# RS Aggarwal Solutions Chapter 4 Triangles Exercise - 4D Class 10 Maths

 Chapter Name RS Aggarwal Chapter 4 Triangles Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 4AExercise 4BExercise 4CExercise 4E Related Study NCERT Solutions for Class 10 Maths

### Exercise 4D Solutions

1. The sides of certain triangles are given below. Determine which of them right triangles are.

(i) 9cm, 16cm, 18cm

(ii) 7cm, 24cm, 25cm

(iii) 1.4cm, 4.8cm, 5cm

(iv) 1.6cm, 3.8cm, 4cm

(v) (a – 1) cm, 2√a cm, (a + 1) cm

Solution

For the given triangle to be right-angled, the sum of the two sides must be equal to the square of the third side.

Here, let the three sides of the triangle be a, b and c.

(i) a = 9 cm, b = 16 cm and c = 18 cm

Then,

a2 + b2 = 92 + 162 = 81 + 256

= 337

c2 = 192

= 361

a2 + b2 ≠ c2

Thus, the given triangle is not right-angled.

(ii) a = 7 cm, b = 24 cm and c = 25 cm

Then,

a2 + b2 = 72 + 242

= 49 + 576

= 625

c2 = 252

= 625

a2 + b2 = c2

Thus, the given triangle is a right-angled.

(iii) a = 1.4 cm, b= 4.8 cm and c= 5 cm

Then,

a2 + b2 = (1.4)2 + (4.8)2

= 1.96 + 23.04

= 25

c2 = 52

= 25

a2 + b2 = c2

Thus, the given triangle is right-angled.

(iv) a = 1.6 cm, b = 3.8 cm and c = 4 cm

Then

a2 + b2 = (1.6)2 + (3.8)2

= 2.56 + 14.44

= 16

a2 + b2 ≠ c2

Thus, the given triangle is not right-angled.

(v) p = (a - 1) cm, q = 2 √a cm and r = (a + 1)cm

Then,

p2 + q2 = (a − 1)2 + (2√a)2

= a2 + 1 – 2a + 4a

= a2 + 1 + 2a

= (a + 1)2

r2 = (a + 1)2

p2 + q2 = r2

Thus, the given triangle is right-angled.

2. A man goes 80m due east and then 150m due north. How far is he from the starting point?

Solution

Let the man starts from point A and goes 80 m due east to B.

Then, from B, he goes 150 m due north to c.

We need to find AC.

In right- angled triangle ABC, we have:

AC2 = AB2 + BC2

Hence, the man is 170 m away from the starting point.

3. A man goes 10m due south and then 24m due west. How far is he from the starting point?

Solution

Let the man starts from point D and goes 10 m due south at E. He then goes 24 m due west at F.

In right ∆DEF, we have:

DE = 10 m, EF = 24 m

DF2 = EF2 + DE2

Hence, the man is 26 m away from the starting point.

4. A 13m long ladder reaches a window of a building 12m above the ground. Determine the distance of the foot of the ladder from the building.

Solution

Let AB and AC be the ladder and height of the building.

It is given that:

AB = 13 m and AC = 12 m

We need to find distance of the foot of the ladder from the building, i.e., BC.

In right-angled triangle ABC, we have:

AB2 = AC2 + BC2

Hence, the distance of the foot ladder from the building is 5 m.

5. A ladder is placed in such a way that its foot is at a distance of 15m from a wall and its top reaches a window 20m above the ground. Find the length of the ladder.

Solution

Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, respectively.

We have:

AB = 20 m and BC = 15 m

Applying Pythagoras theorem in right-angled ABC, we get:

AC2 = AB2 + BC2

Hence, the length of the ladder is 25 m.

6. Two vertical poles of height 9m and 14m stand on a plane ground. If the distance between their feet is 12m, find the distance between their tops.

Solution

Let the two poles be DE and AB and the distance between their bases be BE.

We have:

DE = 9 m, AB = 14 m and BE = 12 m

Draw a line parallel to BE from D, meeting AB at C.

