# RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D

## RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles

### Exercise 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

D is mid point of BC

BD = DC = (frac { 1 }{ 2 }) BC

= (frac { 1 }{ 2 }) x 2a = a

AB² = AD² + BD² (Pythagoras Theorem)

⇒ 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 = (frac { 1 }{ 2 }) BC = (frac { 1 }{ 2 }) x 12 = 6cm

⇒ (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 = (frac { 1 }{ 2 }) AC.

and BO = OD = (frac { 1 }{ 2 }) BD

BD = 24 cm and AC = 10 cm

BO = (frac { 1 }{ 2 }) x BD = (frac { 1 }{ 2 }) x 24 = 12 cm

AO = (frac { 1 }{ 2 }) x AC = (frac { 1 }{ 2 }) 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.

To prove: AB² = AD² – BC x DE + (frac { 1 }{ 4 }) BC2

Proof: In ∆AEB, ∠E = 90°

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

In ∆AED, ∠E = 90°

⇒ 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 =

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 = 1(frac { 1 }{ 2 }) hours

Distance covered by the first plane in 1(frac { 1 }{ 2 }) hours = 1000 x (frac { 3 }{ 2 }) km = 1500 km

and distance covered by another plane in 1(frac { 1 }{ 2 }) hours = 1200 x (frac { 3 }{ 2 }) 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

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

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles are helpful to complete your math homework.

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.

You might also like