Class 9 SELINA Solutions Maths Chapter 20 - Area and Perimeter of Plane Figures

Correct option: (iii) four times

If length and corresponding altitude is doubled, then

Correct option: (iv) one-fourth

If each side is halved, then

Correct option: (iii) 6 cm

Correct option: (iii) 72 cm²

The area of a right-angled triangle is equal to half the product of the sides containing the right angle.

Therefore,

Correct option: (i) 54 cm²

Since the sides of the triangle are 18cm,24cm and 30cm respectively.

Hence area of the triangle is

Again

Hence

Let the sides of the triangle are

a=3x

b=4x

c=5x

Given that the perimeter is 144 cm.

hence

The sides are a=36 cm, b=48 cm and c=60 cm

Area of the triangle is

(i)

Area of the triangle is given by

(ii)

Again area of the triangle

Since the perimeter of the isosceles triangle is 36cm and base is 16cm. hence the length of each of equal sides are

Here

It is given that

Let 'h' be the altitude of the isosceles triangle.

Since the altitude from the vertex bisects the base perpendicularly, we can apply Pythagoras Theorem.

Thus we have,

It is given that

Let 'a' be the length of an equal side.

Hence perimeter=

From ,

Area of

Area of

Now

Given , AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and DBC = 90^o

Hence perimeter=8+10+13+5=36cm

Area of

Area of

Now

(i) 

(ii) Consider the following figure.

(iii)

Let the height of the triangle be x cm.

Equal sides are (x+4) cm.

According to Pythagoras theorem,

Hence perimeter=

Area of the isosceles triangle is given by

Here a=20cm

b=24cm

hence

Each side of the triangle is

Hence the area of the equilateral triangle is given by

The height h of the triangle is given by

The area of the triangle is given as 150sq.cm

Hence AB=15cm,AC=20cm and

Let the two sides be x cm and (x-3) cm.

Now

Hence the sides are 12cm, 9cm and

The required perimeter is 12+9+15=36cm.

Correct option: (ii)

When two diagonals of a quadrilateral cut each other at right angles, then

Correct option: (i)

Correct option: (iii) 256 cm²

Therefore,

Correct option: (iii) 324 cm²

Correct option: (iv) 36 cm

Consider the figure:

From the right triangle ABD we have

The area of right triangle ABD will be:

Again from the equilateral triangle BCD we have

Therefore the area of the triangle BCD will be:

Hence the area of the quadrilateral will be:

The figure can be drawn as follows:

Here ABD is a right triangle. So the area will be:

Again

Now BCD is an isosceles triangle and BP is perpendicular to BD, therefore

From the right triangle DPC we have

So

Hence the area of the quadrilateral will be:

Let the width be x and length 2x km.

Hence

Hence the width is 100m and length is 200m

The required area is given by

Length of the laid with grass=85-5-5=75m

Width of the laid with grass=60-5-5=50m

Hence area of the laid with grass is given by

Area of the rectangle is given by

Let h be the height of the triangle ,then

Consider the following figure.

Thus the required area = area shaded in blue + area shaded in red

= Area ABPQ +  Area TUDC +  Area A'PUD' +  Area QB'C'T

= 2Area ABPQ + 2Area QB'C'T

=2(Area ABPQ +Area QB'C'T)

Perimeter of the garden

Again, length of the garden is given to be 120 m. hence breadth of the garden b is given by

Hence area of the field

Length of the rectangle=x

Width of the rectangle=

_{Hence its perimeter is given by}

_{Again it is given that the perimeter is 4400cm.}

_Hence

 _{Length of the rectangle=1400 cm = 14 m}

(i)

Breadth of the verandah=x

Length of the verandah=x+3

According to the question

(ii)

From the above equation

Hence breadth=3m

Length =3+3=6m

Consider the following figure.

Thus, the area of the shaded portion

=Area(ABCD) + Area(EFGH) - Area(IJKL)…(1)

First we have to calculate the area of the hall.

We need to find the cost of carpeting of 80 cm = 0.8 m wide carpet, if the rate of carpeting is Rs. 25. Per metre.

Then

Let a be the length of each side of the square.

Hence

Hence

And

Consider the following figure.

(i)

The length of the lawn = 30 - 2 - 2 = 26 m

The breadth of the lawn = 12 - 2 = 10 m

(ii)

The orange shaded area in the figure is the required area.

Area of the flower bed is calculated as follows:

Number of tiles required

Fraction of floor uncovered=

Since

Hence the distance between the shorter sides is 16m.

