# 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(a)

Correct option: (ii)
45^{o}

ABC is an isosceles right-angled triangle.

⇒ AB = BC

⇒ ∠BAC = ∠BCA (angles opposite to equal sides are equal)

Given, ∠ABC = 90^{o}

So, in isosceles right-angled ΔABC,

∠ABC + ∠BAC + ∠BCA = 180^{o}

⇒ 90^{o} + 2∠BAC = 180^{o}

⇒ 2∠BAC = 90^{o}

⇒ ∠BAC = 45^{o}

### Solution 1(b)

Correct option: (iii)
^{}

Given, ∠ACB = 90^{o} and ∠CAB = 60^{o}

So, in right-angled ΔACB,

∠ABC + ∠CAB + ∠ACB = 180^{o}

⇒ ∠ABC + 60^{o} + 90^{o}
= 180^{o}

⇒ ∠ABC = 30^{o}

Therefore,

cot ∠ABC = cot 30^{o}
=

### Solution 1(c)

Correct option: (iii)
8 m^{}

### Solution 1(d)

Correct option: (i) 7.32 m^{}

Now,

### Solution 1(e)

Correct option: (iv) 40 cm

AC = CD (given)

⇒ ∠ADC = ∠DAC (angles opposite to equal sides are equal)

Let ∠ADC = ∠DAC = x

In right-angled ΔABC,

∠ABC + ∠ACB + ∠BAC = 180^{o}

⇒ 90^{o} + 60^{o} + ∠BAC = 180^{o}

⇒ ∠BAC = 30^{o}

In right-angled ΔABD,

∠ABD + ∠BAD + ∠ADB = 180^{o}

⇒ 90^{o} + x + 30^{o}
+ x = 180^{o}

⇒ 2x = 60^{o}

⇒ x = 30^{o} = ∠ADB

Now,

### Solution 2

(i)

From the figure we have

_{}

(ii)

From the figure we have

_{}

(iii)

From the figure we have

_{}

### Solution 3

(i)

From the figure we have

_{}

(ii)

From the figure we have

_{}

(iii)

From the figure we have

_{}

### Solution 4

The figure is drawn as follows:

The above figure we have

_{}

Again

_{}

Now

_{}

### Solution 5

(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

### Solution 7

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 8

Consider the figure

From right triangle ACF

_{}

From triangle DEB

_{}

Given _{}, So _{}

Therefore

_{}

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

### Solution 9

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 10

From right triangle AQP

_{}

Also from triangle PBR

_{}

Therefore,

AB = AP + PB =_{}.

## Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle] Exercise Test Yourself

### Solution 1

### Solution 2

### Solution 3

(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 4

In ∆ABE,

(i) In ∆ABE, m∠AEB = 90^{°}

∴ By Pythagoras Theorem, we get

BE^{2} = AB^{2} - AE^{2}

⇒ BE^{2} = (16)^{2} - ()^{2}

⇒ BE^{2} = 256 - 192

⇒ BE^{2} = 64

⇒ BE = 8cm

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

In ∆AEC, m∠AEC = 90^{°}

∴ By Pythagoras Theorem, we get

AC^{2} = AE^{2} + EC^{2}

⇒ AC^{2} = ()^{2} + (15)^{2}

⇒ AC^{2} = 192 + 225

⇒ AC^{2} = 147

⇒ AC = 20. 42 cm

### Solution 5

### Solution 6

Consider the figure

(i) Here AB is _{}times of BC means

_{}

(ii)

Again from the figure

_{}

Therefore, magnitude of angle A is 30

### Solution 7

Given that the ladder makes an angle of 30^{o} 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 8

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 9

(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* = 27.32*cm*, *BC* = 27.32*cm*

(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* = 10*cm*

Therefore,

*AB* = 17.32*cm*, *BC* = 10*cm*

(iii)

Let * BC = xm*

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

In ,

x + 20 = *AB *...(1)

In

From (1)

_{}

From (2)

_{}

_{}

Therefore,

_{}

### Solution 10

(i) From _{}

_{}

Also, from *ABQ*

_{}

Therefore,

_{}

(ii) From _{}

_{}

Also, from *ABQ*

_{}

Therefore,

_{}

### Solution 11

Given tan x^{o} =_{}tan t^{o} = _{}and AB = 48 m;

Let length of *BC = xm*

From *ADC*

_{}

Also, from *BDC*

_{}

From (1)

_{}

Therefore, length of CD is 45 *m*.

### Solution 12

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 _{}