Selina Chapter 24 Solutions of Right Triangles ICSE Solutions Class 9 Maths

ICSE Solutions for Selina Concise Chapter 24 Solutions of Right Triangles Class 9 Maths

Exercise 24(A)

1. Fin ‘x’ if:

(i) 

(ii)

(iii)

Answer 

(i) From the figure we have 

2. Find angle ‘A’ if:

(i)

(ii)

(iii)

Answer 

(i) From the figure we have 
cos A = 10/20 
⇒ cos A = 1/2
⇒ cos A = cos 60°
⇒ A = 60°

(ii) From the figure we have 

sin A = sin45°
⇒ A = 45°

(iii) From the figure we have 
tan A = 10√3/10
⇒ tan A = √3
⇒ tan A = sin 60°
⇒ A = 60°

3. Find angle ‘x’ if:

Answer 

The figure is drawn as follows : 

The above figure we have 
tan 60° = 30/AD 
⇒ √3 = 30/AD 
⇒ AD = 30/√3 
Again,
sin x = AD/20 
⇒ AD = 20 sin x
Now, 
20 sin x = 30/√3 
⇒ sin x = 30/20√3 
⇒ sin x = √3/2
⇒ sin x = sin 60° 
⇒ x = 60°

4. Find AD, if:

(i)

(ii)

Answer 

(i) From the right triangle ABE 
tan45° = AE/BE 
⇒ 1 = AE/BE
⇒ AE = BE
Therefore, AE = BE = 50 m. 
Now, from the rectangle BCDE we have
DE = BC = 10 m.
Therefore, the length of AD will be: 
AD = AE + DE = 50 + 10 = 60 m. 

(ii) From the triangle ABD we have
sin B = AD/AB 
⇒ sin 30 = AD/100 [Since ∠ACD is the exterior angle of the triangle ABC]
⇒ 1/2 = AD/100
⇒ AD = 50 m

5. Find the length of AD.
Given: ABC = 60°.
DBC = 45° and BC = 40 cm

Answer 

From right triangle ABC, 
tan60° = AC/BC 
⇒ √3 = AC/40
⇒ AC = 40√3 cm

From right triangle BDC, 
tan45° = DC/BC
⇒ 1 = DC/40 
⇒ DC = 40 cm 

From the figure, it is clear that AD = AC - DC 
⇒ AD = 40√3 - 40 
⇒ AD = 40(√3 - 1)
⇒ AD = 29.28 cm 

6. Find the lengths of diagonals AC and BD. Given AB = 60 cm and BAD = 60^o.

Answer 

We know, diagonals of a rhombus bisect each other at right angles and also bisect the angle of vertex . 
The figure is shown below : 

Now,

7. Find AB.

Answer 

Consider the figure:

Given FC = 20, ED = 30,  So EP = 10cm 
Therefore, 
tan 60° = FP/EP 
⇒ √3 = FP/10 
⇒ FP = 10√3 = 17.32 cm
Thus, AB = AC + CD + BD = 54.64 cm.

8. In trapezium ABCD, as shown, AB∥DC, AD = DC = BC = 20 cm and A = 60^o. Find:

(i) length of AB

(ii) distance between AB and DC.

Answer 

First draw two perpendiculars to AB from the point D and C respectively. Since AB|| CD therefore PMCD will be a rectangle.
Consider the figure:

(i) From right triangle ADP we have
cos 60° = AP/AD 
⇒ 1/2 = AP/20
⇒ AP = 10 
Similarly from the right triangle BMC we have BM = 10cm . 
Now from the rectangle PMCD we have CD = PM = 20 cm . 
Therefore,
AB = AP + PM + MB = 10 +20 +10 = 40 cm.

(ii) Again from the right triangle APD we have,
sin 60° = PD/20 
⇒ √3/2 = PD/20 
⇒ PD = 10√3
Therefore, the distance between AB and CD is 10√3. 

