ICSE Solutions for Selina Concise Chapter 20 Area and Perimeter of Plane Figure Class 9 Maths
Exercise 20(A)
1. Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm.
AnswerSince the sides of the triangle are 18 cm, 24 cm and 30 cm respectively.
s = 18+24+30218+24+302 = 36
Hence the area of the triangle is
= 216 sq. cm
Again,
A =1/2 base x height
⇒ 216 = 1/2 base ×height
⇒ 216 = 1/2 ×30×h
⇒ h = 14.4 cm
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.
AnswerLet 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
3. ABC is a triangle in which AB = AC = 4 cm and A = 90o. Calculate:
(i) The area of ABC,
(ii) The length of perpendicular from A to BC.
Answer
(i) Area of the triangle is given by
A = (1/2)×AB×AC
= (1/2)×4×4
= 8 sq.cm
(ii) Again area of the triangle
A = (1/2)×BC× h
⇒ h = 2.83 cm
4. The area of an equilateral triangle is IMAGE sq. cm. Find its perimeter.
Answer
Area of an equilateral triangle is given by
side = 12cm
Hence,
Perimeter = 3×(its side)
= 3×12
= 36 cm
5. Find the area of an isosceles triangle with perimeter is 36 cm and base is 16 cm.
Answer
Since the perimeter of the isosceles triangle is 36cm and base is 16cm. hence the length of each of equal sides are
(36-16)/2 = 10 cm
Here,
It is given that
a = equal sides = 10cm
b = base = 16cm
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,
6. The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.
Answer
It is given that
Area = 192 sq. cm
base = 24cm
Knowing the length of equal side, a , and base, b, of an isosceles triangle,
the area can be calculated using the formula,
⇒ a = 20cm
Hence, perimeter = 20 + 20 +24 = 64 cm
7. The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.
\Answer
From ,
In △ABC,
Area of △ABC
△ABC = (1/2)×8√192
= 4√192
Area of △BCD
△BCD = (√3/4)×82
= 16√3
Now
△ABD = △ABC - △BDC
= 4√192 - 16√3
= 32√3 - 16√3
= 16√3 sq.cm
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 = 90o
Answer
Given , AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and DBC = 90o
Hence perimeter=8+10+13+5=36cm
Area of △ABD
Area of △DBC
△DBC = (1/2)×12×5 = 30
Now,
Area of ABCD = area of △ABD + area of △BDC
= 39.7 +30
= 69.7 sq.cm
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.
Answer
Area of the rectangular field = 49572/36.72 = 135000
Let the height of the triangle be x
135000 = 1/2× x ×3x
⇒ x2 = 90000
⇒ x = 300
Height = 300 m
Base = 900 m
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.
Thus, we have, 5x + 4x + 3x = 180
⇒ 12x = 180
⇒ x = 180/12
⇒ x =15
Thus, the sides are 5×15, 3×15 and 4×15.
That is the sides are 75m, 45m and 60m.
Since the sides are in the ratio, 5:3:5, it is a Pythagorean triplet.
Therefore, the triangle is a right angled triangle.
Area of a right angled triangle = (1/2)×base×altitude
⇒ (1/2)×45×60
⇒ 45 ×30 = 1350 m2
(ii) Consider the following figure.
In the above figure,
The largest side is AC = 75m.
The altitude corresponding to AC is BD.
We need to find the value of BD.
Now consider the triangles △BCD and △BAD.
we have,
∠B =∠B [common]
BD = BD [common]
∠D = ∠D = 90°
Similar triangles have similar proportionality.
Thus, we have,
CD/BD = BD/AD
⇒ BD2 = AD × CD ...(1)
From subpart (i), the sides of the triangle are AC = 75m, AB = 60m and BC =45 m
Let AD = x m
Thus applying Pythagoras theorem, from right triangle △BCD, we have
452 = (75-x)2 + BD2
⇒ BD2 = 452 -(75-x)2
⇒ BD2 = 2025 - (5625 +x2 - 150x)
⇒ BD2 = 2025 - 5625 - x2 + 150x
⇒ BD2 = -3600 - x2 + 150x ...(2)
Now applying Pythagoras Theorem, from right triangle △ABD, we have
= 602 = x2 + BD2
⇒ BD2 = 602 - x2
⇒ BD2 = 3600 - x2 ...(3)
From equations (2) and (3), we have,
-3600 - x2 + 150x = 3600 -x2
⇒ 150x = 3600 + 3600
⇒ 150x = 7200
⇒ x = 7200/150
⇒ x = 48
Thus, AD = 48 and CD = 75 - 48 = 27
Substituting the values AD = 48m
and CD = 27 m in equation (1) , we have
BD2 = 48×27
⇒ BD2 = 1296
⇒ BD = 36m
The area of the triangular field from subpart (i) is 1350 m2 .