Then, DC = 12 m and AC = 5 m

We need to find AD, the distance between their tops.

Applying Pythagoras theorem in right-angled ACD, we have:

⇒ AD2 = 52 + 122 = 25 + 144 = 169

Hence, the distance between the tops to the two poles is 13 m.

7. A guy wire attached to a vertical pole of height 18 m is 24m 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

Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.

Now, In right triangle ABC

By using Pythagoras theorem, we have

AB2 = BC2 + CA2

⇒ 242 = 182 + CA2

⇒ CA2 = 576 – 324

⇒ CA2 = 252

⇒ CA = 6√7m

Hence, the stake should be driven 6√7m far from the base of the pole.

8. In the given figure, O is a point inside a ∆PQR such that ∠PQR such that ∠POR = 90°, OP = 6cm and OR = 8cm. If PQ = 24cm and QR = 26cm, prove that ∆PQR is right-angled.

Solution

Applying Pythagoras theorem in right-angled triangle POR, we have:

PR2 = PO2 + OR2

⇒ PR2 = 62 + 82 = 36 + 64 = 100

In ∆ PQR,

PQ2 + PR2 = 242 + 102 = 576 + 100 = 676

And QR2 = 262 = 676

∴ PQ2 + PR2 = QR2

Therefore, by applying Pythagoras theorem, we can say that ∆PQR is right-angled at P.

9. ∆ABC is an isosceles triangle with AB = AC = 13cm. The length of altitude from A on BC is 5cm. Find BC.

Solution

It is given that ∆ ABC is an isosceles triangle.

Also, AB = AC = 13 cm

Suppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.

Applying Pythagoras theorem, we have;

⇒ BD2 = AB2 − AD2 = 132 − 52

⇒ BD2 = 169 − 25 = 144

Hence,

BC = 2(BD) = 2×12 = 24 cm

10. Find the length of altitude AD of an isosceles ∆ABC in which AB = AC = 2a units and BC = a units.

Solution

In isosceles ∆ ABC, we have:

AB = AC = 2a units and BC = a units

Let AD be the altitude drawn from A that meets BC at D.

Then, D is the midpoint of BC.

BD = BC = a/2 units

Applying Pythagoras theorem in right-angled ∆ABD, we have:

⇒ AD2 = AB2 − BD2 = (2a)2 − (a/2)2

⇒ AD2 = 4a2 – a2/4 = 15a2/4

11. ∆ABC is an equilateral triangle of side 2a units. Find each of its altitudes.

Solution

Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, respectively.

Then, D, E and F are the midpoint of BC, AC and AB, respectively.

In right-angled ∆ABD, we have:

AB = 2a and BD = a

Applying Pythagoras theorem, we get:

⇒ AD2 = AB2 − BD2 = (2a)2 − a2

⇒ AD2 = 4a2 − a2 = 3a2

Similarly,

BE = a√3 units and CF = a√3 units

12. Find the height of an equilateral triangle of side 12cm.

Solution

Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. Then, D will be the midpoint of BC.

Applying Pythagoras theorem in right-angled triangle ABD, we get:

⇒ AD2 = 122 − 62 (∵ BD = 1/2 BC = 6)

⇒ AD2 = 144 − 36 = 108

Hence, the height of the given triangle is 6√3 cm.

13. Find the length of a diagonal of a rectangle whose adjacent sides are 30cm and 16cm.

Solution

Let ABCD be the rectangle with diagonals AC and BD meeting at O.

According to the question:

AB = CD = 30 cm and BC = AD = 16 cm

Applying Pythagoras theorem in right-angled triangle ABC, we get:

AC2 = AB2 + BC2 = 302 + 162 = 900 + 256 = 1156

Diagonals of a rectangle are equal.

Therefore, AC = BD = 34 cm

14. Find the length of each side of a rhombus whose diagonals are 24cm and 10cm long.

Solution

Let ABCD be the rhombus with diagonals (AC = 24 cm and BD = 10 cm) meeting at O. We know that the diagonals of a rhombus bisect each other at angles.