9. Use the information given to find the length of AB.

Answer 

From right triangle AQP 

10. Find the length of AB.

Answer 

From right triangle ADE 
tan 45° = AE/DE 
⇒ 1 = AE/30 
⇒ AE = 30 cm 
Also, from triangle DBE 
tan 60° = BE/DE
⇒ √3 = BE/30 
⇒ BE = 30√3  cm 
Therefore, AB = AE + BE = 30 + 30√3 = 30(1 + √3) cm 

11. In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB. Given that AED = 60^o and ACD = 45^o. Calculate:

(i) AB

(ii) AC

(iii) AE

Answer 

(i) From the triangle ADC  we have
tan45° = AD/DC 
⇒ 1 = 2/DC 
⇒ DC = 2 
Since, AD॥DC and AD⊥EC, ABCD is a parallelogram and hence opposite sides are equal. 
Therefore, AB = DC = 2cm 

(ii) Again,
sin 45° = AD/AC
⇒ 1/√2 = 2/AC 
⇒ AC = 2√2

(iii) From the right triangle ADE we have 
sin 60° = AD/AE
⇒ √3/2 = 2/AE
⇒ AE = 4/√3

12. In the given figure, ∠B = 60^° , AB = 16 cm and BC = 23 cm,

Calculate:

(i) BE 

(ii) AC

Answer 

Let BE = x, and EC = 25 -x 
In △ABC, 
sin 60° = AE/AB 

13. Find

(i) BC

(ii) AD

(iii) AC

Answer 

14. In right-angled triangle ABC; B = 90^o. Find the magnitude of angle A, if:

(i) AB is √3 times of BC.

(ii) BC is √3 times of AB.

Answer 

Consider the figure 

15. A ladder is placed against a vertical tower. If the ladder makes an angle of 30^o with the ground and reaches upto a height of 15 m of the tower; find length of the ladder.

Answer 

Given that the ladder makes an angle of 30° with the ground and reaches upto a height of 15 m of the tower which is shown in the figure below:

16. A kite is attached to a 100 m long string. Find the greatest height reached by the kite when its string makes an angles of 60^o with the level ground.

Answer 

17. Find AB and BC, if:

(i)

(ii)

(iii)

Answer 

(i) Let BC = xm 
BD = BC + CD = (x + 20)cm 
In △ABD, 

AB = x = 27.32 cm 
Therefore, BC = x = AB = 27.32 cm
Therefore, AB = 27.32 cm, BC = 27.32 cm 

(ii) Let BC = x m
BD = BC + CD = (x+20)cm 
In △ABD, 

Therefore BC = x = 10cm 
Therefore, 
AB = 17.32 cm, BC = 10 cm 

(iii) Let BC =  x m 
BD = BC + CD = (x+20) cm 
In △ABD,  

∴ BC = x = 27.32 cm 
Therefore, AB = 47.32 cm, BC =27.32 cm 

18. Find PQ, if AB = 150 m, P = 30^o and Q = 45^o.

(i)

(ii)

Answer 

19. If tan x^o = 5 /12, tan y^o = 3/4 and AB = 48 m; find the length of CD

Answer 

⇒ 720 + 20 CD = 36 CD
⇒ 16 CD = 720
⇒ CD = 45
Therefore, length of CD is 45 m.

20. The perimeter of a rhombus is 96 cm and obtuse angle of it is 120^o. Find the lengths of its diagonals.

Answer 

Since in a rhombus all sides are equal.
The diagram is shown below : 

Therefore PQ = 96/4 = 24 cm, Let ∠PQR = 120°
We also know that in rhombus diagonals bisect each other perpendicularly and diagonal bisect the angle at vertex.
Hence POR is a right angle triangle and

Therefore, SQ = 2× OQ  = 2×12 = 24 cm 
So, the length of the diagonal PR = 41.568 cm and SQ = 24 cm.