The cost of levelling the field is Rs. 10 per square metre.
Thus, the total cost of levelling the field = 1350×10 = Rs. 13,500.
11. Each of equal sides of an isosceles triangle is 4 cm greater than its height. IF the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.
Answer
Let the height of the triangle be x cm.
Equal sides are (x+4) cm.
According to Pythagoras theorem,
(x+4)2 = x2 + 122
⇒ 8x = 128
⇒ x = 16cm
Hence, perimeter = 20+20+24 = 64cm
Area of the isosceles triangle is given by
Here a = 20cm
b = 24cm
Hence,
12. Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm.
Answer
Each side of the triangle is 60/3 = 20 cm
Hence the area of the equilateral triangle is given by
A = (√3/4) ×202
= 100√3
= 173.2sq.cm
⇒ h = 17.32cm.
13. In triangle ABC; angle A = 90o, side AB = x cm, AC = (x + 5) cm and area = 150 cm2. Find the sides of the triangle.
Answer
The area of the triangle is given as 150sq.cm
(1/2)×x×(x+5)= 150
⇒ x2 + 5x - 300 = 0
⇒ (x+20)(x-15) = 0
⇒ x = 15
Hence, AB = 15cm, AC = 20cm and
14. If the difference between the sides of a right angled triangle is 3 cm and its area is 54 cm2; find its perimeter.
Answer
Let the two sides be x cm and (x-3) cm.
Now
(1/2)×x×(x-3) = 54
⇒ x2 - 3x - 108 = 0
⇒ (x-12)(x+9) = 0
⇒ x = 12cm
Hence the sides are 12cm, 9cm
and
The required perimeter is 12+9+15=36cm.
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 = 90o. Find the area of quadrilateral ABOC.
= (1/4)×36×48
= 432
∠BOD = ∠COD = 45°
Hence, ∠OBD = 45° and OD = BD = 18cm
Now,
Area of △BOC = (1/2)×36×18
= 324
Area of ABOC = Area of △ABC - Area of △BOC
= 432 - 324
= 108 sq.cm
Exercise 20(B)
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.
Answer
Area = (1/2)× one diagonal × sum of the lengths of the perpendiculars draw from it on the remaining two vertices.
= (1/2)×30×(11+19)
= 450 sq.cm
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.
= (1/2)×16×13
= 104cm2
3. Calculate the area of quadrilateral ABCD, in which ABD = 90o, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.
Answer
Consider the figure:
From the right triangle ABD we have
= 2(5)
= 10
The area of right triangle ABD will be:
△ABD = (1/2)(AB)(BD)
= (1/2)×10×24
= 120
Again from the equilateral triangle BCD we have CP ⊥ BD
Therefore, the area of the triangle BCD will be:
△BCD = (1/2)(BD)(PC)
= (1/2)×24×(12√3)
= 144√3
Hence, the area of the quadrilateral will be:
△ABD + △BCD = 120+ 144√3
= 369.41 cm2
4. Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm A = 90o and BC = CD = 52 cm.
Answer
The figure can be drawn as follows:
Here ABD is a right triangle. So the area will be.
△ABD = (1/2)×24×32
= 384
Again
= 8×5
= 40
Now BCD is an isosceles triangle and BP is perpendicular to BD, therefore
DP = (1/2)×BD
= (1/2)×40
= 20
From the right triangle DPC we have
= 4×12
= 48
So,
= 960
Hence the area of the quadrilateral will be :
△ABD + △DPC = 960 + 384
= 1344 cm2
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.
Answer
Let the width be x and length 2x km.
