In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 4 Triangles EX 4D Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 4 Triangles EX 4D Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 4 Triangles EX 4D Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 4 |

Chapter Name | Triangles |

Exercise | 4D |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 4 Triangles EX 4D**

**Question 1.****Solution:**

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

(i) 9 cm, 16 cm, 18 cm

Longest side = 18

Now (18)² = 324

and (9)² + (16)² = 81 + 256 = 337

324 ≠ 337

It is not a right triangle.

(ii) 1 cm, 24 cm, 25 cm

Here longest side = 25 cm

(25)² = 625

and (7)² x (24)² = 49 + 576 = 625

625 = 625

It is a right triangle

(iii) 1.4 cm, 4.8 cm, 5 cm

Here longest side = 5 cm

(5)² = 25

and (1.4)² + (4.8)² = 1.96 + 23.04 = 25.00 = 25

25 = 25

It is a right triangle

(iv) 1.6 cm, 3.8 cm, 4 cm

Here longest side = 4 cm

(4 )² = 16

and (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00 = 17

16 ≠ 17

It is not a right triangle

(v) (a- 1) cm, 2√a cm, (a + 1) cm

Here longest side = (a + 1) cm

(a + 1)² = a² + 2a + 1

and (a – 1)² + (2 √a )² = a² – 2a + 1 + 4a = a² + 2a + 1

a² + 2a + 1 = a² + 2a + 1

It is a right triangle.

**Question 2.****Solution:**

A man goes 80 m from O to east side and reaches A, then he goes 150 m due north from A and reaches B.

Join OB.

In right ∆OAB,

OB² = OA²+² (Pythagoras Theorem) = (80)² + (150)² = 6400 + 22500 = 28900

⇒ OB = √28900 = 170

He is 170 m away from the starting point.

**Question 3.****Solution:**

A man goes 10 m due south from O and reaches A and then 24 m due west from A and reaches B.

Join OB.

**Question 4.****Solution:**

Length of a ladder = 13 m

Height of the window = 12 m

Distance between the foot of the ladder and building.

In the figures,

AB is ladder, A is window of building AC

AB² = AC² + BC² (Pythagoras Theorem)

⇒ (13)² = (12)² + x²

⇒ 169 = 144 + x²

⇒ x² = 169 – 144 = 25 = (5)²

x = 5

Hence, distance between foot of ladder and building = 5 m.

**Question 5.****Solution:**

Let length of ladder AB = x m

Height of window AC = 20 m

and distance between the foot of the ladder and the building (BC) = 15 m

AB² = AC² + BC² (Pythagoras Theorem)

⇒ x² = 20² + 15² = 400 + 225 = 625 = (25)²

x = 25

Length of ladder = 25 m

**Question 6.****Solution:**

Height of first pole AB = 9 m

and of second pole CD = 14 m

Let distance between their tops CA = x m

From A, draw AE || BD meeting CD at E.

Then EA = DB = 12 m CE = CD – ED = CD – AB = 14-9 = 5 m

In right ∆AEC,

AC² = AE² + CE² = 122 + 52 = 144 + 25 = 169 = (13)²

AC = 13

Distance between their tops = 13 m

**Question 7.****Solution:**

Height of the pole AB = 18 m

and length of wire AC = 24 m

Distance between the base of the pole and other end of the wire

BC = x m (suppose)

In right ∆ABC,

AC² = AB² + BC² (Pythagoras Theorem)

(24)² = (18)² + x²

⇒ 576 = 324 + x²

⇒ x² = 576 – 324 = 252

**Question 8.****Solution:**

In ∆PQR, O is a point in it such that

OP = 6 cm, OR = 8 cm and ∠POR = 90°

PQ = 24 cm, QR = 26 cm

To prove : ∆PQR is a right angled.

In ∆POR, ∠O = 90°

PR² = PO² + OR² = (6)² + (8)² = 36 + 64 = 100 = (10)²

PR = 10

Greatest side QR is 26 cm

QR² = (26)² = 676

and PQ² + PR² = (24)² + (10)² = 576 + 100 = 676

676 = 676

QR² = PQ² + PR²

∆PQR is a right angled triangle and right angle at P.

