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 5
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, m∠AEB = 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, m∠AEC = 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 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,
Solution 18
(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
From ADC
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 
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