Hence,
2(x+2x) = 3/5
⇒ x = 1/10 km = 100m
Hence, the width is 100m and length is 200m
The required area is given by
A = length ×width
= 100×200
= 20,000 sq.m
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.
Answer
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
A = 75×50 = 3750 sq. m
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.
Answer
Area of the rectangle is given by
A = l ×b
= 6×4
= 24 sq. m
Let h be the height of the triangle ,then
(1/2)×base ×h = 3A
⇒ (1/2)×6×h = 3×24
⇒ h = 24cm
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
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)
Area of the footpath is given by
A = 2×(25+25+17+17)
= 168 sq. m
= 1680000 sq.cm
Hence number of tiles required = 1680000/400 = 4200.
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.
Answer
Perimeter of the garden
s = 300/0.75
= 400 sq.m
Again, length of the garden is given to be 120 m. hence breadth of the garden b is given by
2(l+ b) = S
⇒ 2(120 + b) = 400
⇒ b =80 m
Hence, area of the field
A = 120×80
= 9600 sq. m
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.
Answer
Length of the rectangle = x
Width of the rectangle = 4/7 x
Hence its perimeter is given by2[x + (4/7)x] = y
⇒ 2(11x/7) = y
⇒ 22x/7 = y
Again it is given that the perimeter is 4400cm.
Hence
22x/7 = 4400
⇒ x = 1400
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.
Answer
(i) Breadth of the verandah=x
Length of the verandah=x+3
According to the question
2(x+ (x+3))= x (x+3)
⇒ 4x + 6 = x2 + 3x
⇒ x2 - x - 6 = 0
(ii) From the above equation
x2 - x -6= 0
⇒ (x-3)(x+2) = 0
⇒ x= 3
Hence,
breadth=3m
Length =3+3=6m
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.
Answer
Consider the following figure.
Thus, the area of the shaded portion
= Area(ABCD) + Area(EFGH) – Area(IJKL) …(1)
Dimensions of ABCD: 45 m ×15m
Thus, the area of ABCD = 45×15 = 675 m2
Dimensions of EFGH : 80m ×8m
Thus, the area of EFGH = 80×8 = 640 m2
Dimensions of IJKL : 15 m ×8m
Thus, the area of IJKL = 80×8 = 120 m2
Therefore, from equation (1),
The area of the shaded portion = 675 + 640 - 120 = 1195 m2 .
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.
Answer
First we have to calculate the area of the hall.
Area = 45×32
= 1440 m2
cost = (40/1.20)×1440
= 48, 000
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
Cost = (25/0.8)×1440 = Rs. 45,000
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.
Answer
Let a be the length of each side of the square.
Hence
2a2 = (diagonal)2
⇒ a2 = 152/2
⇒ a2 = 112.5
⇒ a = 10.60
Hence,
Area = a2
= 112.5 sq.m
And
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.
Answer
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:
A = 10×2+10×2+30×2
= 20+20+60 = 100 sq. m
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?
Answer
Area of the floor = 15×8 =120sq.m
Area of one tiles = 0.50×0.25 = 0.125 sq. m
Number of tiles required
n = (Area of floor)/(Area of tiles)
= 120/0.125
= 960
Area of carpet uncovered = 2(1×15 + 1×6)= 42 sq.m
Fraction of floor uncovered= 42/120 = 7/20
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
We know that,
Area = Base × Height
= 24×12
⇒ 18×h = 24×12
⇒ h = (24×12)/18
⇒ h = 16
18. Two adjacent sides of parallelogram are 28cm and 26cm. If one diagonal of it is 30cm long : Find the area of the parallelogram. Also, find the distance between its shorter sides.
Solution
At first we have to calculate the area of the triangle having sides, 28cm, 26cm and 30cm.
Let the area be S.
S = (28+26+ 30)/2
= 84/2
= 42 cm
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.
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.
Answer
(i) We know that
Area of Rhombus = (1/2)×AC ×BD
Here,
A = 216sq.cm
AC = 24cm
BD = ?
Now
A = (1/2)×AC ×BD
⇒ 216 = (1/2)×24×BD
⇒ BD = 18cm
(ii) Let a be the length of each side of the rhombus.
a2 = (AC/2)2 + (BD/2)2
⇒ a2 = 122 + 92
⇒ a2 = 225
⇒ a = 15cm
(iii) Perimeter of the rhombus=4a=60cm.