At first we have to calculate the area of the triangle having sides, 28cm, 26cm and 30cm.

Let the area be S.

By Heron's Formula,

Area of a Parallelogram = 2 × Area of a triangle

 = 2 × 336

 = 672 cm²

We know that,

Area of a parallelogram = Height × Base

⇒ 672 = Height × 26

⇒ Height = 25.84 cm

∴ the distance between its shorter sides is 25.84 cm.

(i)

We know that

Here A=216sq.cm

AC=24cm

BD=?

Now

(ii)

Let a be the length of each side of the rhombus.

(iii)

Perimeter of the rhombus=4a=60cm.

Let a be the length of each side of the rhombus.

(i)

It is given that,

AC=24cm

We have to find BD.

Now

Hence the other diagonal is 10cm.

(ii)

Let a be the length of each side of the rhombus.

We know that,

The diagram is redrawn as follows:

Here

AF=1.2m,EF=0.3m,DC=0.6m,BK=1.8-0.6-0.3=0.9m

Hence

Here we found two geometrical figure, one is a triangle and other is the trapezium.

Now

hence area of the whole figure=150+240=630sq.m

We can divide the field into three triangles and one trapezium.

Let A,B,C be the three triangular region and D be the trapezoidal region.

Now

Area of the figure=Area of A+ Area of B+ Area of C+ Area of D

=3900+1250+312.5+2062.5

=7525sq.m

Correct option: (i) 77 cm²

In right-angled ΔABC,

Diameter of semi-circle BPC = BC = 14 cm

⇒ Radius of semi-circle BPC, r = 7 cm

Correct option: (ii) doubled

Diameter of a circle = 14 cm

If diameter is doubled, then

Correct option: (iii) 42 cm²

Radius of a quadrant OCA = Side of a square = 14 cm

Then,

Now, area of shaded portion = Area of square OCBA - Area of quadrant OCA

= (196 - 154) cm²

= 42 cm²

Correct option: (iii) four times

Let the radius of a circle = r

Then,

If radius is doubled, then

Correct option: (i) 1848 cm²

Therefore,

area of shaded portion = Area of circle with radius 28 cm - Area of circle with radius 14 cm

= (2464 - 616) cm²

= 1848 cm²

Let r be the radius of the circle.

(i)

                              = 88 cm

(ii)

         = 616 cm²

Let r be the radius of the circle.

Circumference of the first circle

Circumference of the second circle

Let r be the radius of the resulting circle.

Circumference of the first circle

Circumference of the second circle

Let r be the radius of the resulting circle.

Hencearea of the circle

Let the area of the resulting circle be r.

Hence the radius of the resulting circle is 20cm.

Area of the circle having radius 85m is

Let r be the radius of the circle whose area is 49times of the given circle.

Hence circumference of the circle

Area of the rectangle is given by

For the largest circle, the radius of the circle will be half of the sorter side of the rectangle.

Hence r=21cm.

Hence

Area of the square is given by

Since there are four identical circlesinside the square.

Hence radius of each circle is one fourth of the side of the square.

Area remaining in the cardboard is

Let the radius of the two circles be 3r and 8r respectively.

According to the question

Hence radius of the smaller circle is

Area of the smaller circle is given by

Let the diameter of the three circles be 3d, 5d and 6d respectively.

Now

Let r be the radius of the circular park.

Radius of the outer circular region including the path is given by

Area of that circular region is

Hence area of the path is given by

Time interval is 

Area covered in one 60 minutes=

Hence area swept in 35 minutes is given by

Let R and r be the radius of the big and small circles respectively.

Since AB=AC and

Now

Let x be the width of the footpath.

Then

Again it is given that area of the footpath is 360sq.m.

Hence

Hence width of the footpath is 3m.

Area of the square is 484.

Let a be the length of each side of the square.

Now

Hence length of the wire is=4x22=88m.

(i)

Now this 88m wire is bent in the form of an equilateral triangle.

(ii)

Let x be the breadth of the rectangle.

Now

Hence area=16x28=448m^₂

(i)

Again

(ii)

(iii)

For the triangle EBC,

S=19cm

Let h be the height.

(iv)

In the given figure, we can observe that the non-parallel sides are equal and hence it is an isosceles trapezium.

Therefore, let us draw DE and CF perpendiculars to AB.

Thus, the area of the parallelogram is given by

Let b be the breadth of rectangle. then its perimeter

Again

Hence its length is 20cm and width is 15cm.

