SELINA Solutions for Class 9 Maths Chapter 24 - Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle]
Chapter 24 - Solution of Right Triangles [Simple 2-D Problems Involving One Right-angled Triangle] Exercise Ex. 24
(i)
From the figure we have
(ii)
From the figure we have
(iii)
From the figure we have
(i)
From the figure we have
(ii)
From the figure we have
(iii)
From the figure we have
The figure is drawn as follows:
The above figure we have
Again
Now
(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
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,
.
.
Consider the figure
From right triangle ACF
From triangle DEB
Given
, So
Therefore
Thus AB = AC + CD + BD = 54.64 cm.
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 .
From right triangle AQP
Also from triangle PBR
Therefore,
AB
= AP + PB =.
(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 ∆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
Consider the figure
(i) Here AB is times of BC means
(ii)
Again from the figure
Therefore, magnitude of angle A is 30
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.
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 .
(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,
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.
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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