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.
Answer
Let a be the length of each side of the rhombus.
4a = perimeter
⇒ 4a = 52
⇒ a = 13cm
(i) It is given that,
AC=24cm
We have to find BD.
Now
a2 = (AC/2)2 + (BD/2)2
⇒ 132 = 122 + (BD/2)2
⇒ (BD/2)2 = 52
⇒ BD = 10cm
Hence, the other diagonal is 10cm.
(ii) Area of the rhombus = (1/2)×AC×BD
= (1/2)×24×10
= 120 sq.cm
21. The perimeter of a rhombus is 46 cm. If the height of the rhombus is 8 cm; find its area.
Answer
Let a be the length of each side of the rhombus.
4a = perimeter
⇒ 4a = 46
⇒ a = 11.5cm
We know that,
Area = Base ×Height
= 11.5×8
= 92 sq.cm
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.
Answer
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,
Area of ABCDEF = Area of AHEF + Area of HKCD + ΔKBC
= 1.2×0.3 + 2×0.6 + (1/2)×2×0.9
= 2.46 sq. m
23. Calculate the area of the figure given below: which is not drawn scale.
Answer
Here we found two geometrical figure, one is a triangle and other is the trapezium.
Now
Area of the triangle = (1/2)×12×25
= 150 sq. m
= 20×24
= 480 sq. m
Hence, area of the whole figure=150+240=630sq.m
Answer
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 A = (1/2)×AD×GE
= (1/2)×(50+40+15+25)×60
= 3900 sq. m
Area of B = (1/2)×AF×BF
= (1/2)×50×50
= 1250 sq. m
Area of B = (1/2)×HD ×CH
= (1/2)×25×25
= 312.5 sq. m
Area of D = (1/2)×(BF + CH)×(FG + GH)
= (1/2)×(50+25)×(40+15)
= (1/2)×75×55
= 2062.5 sq. m
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
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.
Answer
Let x be the width of the footpath.
Then
Area of footpath = 2×(30+24)x + 4x2 .
= 4x2 + 108x
Again it is given that area of the footpath is 360sq.m.
Hence,
4x2 + 108x = 360
⇒ x2 + 27x - 90 = 0
⇒ (x-3)(x+30) = 0
⇒ x = 3
Hence, width of the footpath is 3m.
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.
Answer
Area of the square is 484.
Let a be the length of each side of the square.
Now
a2 = 484
⇒ a = 22m
Hence length of the wire is=4×22=88m.
(i) Now this 88m wire is bent in the form of an equilateral triangle.
Side of the triangle = 88/3
= 29.3 m
Area of the triangle = (√3/4)×(side)2
= (√3/4)×(29.3)2
= 372.58 m2 .
(ii) Let x be the breadth of the rectangle.
Now
2(l+ b) = 88
⇒ 16 + x = 44
⇒ x = 28 m
Hence, area=16×28=448m2 .
27. For each trapezium given below; find its area.
(i)
(ii)
(iii)
(iv)
Answer
(i)
= (1/4)×8×18.3
= 36.6 cm2
Again,
Area of △ EBC = (1/2)×8×h
36.6 =4h
⇒ h =9.15
Area of ABCD = (1/2)×(12+20)×9.15
= (1/2)×32×9.15
= 146.64 sq.cm
(ii)
= (1/2)×22×8
= 88 sq. cm
(iii)
For the triangle EBC,
S = 19cm
Let h be the height.
Area of △EBC = (1/2)×12×h
⇒ 59.9 = 6h
⇒ h = 59.9/6 = 9.98cm
Area of ABCD = (1/2)×(20+32)×9.98
= (1/2)×52×9.98
= 259.48 cm2
(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
Since AB = AE + EF + FB and CD = EF = 18cm, we have
30 = AE +18+ FB
⇒ 2AE + 18 = 30
⇒ 2AE = 30 - 18
⇒ 2AE = 12
⇒ AE = 6 cm
Now, consider the right angled triangle ADE.