Let b be the width of the rectangle.

Again perimeter of the rectangle is 104m.

Hence

Hence

length=32m

width=20m.

Let a be the length of the sides of the square.

According to the question

Hence sides of the square are 12cm each and

Length of the rectangle =2a=24cm

Width of the rectangle=a+6=18cm.

Length of the wall=45+2=47m

Breath of the wall=30+2=32m

Hence area of the inner surface of the wall is given by

Let a be the length of each side.

Hence length of the wire=96cm

(i)

For the equilateral triangle,

(ii)

Let the adjacent side of the rectangle be x and y cm.

Since the perimeter is 96 cm, we have,

Hence

Hence area of the rectangle is = 26 x 22 = 572 sq.cm

Let 'y' and 'h' be the area and the height of the first parallelogram respectively.

Let 'height' be the height of the second parallelogram

base of the first parallelogram=cm

base of second parallelogram=cm

Let the radius of the field is r meter.

Therefore circumference of the field will be: meter.

Now the cost of fencing the circular field is 52,800 at rate 240 per meter.

Therefore

Thus the radius of the field is 35 meter.

Now the area of the field will be:

Thus the cost of ploughing the field will be:

Let r and R be the radius of the two circles.

Putting the value of r in (2)

Hence the radius of the two circles is 3cm and 7cm respectively.

Since the diameter of the circle is the diagonal of the square inscribed in the circle.

Let a be the length of the sides of the square.

Hence

Hence the area of the square is 98sq.cm.

Let the sides of the triangle be

a = 12x

b = 5x

c = 13x

Given that the perimeter = 450 cm

⇒ 12x + 5x + 13x = 450

⇒ 30x = 450

⇒ x = 15

Hence, the sides of a triangle are

a = 12x = 12(15) = 180 cm

b = 5x = 5(15) = 75 cm

c = 13x = 13(15) = 195 cm

Now,

Let the sides of the triangle be

a = 26 cm, b = 28 cm and c = 30 cm

Now,

Construction: Draw CM ⊥ AB

In right-angled triangle CMB,

BM² = BC² - CM² = (15)² - (9)² = 225 - 81 = 144

⇒ BM = 12 m

Now, AB = AM + BM = 23 + 12 = 35 m

Let the sides of two squares be a and b respectively.

Then, area of one square, S₁ = a²

And, area of second square, S₂ = b²

Given, S₁ + S₂ = 400 cm²

⇒ a² + b² = 400 cm² …..(1)

Also, difference in perimeter = 16 cm

⇒ 4a - 4b = 16 cm

⇒ a - b = 4

⇒ a = (4 + b)

Substituting the value of 'a' in (1), we get

(4 + b)² + b² = 400

⇒ 16 + 8b + b² + b² = 400

⇒ 2b² + 8b - 384 = 0

⇒ b² + 4b - 192 = 0

⇒ b² + 16b - 12b - 192 = 0

⇒ b(b + 16) - 12(b + 16) = 0

⇒ (b +16)(b - 12) = 0

⇒ b + 16 = 0 or b - 12 = 0

⇒ b = -16 or b = 12

Since, the side of a square cannot be negative, we reject -16.

Thus, b = 12

⇒ a = 4 + b = 4 + 12 = 16

Hence, the sides of a square are 16 cm and 12 cm respectively.

Diagonal of a square = 24 cm

Let the radii of two circles be r₁ and r₂ respectively.

Sum of the areas of two circles = 130π sq. cm

⇒ πr₁² + πr₂² = 130π

⇒ r₁² + r₂² = 130 ….(i)

Also, distance between two radii = 14 cm

⇒ r₁ + r₂ = 14

⇒ r₁ = (14 - r₂)

Substituting the value of r₁ in (i), we get

(14 - r₂)² + r₂² = 130

⇒ 196 - 28r₂ + r₂² + r₂² = 130

⇒ 2r₂² - 28r₂ + 66 = 0

⇒ r₂² - 14r₂ + 33 = 0

⇒ r₂² - 11r₂ - 3r₂ + 33 = 0

⇒ r₂ (r₂ - 11) - 3 (r₂ - 11) = 0

⇒ (r₂ - 11) (r₂ - 3) = 0

⇒ r₂ = 11 or r₂ = 3

⇒ r₁ = 14 - 11 = 3 or r₁ = 14 - 3 = 11

Thus, the radii of two circles are 11 cm and 3 cm respectively.