Concise Class-9 Area and Perimeter of Plane Figure ICSE Maths Selina Solutions

Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions Chapter-20. We provide step by step Solutions of Exercise / lesson-20 Area and Perimeter of Plane Figure for ICSE Class-9th Concise Selina Mathematics by R K Bansal.

Our Solutions contain all type Questions with Exe-20 A, Exe-20 B , Exe-20 C and Exe-20 D to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9th Mathematics .

Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions Chapter-20

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-: Select Topics :-

Exe-20 A,

Exe-20 B,

 Exe-20 C,

 Exe-20 D,

 

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Exe-20 A, Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions

Question 1

Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm.

Answer 1

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

s = $\frac{18+24+30}{2}$
= 36

Hence the area of the triangle is

= 216 sq cm

again

A =1/2 base x height

hence

216 = 1/2 base x height

216 = 1/2 x 30 x h

h = 14.4 cm

Question 2

The length of the sides of a triangle are in the ratio 3: 4: 5. Find the area of the triangle if its perimeter is 144 cm.

Answer 2

Let the sides of the triangle are

a=3x

b=4x

c=5x

Given that the perimeter is 144 cm.

hence

⇒ 3x + 4x + 5x =144

⇒ x = 144/ 12

s = (a+b+c )/2

⇒ 12x/2

⇒ 6x =72

⇒ x = 12

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

Area of the triangle is

= 864 sq cm

Question 3

ABC is a triangle in which AB = AC = 4 cm and A = 90^o. Calculate:

(i) The area of ABC,

(ii) The length of perpendicular from A to BC.

Answer 3

(i)

Area of the triangle is given by

(ii)

Again area of the triangle

Question 4

The area of an equilateral triangle is    sq. cm. Find its perimeter.

Solution 4

Question 5

Find the area of an isosceles triangle with perimeter is 36 cm and base is 16 cm.

Solution 5

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,

Question 6

The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.

Solution 6

It is given that

Let ‘a’ be the length of an equal side.

Hence perimeter=

Question 7

The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.

Solution 7

From ,

Area of

Area of

Now

Question 8

Find the area and the perimeter of quadrilateral ABCD, given below; if AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and DBC = 90^o

^{Solution 8}

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

Question 9

The base of a triangular field is three times its height. If the cost of cultivating the field at 36.72 per 100 m² is 49,572; find its base and height.

Solution 9

Question 10

The sides of a triangular field are in the ratio 5 : 3 : 4 and its perimeter is 180 m. Find:

(i) its area.

(ii) altitude of the triangle corresponding to its largest side.

(iii) the cost of leveling the field at the rate of Rs. 10 per square metre.

Question 14

If the difference between the sides of a right angled triangle is 3 cm and its area is 54 cm²; find its perimeter.

Solution 14

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.

Question 15

AD is altitude of an isosceles triangle ABC in which AB = AC = 30 cm and BC = 36 cm. A point O is marked on AD in such a way that BOC = 90^o. Find the area of quadrilateral ABOC.

Now

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Concise Class-9 Area and Perimeter of Plane Figure ICSE Mathematics Solutions  Ex. 20(B)

Question 1

Find the area of a quadrilateral one of whose diagonals is 30 cm long and the perpendiculars from the other two vertices are 19 cm and 11 cm respectively.

Solution 1

Question 2

The diagonals of a quadrilateral are 16 cm and 13 cm. If they intersect each other at right angles; find the area of the quadrilateral.

Question 3

Calculate the area of quadrilateral ABCD, in which ABD = 90^o, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.

Solution 3

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:

Question 4

Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm A = 90^o and BC = CD = 52 cm.

Solution 4

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:

Question 5

The perimeter of a rectangular field is 3 / 5 km. If the length of the field is twice its width; find the area of the rectangle in sq. metres.

Solution 5

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

Question 6

A rectangular plot 85 m long and 60 m broad is to be covered with grass leaving 5 m all around. Find the area to be laid with grass.

Solution 6

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

Question 7

The length and the breadth of a rectangle are 6 cm and 4 cm respectively. Find the height of a triangle whose base is 6 cm and area is 3 times that of the rectangle.

Solution 7

Area of the rectangle is given by

Let h be the height of the triangle ,then

Question 8

How many tiles, each of area 400 cm², will be needed to pave a footpath which is 2 m wide and surrounds a grass plot 25 m long and 13 m wide?

Solution 8

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)

Question 9

The cost of enclosing a rectangular garden with a fence all round, at the rate of 75 paise per metre, is Rs. 300. If the length of the garden is 120 metres, find the area of the field in square metres.

Solution 9

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

Question 10

The width of a rectangular room is 4 / 7 of its length, x, and its perimeter is y. Write an equation connecting x and y. Find the length of the room when the perimeter is 4400 cm.

Solution 10

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}

Question 11

The length of a rectangular verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.

(i) Taking x as the breadth of the verandah, write an equation in x that represents the above statement.

(ii) Solve the equation obtained in (i) above and hence find the dimensions of the verandah.

Solution 11

(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

Question 12

The diagram, given below, shows two paths drawn inside a rectangular field 80 m long and 45 m wide. The widths of the two paths are 8 m and 15 m as shown. Find the area of the shaded portion.

Solution 12

Consider the following figure.

Thus, the area of the shaded portion

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

Question 13

The rate for a 1.20 m wide carpet is Rs. 40 per metre; find the cost of covering a hall 45 m long and 32 m wide with this carpet. Also, find the cost of carpeting the same hall if the carpet, 80 wide, is at Rs. 25. Per metre.