AD2 = AE2 + DE2
⇒ 122 = 62 + DE2
⇒ 144 = 36 + DE2
⇒ DE2 = 144 - 36
⇒ DE2 = 108
⇒ DE = √(36×3)
⇒ DE = 6√3
Area(◻️ABCD) = Area (△ADE) + Area (◻️DEFC) + Area(△CFB)
⇒ Area(◻️ABCD) = (1/2)×6×6√3 + 18×6√3 + (1/2)×6×6√3
⇒ Area(◻️ABCD) = 6×6√3 + 18×6√3
⇒ Area(◻️ABCD) = 144√3 = 249.41 cm2 .
If the area of the rectangular board is 300 cm2; find its dimensions.
Let b be the breadth of rectangle. then its perimeter
2(x+ b) = 70
⇒ x + b = 35
⇒ b = 35 - x
Again,
x × b = 300
⇒ x(35-x) = 300
⇒ x2 - 35x + 300 = 0
⇒ (x-15)(x-20) = 0
⇒ x =15, 20
Hence, its length is 20cm and width is 15cm.
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.
Answer
Let b be the width of the rectangle.
x × b = 640
⇒ b = 640/x
Again perimeter of the rectangle is 104m.
Hence
2[x + (640/x)] = 104
⇒ x2 - 52x + 640 = 0
⇒ (x-32)(x-20) = 0
⇒ x = 32, 20
Hence,
length= 32m
width= 20m.
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.
Answer
Let a be the length of the sides of the square.
According to the question
2a×(a+6) = 3a2
⇒ 2a2 + 12a = 3a2
⇒ a = 12
Hence, sides of the square are 12cm each and
Length of the rectangle = 2a = 24cm
Width of the rectangle=a + 6=18cm.
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.
Answer
The figure is shown below :
(Area of ΔABP)/(Area of trapezium APCD) = 1/5
⇒ 60 - 5CP = 12 + CP
⇒ 6CP = 48
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.
Answer
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
A = (47×2×2.4) + (32×2×2.4)
= 225.6 + 153.6
= 379.2 m2
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.
Answer
Let a be the length of each side.
a2 = 576
⇒ a = 24cm
⇒ 4a = 96cm
Hence length of the wire=96cm
(i) For the equilateral triangle,
side = 96/3 = 32cm
Area = (√3/4)(side)2
= (√3/4)×322
= 256√3 sq. m
(ii) Let the adjacent side of the rectangle be x and y cm.
Since the perimeter is 96 cm, we have,
2(x+ y) = 96
Hence,
x+ y = 48
x- y = 4
⇒ x = 26
⇒ y = 22
Hence, area of the rectangle is = 26×22 = 572 sq.cm
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.
Answer
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= y/h cm
base of second parallelogram= [(y/h) +x] cm
⇒ [(y/h) +x]×height = 2y
⇒ height = 2hy/(y+hx)
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.
EF = (1/2)×(AD + BC) = 26cm
Area of the trapezium = (1/2)×(AD + BC) ×h
= 26×15
= 390 cm2
36. The diagonal of a rectangular plot is 34 m and its perimeter is 92 m. Find its area.
Since the perimeter is 92 m, we have,
2(a+ b) = 92
⇒ a+ b = 46 m ....(1)
⇒ a2 + b2 = 342 ...(2)
we know that
(a+ b)2 - a2 - b2 = 2ab
From equations (1) and (2), we have,
462 - 342 = 2ab
⇒ 2ab = 960
⇒ ab = 960/2
⇒ ab = 480 m2
Exercise 20(C)
1. The diameter of a circle is 28 cm. Find its:
(i) Circumference
(ii) Area.
Answer
Let r be the radius of the circle.
(i) 2r = 28cm
circumference = 2Ï€r
= 28Ï€ cm
= 88 cm
(ii) area = πr2
= π(28/2)2
= 196Ï€cm2
= 616 cm2
2. The circumference of a circular field is 308 m. Find is:
(i) Radius
(ii) Area.
(i) 2Ï€r = 308
⇒ r = 308/2Ï€
⇒ r = (308/2)×(7/22)
⇒ r = 49m
= (22/7)×(49)2
= 7546 m2
3. The sum of the circumference and diameter of a circle is 116 cm. Find its radius.
Answer
Let r be the radius of the circle.
