAD,

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA = 90°

So, by AAA rule

ΔCBA ~ ΔCAD

So, AC /DC = BC/AC

⇒ AC

^{2}= DC × BC

(iii)In ΔDCA & ΔDAB∠DCA = ∠DAB (both are equal to 90°)

∠CDA = ∠ADB (common angle)

∠DAC = ∠DBA (remaining angle)

So, by AAA rule

ΔDCA ~ ΔDAB

so, DC/DA = DA/DB

⇒ AD

^{2}= BD × CD

(iv)From part (i) AB^{2}= CB x BDFrom part (ii) AC

^{2}= DC × BCSo, AB

^{2}/AC^{2}= CB x BD/DC x BCAB

^{2}/AC^{2 }= BD/DCHence 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 △ABC

By using Pythagoras theoremBC

^{2}= AB^{2}+ AC^{2}24

^{2 }= AB^{2 }+ 18^{2}AB

^{2 }= 576 − 324= 252

So, AB = 6√7 m

Hence, the stake has to be 6√7 m from base A.

### Question 21. Determine whether the triangle having sides (a – 1) cm, 2 √a cm and (a + 1) cm is a right-angled triangle.

**Solution:**

Given,

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

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

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

AB² = (a – 1)

^{2 }By using (a – b)

^{2 }= a^{2}+ b^{2}– 2ab= a

^{2}+ 1^{2}– 2 × a × 1AB

^{2 }= a^{2}+ 1 -2aBC

^{2}= (2√a)^{2}∴ BC = 4a

CA

^{2}= (a + 1)^{2}By using (a + b)

^{2 }= a^{2 }+ b^{2}+ 2ab= a

^{2}+ 1^{2}+ 2 × a × 1CA

^{2}= a^{2}+ 1 + 2aBy using Pythagoras Theorem

AC

^{2}= AB^{2}+ BC^{2}On putting value of AC

^{2}, AB^{2}and BC^{2}in the above equation,a

^{2}+ 1 + 2a = a^{2}+ 1 – 2a + 4aa

^{2}+ 1 + 2a = a^{2}+ 1 + 2aAC

^{2}= AB^{2}+ BC^{2}∆ABC is right-angled ∆ at B.

Hence proved

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

**Solution:**

Given,

In Δ ABC AD is median.

Construction: AE ⊥ BC

Now,

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

In Δ AED,

By using Pythagoras Theorem

AD

^{2}= AE^{2}+ DE^{2}⇒ AE

^{2}= AD^{2}– DE^{2}…..(2)In Δ AEB,

AB

^{2}= AE^{2 }+ BE^{2}⇒ AD

^{2}– DE^{2}+ BE^{2}[From eq(2)]= (BD + DE)

^{2}+ AD^{2}– DE^{2}[∴ BE = BD + DE]BD

^{2}+ DE^{2}+ 2BD x DE + AD^{2}– DE^{2}= BD

^{2}+ AD^{2}+ 2BD x DE= (1/2BC)

^{2}+ AD^{2 }+ (2 × 1/2BC × DE) [From eq(1)]= (1/4BC)

^{2}+ AD^{2 }+ BC x DE ….(3)In Δ AED,

By using Pythagoras Theorem

AC

^{2}= AE^{2 }+ EC^{2}= AD

^{2}– DE^{2}+ EC^{2}= AD

^{2}– DE^{2}+ (DC – DE)^{2}= AD

^{2}– DE^{2}+ DC^{2}+ DE^{2}– 2DC x DEAD

^{2}+ DC^{2}– 2DC x DE= AD

^{2}+ (1/2BC)^{2}– (2 × 1/2BC x DE)= AD

^{2}+ (1/4BC)^{2}– BC x DE ….(4)On adding eq(3) and (4), we get

AB

^{2 }+ AC^{2}= 1/4BC^{2 }+ AD^{2}+ BC x DE + AD^{2}+ 1/4BC^{2}– BC x DE= 1/2BC

^{2 }+ 2AD^{2}2(AB

^{2}+ AC^{2}) = BC^{2}+ 4AD^{2}2AB

^{2}+ 2AC^{2}= BC^{2}+ 4AD^{2}Hence Proved

### Question 23. In right-angled triangle ABC in which ∠C = 90°, if D is the mid point of BC, prove that AB^{2} = 4AD^{2 }– 3AC^{2}.

**Solution:**

Given :

