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

Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.

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

Fin 'x' if:

(i)

(ii)

(iii)

(i)

From the figure we have

_{}

(ii)

From the figure we have

_{}

(iii)

From the figure we have

_{}

Find angle 'A' if:

(i)

(ii)

(iii)

(i)

From the figure we have

_{}

(ii)

From the figure we have

_{}

(iii)

From the figure we have

_{}

Find angle 'x' if:

The figure is drawn as follows:

The above figure we have

_{}

Again

_{}

Now

_{}

Find AD, if:

(i)

(ii)

(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

Find the length of AD.

Given: _{}ABC = 60^{o}.

_{}DBC = 45^{o}

And BC = 40 cm.

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

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,

_{}.

_{}.

Find AB.

Consider the figure

From right triangle ACF

_{}

From triangle DEB

_{}

Given
_{}, So _{}

Therefore

_{}

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

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.

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 _{}.

Use the information given to find the length of AB.

From right triangle AQP

_{}

Also from triangle PBR

_{}

Therefore,

AB
= AP + PB =_{}.

Find the length of AB.

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

(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

_{}

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

Calculate:

(i) BE (ii) AC

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

Find

(i) BC

(ii) AD

(iii) AC

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

(i) AB is _{}times of BC.

(ii) BC is _{}times of AB.

Consider the figure

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

_{}

(ii)

Again from the figure

_{}

Therefore, magnitude of angle A is 30

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.

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.

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.

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 _{}.

Find AB and BC, if:

(i)

(ii)

(iii)

(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,

_{}

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

(i)

(ii)

(i) From _{}

_{}

Also, from *ABQ*

_{}

Therefore,

_{}

(ii) From _{}

_{}

Also, from *ABQ*

_{}

Therefore,

_{}

If tan x^{o} =_{},

tan y^{o} = _{}and AB = 48 m; find the length of CD.

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*.

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

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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