2Ï€r + 2r = 116
⇒ 2r(Ï€+1) = 116
⇒ r = 116/2(Ï€+1) = 14 cm
4. The radii of two circles are 25 cm and 18 cm. Find the radius of the circle which has circumference equal to the sum of circumferences of these two circles.
Answer
Circumference of the first circle
S1 = 2Ï€×25 = 50Ï€ cm
Circumference of the second circle
S2 = 2Ï€×18 = 36Ï€ cm
Let r be the radius of the resulting circle.
2Ï€r = 50Ï€ + 36Ï€
⇒ 2Ï€r = 86Ï€
⇒ r = 86Ï€/2Ï€ = 43 cm
5. The radii of two circles are 48 cm and 13 cm. Find the area of the circle which has its circumference equal to the difference of the circumferences of the given two circles.
Answer
Circumference of the first circle
S1 = 2Ï€ ×48 = 96Ï€ cm
Circumference of the second circle
S2 = 2Ï€ ×13 = 26Ï€ cm
Let r be the radius of the resulting circle.
2Ï€r = 96Ï€ - 26Ï€
⇒ 2Ï€r = 70Ï€
⇒ r = 70Ï€/2Ï€ = 35 cm
Hence, area of the circle
A = πr2
= Ï€×352
= 3850 cm2
6. The diameters of two circles are 32 cm and 24 cm. Find the radius of the circle having its area equal to sum of the areas of the two given circle.
Answer
Let the area of the resulting circle be r.
Ï€×(16)2 + Ï€×(12)2 = Ï€ ×r2
⇒ 256Ï€ + 144Ï€ = Ï€×r2
⇒ 400Ï€ = Ï€×r2
⇒ r2 = 400
⇒ r = 20cm
Hence, the radius of the resulting circle is 20cm.
7. The radius of a circle is 5 m. Find the circumference of the circle whose area is 49 times the area of the given circle.
Answer
Area of the circle having radius 85m is
A = Ï€×(85)2 = 7225Ï€m2
Let r be the radius of the circle whose area is 49times of the given circle.
Ï€r2 = 49×(Ï€×52 )
⇒ r2 = (7×5)2
⇒ r = 35
Hence, circumference of the circle
S = 2Ï€r
= 2Ï€ ×35
= 220 m
8. A circle of largest area is cut from a rectangular piece of card-board with dimensions 55 cm and 42 cm. Find the ratio between the area of the circle cut and the area of the remaining card-board.
Answer
Area of the rectangle is given by
A = 55×42 = 2310 cm2
For the largest circle, the radius of the circle will be half of the sorter side of the rectangle.
Hence, r = 21cm.
Area of the circle = Ï€×(21)2 = 1384.74 cm2
Area remaining = 2310 - 1384.74 = 925 .26
Hence,
the volume of the circle area remaining = 1384.74 : 915.26
= 3 : 2
9.The following figure shows a square cardboard ABCD of side 28 cm. Four identical circles of largest possible size are cut from this card as shown below.
Find the area of the remaining card-board.
Answer
Area of the square is given by
A = 282 = 784 cm2
Since there are four identical circles inside the square.
Hence radius of each circle is one fourth of the side of the square.
Area of one circle = Ï€× 72 = 154 cm2
Area of four circle = 4×154 cm2 = 616 cm2
Area remaining = 784 - 616 = 168 cm2
Area remaining in the cardboard is = 168 cm2
10. The radii of two circles are in the ratio 3 : 8. If the difference between their areas is 2695 cm2, find the area of the smaller circle.
Answer
Let the radius of the two circles be 3r and 8r respectively.
area of the circle having radius 3r = π(3r)2
= 9Ï€r2
area of the circle having radius 8r = π(8r)2
= 64 πr2
According to the question
64Ï€r2 - 9Ï€r2 = 2695Ï€
⇒ 55r2 = 2695
⇒ r2 = 49
⇒ r =7 cm
Hence radius of the smaller circle is
Area of the smaller circle is given by
A = Ï€r2 = (22/7)×212 = 1386 cm2
11. The diameters of three circles are in the ratio 3 : 5 : 6. If the sum of the circumferences of these circles be 308 cm; find the difference between the areas of the largest and the smallest of these circles.