**Question 9.****Solution:**

In isosceles ∆ABC, AB = AC = 13 cm

AL is altitude from A to BC

and AL = 5 cm

Now, in right ∆ALB

AB² = AL² + BL²

(13)² = (5)² + BL²

⇒ 169 = 25 + BL²

⇒ BL² = 169 – 25 = 144 = (12)²

BL = 12 cm

L is mid point of BC

BC = 2 x BC = 2 x 12 = 24 cm

**Question 10.****Solution:**

In an isosceles ∆ABC in which

AB = AC = 2a units, BC = a units

**Question 11.****Solution:**

∆ABC is an equilateral triangle

and AB = BC = CA = 2a

AD ⊥ BC

D is mid point of BC

BD = DC = 12 BC

= 12 x 2a = a

Now, in right ∆ADB,

AB² = AD² + BD² (Pythagoras Theorem)

(2a)² = AD² + a²

⇒ 4a² – a² = AD²

⇒ AD² = 3a² = (√3 a)²

AD = √3 a = a√3 units

**Question 12.****Solution:**

∆ABC is an equilateral triangle in which

AB = BC = CA = 12 cm

AD ⊥ BC which bisects BC at D

BD = DC = 12 BC = 12 x 12 = 6cm

Now, in right ∆ADB,

AB² = AD² + BD²

⇒ (12)² = AD² + (6)²

⇒ 144 = AD² + 36

AD² = 144 – 36 = 108

AD = √108 = √(36 x 3) = 6√3 cm

**Question 13.****Solution:**

Let ABCD is a rectangle in which adjacent sides.

AB = 30 cm and BC = 16 cm

AC is its diagonal.

In right ∆ABC,

AC² = AB² + BC² (Pythagoras Theorem)

= (30)² + (16)² = 900 + 256 = 1156 = (34)²

Diagonal AC = 34 cm

**Question 14.****Solution:**

ABCD is a rhombus

Its diagonals AC and BD bisect each other at O.

AO = OC = 12 AC.

and BO = OD = 12 BD

BD = 24 cm and AC = 10 cm

BO = 12 x BD = 12 x 24 = 12 cm

AO = 12 x AC = 12 x 10 = 5 cm

Now, in right ∆AOB,

AB² = AO² + BO² = (5)² + (12)² = 144 + 25 = 169 = (13)²

AB = 13

Hence, each side of rhombus = 13 cm

**Question 15.****Solution:**

Given : In ∆ABC, AC > AB.

D is the mid point of BC and AE ⊥ BC.

AD is joined.

To prove: AB² = AD² – BC x DE + 14 BC2

Proof: In ∆AEB, ∠E = 90°

AB² = AE² + BE² …..(i) (Pythagoras Theorem)

In ∆AED, ∠E = 90°

AD² = AE² + DE²

⇒ AE² = AD² – DE² …..(ii)

Now, substitute eq. (ii) in eq. (i)

AB² = AE² + BE²

AB² = AD² – DE² + BE² [from (ii)]

AB² = (AD² – DE²) + (BD – DE)² [BE = BD – DE²]

**Question 16.****Solution:**

**Question 17.****Solution:**

Given : In ∆ABC, D is the mid point of BC, AE ⊥ BC,

BC = a, AC = b, AB = c, ED = x, AD =p and AE =

AD is joined.

To prove :

**Question 18.****Solution:**

Given : In ∆ABC, AB =AC

BC is produced to D and AD is joined.

To prove : (AD² – AC²) = BD x CD

Construction : Draw AE ⊥ BC.

Proof: In ∆ABC,

AB = AC and AE ⊥ BC

BE = EC

Now, in right ∆AED, ∠E = 90°

AD² = AE² + ED² …..(i)

and in right ∆AEC, ∠E = 90°

AC² = AE² + EC² …..(ii)

Now, subtracting (i) and (ii),

AD² – AC² = (AE² + ED²) – (AE² + EC²)

= AE² + ED² – AE² – EC²

= ED² – EC²

= (ED + EC) (ED – EC) (BE = EC proved above)

= BD x CD = BD x CD

AD² – AC² = BD x CD

Hence proved.

**Question 19.****Solution:**

Given : In ∆ABC, AB = BC and ∠ABC = 90°

∆ACD and ∆ABE are similar to each other.

To prove : Ratio between area ∆ABE and area ∆ACD.

Proof: Let AB = BC = x

Now, in right ∆ABC,

⇒ AC² = AB² + BC² = AB² + AB² = 2AB² = 2x²

∆ABE and ∆ACD are similar

Ratio between the areas of ∆ABE and ∆ACD = 1 : 2

**Question 20.****Solution:**

An aeroplane flies from airport to north at the speed of 1000 km/hr.

Another aeroplane flies from the airport to west at the speed of 1200 km/hr.

Period = 112 hours

Distance covered by the first plane in 112 hours = 1000 x 32 km = 1500 km

and distance covered by another plane in 112 hours = 1200 x 32 km = 1800 km

At present, the distance between them

AB² = (BO)² + (AO)²

= (1800)² + (1500)²

= 3240000 + 2250000

= 5490000

**Question 21.****Solution:**

Given : In ∆ABC,

D is the mid point of BC and AL ⊥ BC

AD is joined.

To prove:

**Question 22.****Solution:**

AM is rod and BC is string out of rod.

In ∆BMC,

BC² = BM² + CM² = (1.8)² + (2.4)²

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class10**

**All Subject NCERT Solutions For Class 10**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.