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

(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 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.

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 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.

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