# SELINA Solutions for Class 9 Maths Chapter 24 - Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]

Get free access to Selina Solutions for ICSE Class 9 Mathematics Chapter 24 Solution of Right Triangles, Simple 2D problems involving one right-angled Triangle on TopperLearning. Learn how to calculate the lengths of diagonals of a rhombus. Also, learn how to find the length of side of figure formed using different triangles and a quadrilateral.

In this ICSE Class 9 Maths chapter, understand the ways to use right-angled triangle to find answers involving a complex figure. If you have any Math doubts, you can anytime post your question on our ‘UnDoubt’ platform for explanations and it will be answered by Math experts.

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## Chapter 24 - Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle] Exercise Ex. 24

Question 1

Fin 'x' if:

(i)

(ii)

(iii)

Solution 1

(i)

From the figure we have

(ii)

From the figure we have

(iii)

From the figure we have

Question 2

Find angle 'A' if:

(i)

(ii)

(iii)

Solution 2

(i)

From the figure we have

(ii)

From the figure we have

(iii)

From the figure we have

Question 3

Find angle 'x' if:

Solution 3

The figure is drawn as follows:

The above figure we have

Again

Now

Question 4

(i)

(ii)

Solution 4

(i)

From the right triangle ABE

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

Question 5

Find the length of AD.

Given: ABC = 60o.

DBC = 45o

And BC = 40 cm.

Solution 5

Question 6

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

Solution 6

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

And

Also given

In right triangle AOB

Also

Therefore,

.

.

Question 7

Find AB.

Solution 7

Consider the figure

From right triangle ACF

From triangle DEB

Given , So

Therefore

Thus AB = AC + CD + BD = 54.64 cm.

Question 8

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

(i) length of AB

(ii) distance between AB and DC.

Solution 8

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

Similarly from the right triangle BMC we have BM = 10 cm.

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

Therefore the distance between AB and CD is .

Question 9

Use the information given to find the length of AB.

Solution 9

From right triangle AQP

Also from triangle PBR

Therefore,

AB = AP + PB =.

Question 10

Find the length of AB.

Solution 10

Question 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 = 60o and ACD = 45o. Calculate:

(i) AB(ii) AC(iii) AE

Solution 11

(i)

From the triangle ADC we have

Since AD || DC and,  ABCD is a parallelogram and hence opposite sides are equal.

Therefore AB = DC = 2 cm

(ii)

Again

(iii)

From the right triangle ADE we have

Question 12

In the given figure, B = 60° , AB = 16 cm and BC = 23 cm,

Calculate:

(i) BE (ii) AC

Solution 12

In ∆ABE,

(i) In ∆ABE, mAEB = 90°

By Pythagoras Theorem, we get

BE2 = AB2 - AE2

BE2 = (16)2 - ()2

BE2 = 256 - 192

BE2 = 64

BE = 8cm

(ii) EC = BC - BE = 23 - 8 = 15

In ∆AEC, mAEC = 90°

By Pythagoras Theorem, we get

AC2 = AE2 + EC2

AC2 = ()2 + (15)2

AC2 = 192 + 225

AC2 = 147

AC = 20. 42 cm

Question 13

Find

(i) BC

(iii) AC

Solution 13

Question 14

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

(i) AB is times of BC.

(ii) BC is times of AB.

Solution 14

Consider the figure

(i) Here AB is times of BC means

(ii)

Again from the figure

Therefore, magnitude of angle A is 30

Question 15

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

Solution 15

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

Suppose the length of the ladder is x m

From the figure

Therefore the length of the ladder is 30m.

Question 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 60o with the level ground.

Solution 16

Given that the kite is attached to a 100 m long string and it makes an angle of  60 with the ground level which is shown in the figure below:

Suppose that the greatest height is x m.

From the figure

Therefore the greatest height reached by the kite is .

Question 17

Find AB and BC, if:

(i)

(ii)

(iii)

Solution 17

(i)Let  BC = xm

BD = BC + CD = (x+20)cm

In ABD,

tan 30 =

x+20 = AB        .....(1)

In ABC

tan 45 =

1 =

AB = x       ... (2)

From (1)

AB + 20 = AB

AB(-1) = 20

AB

=

=  = 27.32 cm

From (2)

AB = x = 27.32cm

Therefore BC = x = AB = 27.32cm

Therefore, AB = 27.32cm, BC = 27.32cm

(ii)

Let  BC = xm

BD = BC + CD = (x + 20) cm

In ,

tan 30 =

x + 20 =  AB       ...(1)

In

tan 60 =

x =                    ...(2 )

From (1)

From (2)

Therefore BC = x = 10cm

Therefore,

AB = 17.32cm, BC = 10cm

(iii)

Let  BC = xm

BD = BC + CD = (x + 20)cm

In ,

x + 20 = AB         ...(1)

In

From (1)

From (2)

Therefore,

Question 18

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

(i)

(ii)

Solution 18

(i) From

Also, from ABQ

Therefore,

(ii) From

Also, from ABQ

Therefore,

Question 19

If tan xo =,

tan yo = and AB = 48 m; find the length of CD.

Solution 19

Given tan xo =tan to = and AB = 48 m;

Let length of  BC = xm

Also, from BDC

From (1)

Therefore, length of CD is 45 m.

Question 20

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

Solution 20

Since in a rhombus all sides are equal.

The diagram is shown below:

Therefore ,Let .

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

But

Therefore,

Also,

But

Therefore,

So, the length of the diagonal and

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