Answer
Let the diameter of the three circles be 3d, 5d and 6d respectively.
Ï€×3d + Ï€×5d + Ï€×6d = 308
⇒ 14Ï€d = 308
⇒ d = 7
radius of the smallest circle = 21/2 = 10.5
Area = Ï€×(10.5)2 = 346
radius of the largest circle = 42/2 = 21
Area = Ï€ ×(21)2
= 1385.5
Difference = 1385.5 - 346
= 1039.5
12. Find the area of a ring shaped region enclosed between two concentric circles of radii 20 cm and 15 cm.
Answer
Area of the ring = π(20)2 - π(15)2
= 400Ï€ - 225Ï€
= 175Ï€
= 549.7 cm2
13. The circumference of a given circular park is 55 m. It is surrounded by a path of uniform width 3.5 m . Find the area of the path.
Answer
Let r be the radius of the circular park.
2Ï€r = 55
⇒ r = 55/2Ï€ = 8.75 m
area of the park = Ï€×(8.75)2 = 240.625 m2
Radius of the outer circular region including the path is given by
R = 8.75 + 3.5 = 12.25 m
Area of that circular region is
A = Ï€×(12.25)2 = 471.625 m2
Hence, area of the path is given by
Area of the path = 471.625 - 240.625 = 231 m2
14.There are two circular gardens A and B. The circumference of garden A is 1.760 km and the area of garden B is 25 times the area of garden A. Find the circumference of garden B.
2Ï€r = 1760 m
⇒ r = (1760×7)/(2×22) = 280 m
Area of garden A= Ï€r2 = (22/7) × 2802 m2
Let R be the radius of the circular garden B.
Since the area of garden B is 25 times the area of garden A, we have,
2Ï€R2 = 25 ×Ï€r2
⇒ Ï€R2 = 25 × Ï€ × 2802
⇒ R2 = 1960000
⇒ R = 1400 m
15. A wheel has diameter 84 cm. Find how many completer revolutions must it make to cover 3.168 km.
Answer
Diameter of the wheel = 84cm
Thus, radius of the wheel = 42cm
Circumference of the wheel = 2×(22/7)×42 = 264 cm
In 264 cm, wheel is covering one revolution.
Thus, in 3.168 km = 3.168 ×100000 cm, number of revolutions covered by the wheel = (3.168/264)×100000 = 1200
16. Each wheel of a car is of diameter 80 cm. How many completer revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
= 11 km
= 1100000 cm
Circumference of the wheel = distance covered by the wheel in one revolution
thus, we have,
Circumference = 2×(22/7)×(80/2) = 251.43 cm
Thus, the number of revolutions covered
by the wheel in 1100000 cm = 1100000/251.43 ≈ 4375
17. An express train is running between two stations with a uniform speed. If the diameter of each wheel of the train is 42 cm and each wheel makes 1200 revolutions per minute, find the speed of the train.
Answer
Radius of the wheel = 42/2 = 21 cm
circumference of the wheel = 2Ï€×21 = 132 cm
Distance travelled in one minute = 132×1200
= 158400 cm
= 1.584 km
Hence, the speed of the train = 1.584km/(1/60)hr
= 95.04 km/hr
18. The minute hand of a clock is 8 cm long. Find the area swept by the minute hand between 8.30 a.m. and 9.05 a.m.
Time interval is
9.05 - 8.30 = 35 minute
Area covered in one 60 minutes
= Ï€×82 = 201 cm2
Hence area swept in 35 minutes is given by
19. The shaded portion of the figure, given alongside, shows two concentric circles.
If the circumference of the two circles be 396 cm and 374 cm, find the area of the shaded portion.
Answer
Let R and r be the radius of the big and small circles respectively.
Thus, we have,
2Ï€R = 396 cm
⇒ R = (396×7)/(2×22)
⇒ R = 63 cm
Thus, area of the bigger circle = πR2
= 12474 cm2
⇒ 2Ï€r = 374
⇒ r = (374×7)/(2×22)
⇒ r = 59.5 cm
Thus, the area of the smaller circle = πr2
= (22/7)×59.52
= 11126.5 cm2
20. In the given figure, the area of the shaded portion is 770 cm2. If the circumference of the outer circle is 132 cm, find the width of the shaded portion.