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

To prove : AB

^{2}= 4AD^{2}– 3AC^{2}In ∆ ACD,

By using Pythagoras Theorem

AD

^{2}= AC^{2}+ CD^{2}CD

^{2}= AD^{2}– AC^{2 }……….(1)In ∆ACB,

By using Pythagoras Theorem

AB

^{2 }= AC^{2}+ BC^{2}AB

^{2}= AC^{2}+ (2CD)^{2}[D is the mid-point of BC]AB

^{2}= AC^{2}+ 4CD^{2}∴ AB

^{2}= AC^{2}+ 4(AD^{2}– AC^{2}) [From eq(1)]AB

^{2}= AC^{2 }+ 4AD^{2}– 4AC^{2}AB

^{2}= 4AD^{2}– 4AC^{2}+ AC^{2}∴ AB

^{2}= 4AD^{2}– 3AC^{2}Hence Proved

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

**(i) b ^{2 }= p^{2 }+ a^{2}/4 + ax**

**(ii) c ^{2 }= p^{2}– ax + a^{2}/4 **

**(iii) b ^{2 }+ c^{2 }= 2p^{2 }+ a^{2}/2**

**Solution:**

Given,

D is the midpoint of BC

(i)In ∆ AECAC

^{2 }= AE^{2 }+ EC^{2}b

^{2 }= AE^{2 }+ (ED + DC)^{2}b

^{2 }= AD^{2 }+ DC^{2 }+ 2 x ED x DC [Given BC = 2CD]b

^{2 }= p^{2 }+ (a/2)^{2 }+ 2(a/2)xb

^{2 }= p^{2 }+ a^{2}/4 + ax ………..(i)

(ii)In ∆ AEBAB

^{2 }= AE^{2 }+ BE^{2}c

^{2 }= AD^{2 }– ED^{2 }+ (BD – ED)^{2}c

^{2 }= p^{2 }– ED^{2 }+ BD^{2 }+ ED^{2 }– 2BD x EDc

^{2 }= P^{2 }+ (a/2)^{2 }– 2(a/2)^{2}xc

^{2 }= p^{2 }– ax + a^{2}/4 ……………….(ii)

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

^{2 }+ c^{2 }= 2p^{2 }+ a^{2}/2Hence Proved

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

**(i) AB x AQ = AC x AP**

**(ii) BC ^{2} = (AC x CP + AB x BQ)**

**Solution:**

(i)Given :∠A is obtuse.

PB is perpendicular to AC.

QC is perpendicular to AB.

To Prove :

AB × AQ = AC × AP.

Proof:

In ΔACQ and ΔABP,

⇒ ∠CAQ = ∠BAP [Vertically opposite ∠]

⇒ ∠Q = ∠P [∠Q = ∠P = 90 º]

So, by AA rule

ΔACQ ~ ΔABP

By Property of Similar Triangles,

⇒ CQ/BP = AC/AB = AQ/AP

⇒ AC/AB = AQ/AP

⇒ AB × AQ = AC × AP ……..(i)

Hence Proved.

(ii)To Prove :BC² = AB × BQ + AC × CP

Proof:

By using Pythagoras Theorem

⇒ BC

^{2}= CQ^{2}+ QB^{2}⇒ BC

^{2}= CQ^{2}+ (QA + AB)^{2}⇒ BC

^{2}= CQ^{2}+ QA^{2}+ AB^{2}+ 2 QA × AB⇒ BC

^{2}= CQ^{2}+ QA^{2 }+ AB^{2}+ QA × AB + QA × AB⇒ BC

^{2}= AC^{2 }+ AB^{2 }+ QA × AB + QA × AB [In ΔACQ, CQ^{2}+ QA^{2}= AC^{2}]⇒ BC

^{2}= AC^{2}+ AB^{2}+ QA × AB + AC × CP [By Eq (i)]⇒ BC

^{2}= AC^{2}+ AC × CP + AB² + QA × AB⇒ BC

^{2}= AC × (AC + CP) + AB × (QA + AB)⇒ BC

^{2}= AC × CP + AB × BQ [CP = AC + CP, BQ = AQ + AB]⇒ BC

^{2}= AB × BQ + AC × CP.Hence, Proved.

### Question 26. In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that BC^{2} = 4 (AD^{2} – AC^{2}).

**Solution:**

Given:

∠C = 90°

In ∆ADC

By using Pythagoras Theorem

AD

^{2 }= AC^{2 }+ DC^{2}AD

^{2 }= AC^{2 }+ (1/2BC)^{2}[DC = 1/2BC]AD

^{2 }= AC^{2 }+ 1/4(BC)^{2}4AD

^{2 }= 4AC^{2 }+ (BC)^{2}-(BC)

^{2 }= 4AC^{2 }– 4AD^{2 }Taking minus common

(BC)

^{2 }= 4(AD^{2 }– AC^{2})Hence, proved

### Question 27. In a quadrilateral ABCD, ∠B = 90°, AD^{2} = AB^{2} + BC^{2} + CD^{2}, prove that ∠ACD = 90°.

**Solution:**

Given: ABCD is a quadrilateral, ∠B = 90° and AD

^{2}= AB^{2}+ BC^{2}+ CD^{2}To prove: ∠ACD = 90°

Proof: In right ∆ABC,

By using Pythagoras Theorem

AC

^{2}= AB^{2}+ BC^{2}… (1)Given, AD

^{2}= AB^{2}+ BC^{2}+ CD^{2}⇒ AD

^{2}= AC^{2}+ CD^{2}[Using eq(1)]In ∆ACD,

AD

^{2}= AC^{2}+ CD^{2}So, ∠ACD = 90° [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 × 3/2 km = 1800 km

Now in ΔAOB,

Using Pythagoras Theorem,

AB

^{2}= AO^{2}+ OB^{2}⇒ AB

^{2}= (1500)^{2}+ (1800)^{2}⇒ AB = √(2250000 + 3240000)

= √5490000

⇒ AB = 300√61 km

Hence, in hours the distance between two plane = 300√61 km.