Applying Pythagoras theorem in right-angled AOB, we get:

AB2 = AO2 + BO2 = 122 + 52

⇒ AB2 = 144 + 25 = 169

Hence, the length of each side of the rhombus is 13 cm.

15. In ∆ABC, D is the midpoint of BC and AE⊥BC. If AC>AB, show that AB2 = AD2 + 1/4BC2 − BC.DE

Solution

In right-angled triangle AED, applying Pythagoras theorem, we have:

AB2 = AE2 + ED2 …(i)

In right-angled triangle AED, applying Pythagoras theorem, we have:

⇒ AE2 = AD2 – ED2 ...(ii)

Therefore,

AB2 = AD2 − ED2 + EB2 (from (i) and (ii))

AB2 = AD2 − ED2 + (BD − DE)2

= AD2 − ED2 + (1/2BC − DE)2

= AD2 – DE2 + 1/4 BC2 + DE2 − BC.DE

= AD2 + 1/4BC2 – BC.DE

This completes the proof.

16. In the given figure, ACB = 90° CD ⊥ AB. Prove that BC2/AC2 =  BD/AD

Solution

Given: ∠ACB = 90° and CD ⊥ AB

Proof: In ∆ ACB and ∆ CDB

∠ACB = ∠CDB = 90° (Given)

∠ABC = ∠CBD (Common)

By AA similarity-criterion ∆ACB ~ ∆CDB

When two triangles are similar, then the ratios of the lengths of their corresponding sides are proportional.

∴ BC/BD = AB/BC

⇒ BC2 = BD. AB …(1)

In ∆ ACB and ∆ ADC

∠ACB = ∠ADC = 90° (Given)

∠CAB = ∠DAC (Common)

By AA similarity-criterion ∆ ACB ~ ∆ADC

When two triangles are similar, then the ratios of their corresponding sides are proportional.

⇒ AC2 = AD. AB  ...(2)

Dividing (2) by (1), we get

17. In the given figure, D is the midpoint of side BC and AEBC. If BC = a, AC = b, AB = c, AD = p and AE = h, prove that

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

(ii) c2 = p2 − ax + a2/x

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

(iv) b2 c2 = 2ax

Solution

(i) In right-angled triangle AEC, applying Pythagoras theorem, we have:

AC2 = AE2 + EC2

⇒ b2 = ℎ2 + (x + a/2)2 = ℎ2 + x2 + a2/4 + ax  …(i)

In right – angled triangle AED, we have:

⇒ p2 = h2 + x2  ...(ii)

Therefore,

from (i) and (ii),

b2 = p2 + ax + a2/x

(ii) In right-angled triangle AEB, applying Pythagoras, we have:

AB2 = AE2 + EB2

⇒ c2 = ℎ2 + (a/2 − x)2 (∵ BD = a/2 and BE = BD - x)

⇒ c2 = h2 + x2 – a2/4 (∵ ℎ2 + x2 = p2

⇒ c2 = p2 − ax + a2/x

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

⇒ b2 + c2 = p2 + ax + a2/4 + p2 – ax + a2/4

= 2p2 + ax – ax + (a2 + a2)/4

(iv) Subtracting (ii) from (i), we get:

b2 – c2 = p2 + ax + a2/4 – (p2 – ax + a2/4)

= p2 – p2 + ax + ax + a2/4 – a2/4

= 2ax

18. In ∆ABC, AB = AC. Side BC is produced to D. Prove that AD2 − AC2 = BD. CD

Solution

Draw AE ⊥ BC, meeting BC at D.

Applying Pythagoras theorem in right-angled triangle AED, we get:

Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.

So, BE = CE

And DE + CE = DE + BE = BD

⇒ AE2 = AD2 − DE2 …(i)

In ∆ACE,

AC2 = AE2 + EC2

⇒ AE2 = AC2 − EC2 ...(ii)

Using (i) and (ii),

⇒ AD2 – DE2 = AC2 – EC2

⇒ AD2 – AC2 = DE2 – EC2

= (DE + CE) (DE – CE)

= (DE + BE) CD

= BD.CD

19. ABC is an isosceles triangle, right-angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ∆ABE and ∆ACD.

Solution

We have, ABC as an isosceles triangle, right angled at B.