Solution 13

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

Question 14

Find the area and perimeter of a square plot of land, the length of whose diagonal is 15 metres. Given your answer correct to 2 places of decimals.

Solution 14

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

Hence

Hence

And

Question 15

The shaded region of the given diagram represents the lawn in the form of a house. On the three sides of the lawn there are flowerbeds having a uniform width of 2 m.

(i) Find the length and the breadth of the lawn.

(ii) Hence, or otherwise, find the area of the flower-beds.

Question 15

The shaded region of the given diagram represents the lawn in the form of a house. On the three sides of the lawn there are flowerbeds having a uniform width of 2 m.

(i) Find the length and the breadth of the lawn.

(ii) Hence, or otherwise, find the area of the flower-beds.

Solution 15

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:

Question 16

A floor which measures 15 m 8 m is to be laid with tiles measuring 50 m 25 cm. Find the number of tiles required.

Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?

Solution 16

Number of tiles required

Fraction of floor uncovered=

Question 17

Two adjacent sides of parallelogram are 24 cm and 18 cm. If the distance between the longer sides is 12 cm; find the distance between the shorter sides.

Solution 18

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.

Question 19

The area of a rhombus is 216 sq. cm. If its one diagonals is 24 cm; find:

(i) Length of its other diagonal,

(ii) Length of its side,

(iii) Perimeter of the rhombus.

Solution 19

(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.

Question 20

The perimeter of a rhombus is 52 cm. If one diagonal is 24 cm; find:

(i) The length of its other diagonal,

(ii) Its area.

Solution 20

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)

Question 21

The perimeter of a rhombus is 46 cm. If the height of the rhombus is 8 cm; find its area.

Solution 21

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

We know that,

Question 22

The figure given below shows the cross-section of a concrete structure. Calculate the area of cross-section if AB = 1.8 cm, CD = 0.6 m, DE = 0.8 m, EF = 0.3 m and AF = 1.2 m.

Solution 22

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

Question 23

Calculate the area of the figure given below: which is not drawn scale.

Solution 23

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

Question 24

The following diagram shows a pentagonal field ABCDE in which the lengths of AF, FG, GH and HD are 50 m, 40 m, 15 m and 25 m respectively; and the lengths of perpendiculars BF, CH and EG are 50 m, 25 m ad 60 m respectively. Determine the area of the field.

Solution 24

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

=7525 sq.m

Question 25

A foot path of uniform width runs all around the outside of a rectangular field 30 m long and 24 m wide. If the path occupies an area of 360 m², find its width.

Solution 25

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.

Question 26

A wire when bent in the form of a square encloses an area of 484 m². Find the largest area enclosed by the same wire when bent to from:

(i) An equilateral triangle.

(ii) A rectangle of length 16 m.

Solution 26

Area of the square is 484.

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

Now

Hence length of the wire is=4×22=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=16×28=448m_²

Question 27

For each trapezium given below; find its area.

(i)

(ii)

(iii)

(iv)

Solution 27

(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

Question 29

The area of a rectangular is 640 m². Taking its length as x cm; find in terms of x, the width of the rectangle. If the perimeter of the rectangle is 104 m; find its dimensions.

Solution 29

Let b be the width of the rectangle.

Again perimeter of the rectangle is 104m.

Hence

Hence

length=32m

width=20m.

Question 30

The length of a rectangle is twice the side of a square and its width is 6 cm greater than the side of the square. If area of the rectangle is three times the area of the square; find the dimensions of each.

Solution 30

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.

Question 31

ABCD is a square with each side 12 cm. P is a point on BC such that area of _{ABP: area of trapezium}APCD = 1: 5. Find the length of CP.

Question 32

A rectangular plot of land measures 45 m 30 m. A boundary wall of height 2.4 m is built all around the plot at a distance of 1 m from the plot. Find the area of the inner surface of the boundary wall.

Solution 32

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

Question 33

A wire when bent in the form of a square encloses an area = 576 cm². Find the largest area enclosed by the same wire when bent to form;

(i) an equilateral triangle.

(ii) A rectangle whose adjacent sides differ by 4 cm.

Solution 33

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

Question 34

The area of a parallelogram is y cm² and its height is h cm. The base of another parallelogram is x cm more than the base of the first parallelogram and its area is twice the area of the first. Find, in terms of y, h and x, the expression for the height of the second parallelogram.

Solution 34

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

Question 35

The distance between parallel sides of a trapezium is 15 cm and the length of the line segment joining the mid-points of its non-parallel sides is 26 cm. Find the area of the trapezium.

Question 36

The diagonal of a rectangular plot is 34 m and its perimeter is 92 m. Find its area.

Concise Class-9 Exe-20 C Area and Perimeter of Plane Figure ICSE Mathematics Solutions

Question 1

The diameter of a circle is 28 cm. Find its:

(i) Circumference

(ii) Area.

Solution 1

Let r be the radius of the circle.

(i)

= 88 cm

(ii)

= 616 cm²

Question 2

The circumference of a circular field is 308 m. Find is:

(i) Radius

(ii) Area.

Question 10

Two circles touch each other externally. The sum of their areas is 74π cm² and the distance between their centres is 12 cm. Find the diameters of the circle.

Solution 10

Question 11

If a square in inscribed in a circle, find the ratio of the areas of the circle and the square.

— End of Concise Class-9 Area and Perimeter of Plane Figure Selina Solutions :–

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