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

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Chapter 24 - 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 5

F r o m space r i g h t space t r i a n g l e space A B C comma
tan 60 degree equals fraction numerator A C over denominator B C end fraction
rightwards double arrow square root of 3 equals fraction numerator A C over denominator 40 end fraction
rightwards double arrow A C equals 40 square root of 3 space c m
F r o m space r i g h t space t r i a n g l e space B D C comma
tan 45 degree equals fraction numerator D C over denominator B C end fraction
rightwards double arrow 1 equals fraction numerator D C over denominator 40 end fraction
rightwards double arrow D C equals 40 space c m
F r o m space t h e space f i g u r e comma space i t space i s space c l e a r space t h a t space A D equals A C minus D C
rightwards double arrow A D equals 40 square root of 3 minus 40
rightwards double arrow A D equals 40 open parentheses square root of 3 minus 1 close parentheses
rightwards double arrow A D equals 29.28 space c m

 

 

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 10

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 13


Solution 14

Consider the figure

 

 

(i) Here AB is times of BC means

(ii)

Again from the figure

Therefore, magnitude of angle A is 30degree

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  60degree 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 Syntax error from line 1 column 49 to line 1 column 72. Unexpected '<mstyle '.ABD,

tan 30degree = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

x+20 = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.AB        .....(1)

In Syntax error from line 1 column 49 to line 1 column 72. Unexpected '<mstyle '.ABC

tan 45degree = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

1 = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

AB = x       ... (2)

 

From (1)

AB + 20 = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.AB

AB(Syntax error from line 1 column 49 to line 1 column 72. Unexpected '<mstyle '.-1) = 20

ABSyntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

    = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

    = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. = 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 Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.,

tan 30degree = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

x + 20 = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. AB       ...(1)

 

In Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

tan 60degree = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

x = Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.                   ...(2 )

 

From (1)

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.


From (2)

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Therefore BC = x = 10cm

 

Therefore,

AB = 17.32cm, BC = 10cm

 

(iii)

Let  BC = xm

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

 

In Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.,

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

x + 20 = AB         ...(1)

 

In Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

 

From (1)

From (2)

Therefore,

Solution 18

(i) From

Also, from Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.ABQ

Therefore,

(ii) From

Also, from Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.ABQ

Therefore,

Solution 19

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

Let length of  BC = xm

From Syntax error from line 1 column 49 to line 1 column 72. Unexpected '<mstyle '.ADC

Also, from Syntax error from line 1 column 49 to line 1 column 72. Unexpected '<mstyle '.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