Now, AB = BC

Applying Pythagoras theorem in right-angled triangle ABC, we get:

AC2 = AB2 + BC2 = 2AB2  (∵ AB = AC) … (i)

∵ ∆ACD ~ ∆ABE

We know that ratio of areas of 2 similar triangles is equal to squares of the ratio of their corresponding sides.

∴ ar(∆ABE)/ar(∆ACD) = AB2/AC [from (i)]

= 1/2

1 : 2

20. An aeroplane leaves an airport and flies due north at a speed of 1000km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1.1/2 hours?

Solution

Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr

Distance covered by plane A in 1.1/2 hours = 1000 × 3/2 = 1500 km

Distance covered by plane B in 1 1/2 hours = 1200 × 3/2 = 1800 km

Now, In right triangle ABC

By using Pythagoras theorem, we have

AB2 = BC2 + CA2

= (1800)2 + (1500)2

= 3240000 + 2250000

= 5490000

∴ AB2 = 5490000

⇒ AB = 300√61m

Hence, the distance between two planes after 1/1/2 hours 300√61m

21. In a ΔABC, AD is a median and AL BC.

Prove that

(a) AC2 = AD2 + BC.DL + (BC/2)2

(b) AB2 = AD2 – BC.DL + (BC/2)2

(c) AC2 + AB2 = 2AD2 + ½ BC2

Solution

(a) In right triangle ALD

Using Pythagoras theorem, we have

AC2 = AL2 + LC2

= AD2 − DL2 + (DL + DC)2  [Using (1)]

= AD2 − DL2 + (DL + BC/2) [∵ AD is a median]

= AD− DL+ DL2 + (BC/2)+ BC.DL

∴ AC2 = AD2 + BC.DL + (BC/2) ...(2)

(b) In right triangle ALD

Using Pythagoras theorem, we have

AL2 = AD2 – DL2 …(3)

Again, In right triangle ABL

Using Pythagoras theorem, we have

AB2 = AL2 + LB2

= AD2 – DL2 + LB2  [Using (3)]

= AD2 – DL2 + (BD – DL)2

= AD2 – DL2 + (1/2 BC – DL)2

= AD2 – DL2 + (BC/2)2 – BC. DL + DL2

∴ AB2 = AD2 – BC. DL + (BC/2)2 ...(4)

(c) Adding (2) and (4), we get,

= AC+ AB2 = AD2 + BC.DL + (BC/2)2 + AD2 – BC.DL + (BC/2)2

= 2AD2 + BC2/4 + BC2/4

= 2 AD2 + 1/2 BC2

22. Naman is doing fly-fishing in a stream. The trip fishing rod is 1.8m above the surface of the water and the fly at the end of the string rests on the water 3.6m away from him and 2.4 m from the point directly under the tip of the rod. Assuming that the string ( from the tip of his rod to the fly) is taut, how much string does he have out (see the adjoining figure) if he pulls in the string at the rate of 5cm per second, what will be the horizontal distance of the fly from him after 12 seconds?

Solution

Naman pulls in the string at the rate of 5 cm per second.

Hence, after 12 seconds the length of the string he will pulled is given by

12 × 5 = 60 cm or 0.6 m

Now, in ∆BMC

By using Pythagoras theorem, we have

= (2.4)2 + (1.8)2

= 9

∴ BC = 3 m

Now, BC’ = BC – 0.6

= 3 – 0.6

= 2.4 m

Now, In ∆BC’M

By using Pythagoras theorem, we have

C’M2 = BC’2 – MB2

= (2.4)2 − (1.8)2

= 2.52

∴ C’M = 1.6 m

The horizontal distance of the fly from him after 12 seconds is given by

C’A = C’M + MA

= 1.6 + 1.2

= 2.8 m