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

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

### Solution 1

(i)

From the figure we have

(ii)

From the figure we have

(iii)

From the figure we have

### Solution 2

(i)

From the figure we have

(ii)

From the figure we have

(iii)

From the figure we have

### Solution 3

The figure is drawn as follows:

The above figure we have

Again

Now

### 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

### 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,

.

.

### Solution 7

Consider the figure

From right triangle ACF

From triangle DEB

Given , So

Therefore

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

### 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 .

### Solution 9

From right triangle AQP

Also from triangle PBR

Therefore,

AB = AP + PB =.

### 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

### 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

### 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

### 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.

### 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 .

### 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,

(i) From

Also, from ABQ

Therefore,

(ii) From

Also, from ABQ

Therefore,

### 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.

### 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