Answer
From the given data, we can calculate the area of the outer circle and then the area of inner circle and hence the width of the shaded portion.
Given that the circumference of the outer circle is 132 cm
Thus, we have, 2Ï€r = 132 cm
⇒ R = (132×7)/(2×22)
⇒ R = 21 cm
Area of the bigger circle = πR2
= (22/7)×212
= 1386 cm2
Also given the area of the shaded portion.
Thus the area of the inner circle = Area of the outer circle - Area of the shaded portion
= 1386- 770
= 616 cm2
⇒ Ï€r2 = 616
⇒ r2 =(616×7)/22
⇒ r2 = 196
⇒ r = 14cm
Thus, the width of the shaded portion = 21 - 14 = 7 cm
21.The cost of fencing a circular field at the rate of 240 per meter is 52,800. The field is to be ploughed at the rate of 12.50 per m2. Find the cost of pouching the field.
Answer
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
2Ï€r×240 = 52800
r = (52800×7)/(2×240×22) = 35
Thus the radius of the field is 35 meter.
Now the area of the field will be:
Ï€r2 = (22/7)×(35)2 = 3850 m2
Thus the cost of ploughing the field will be:
22. Two circles touch each other externally. The sum of their areas is 58 cm2 and the distance between their centers is 10 cm. Find the radii of the two circles.
Answer
Let r and R be the radius of the two circles.
y + R = 10 ...(1)
πy2 + πR2 = 58π ...(2)
Putting the value of r in (2)
r2 + R2 = 58
⇒ (10-R)2 + R2 = 58
⇒ 100 - 20R + R2 + R2 = 58
⇒ 2R2 - 20R + 42 = 0
⇒ R2 - 10R + 21 = 0
⇒ (R - 3)(R-7) = 0
⇒ R = 3, 7
Hence, the radius of the two circles is 3cm and 7cm respectively.
23. The given figures shows a rectangle ABCD inscribed in a circle as shown alongside. If AB = 28 cm and BC = 21 cm, find the area of the shaded portion of the given figure.
From the figure:
AB = 28cm
BC = 21cm
Hence diameter of the circle is 35 cm and hence
Area = Ï€×(35/2)2 = 962.5 cm2
Area of the rectangle = 28×21 = 588 cm2
Hence area of the shaded portion is given by
A = 962 - 588 = 374.5 cm2
24. A square is inscribed in a circle of radius 7 cm. Find the area of the square.
Answer
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,
√2a = 2×7
⇒ a = √2 × 7
⇒ a2 = 98
Hence, the area of the square is 98 sq.cm.
25. A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of 484√3 cm2. If the same wire is bent into the form of a circle of largest area, find the area of this circle.
Now, area of equilateral triangle formed = 484√3 cm2
⇒ (√3/4)a2 = 484√3
⇒ a2 = 4×484
⇒ a = 2×22 = 44 cm
Then, perimeter of equilateral triangle = 3a= 3×44 = 132 cm
Now, length of wire = perimeter of equilateral triangle = circumference of circle
⇒ circumference of circle = 132 cm
⇒ 2Ï€r = 132 (r is radius of circle)
∴ Area of circle = Ï€r2 = (22/7)×21×21 = 1386 cm2
Exercise 20 (D)
Answer
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,
2. A triangle and a parallelogram have the same base and the same area. If the side of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Answer
Let the sides of the triangle be
a = 26 cm, b = 28 cm and c = 30 cm
Now,
= 7×4×6×2
= 336 cm2
Base of a parallelogram = 28 cm
Given,
Area of parallelogram = Area of triangle
⇒ Base × Height = 336
⇒ 28× Height = 336
⇒ Height = 12 cm
3. Using the information in the following figure, find its area.
Answer
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
∴ Area of trapezium ABCD = (1/2)×(sum of parallel sides)×Height
= (1/2)×(AB + CD)×AD
= (1/2)×(23+35)×9
= (1/2)×58×9
= 261 m2
4. Sum of the areas or two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of the two squares.
Answer
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.
5. Find the area and the perimeter of a square with diagonal 24 cm.
[Take √2 =1.41]
Solution
Diagonal of a square = 24 cm
⇒ 24 = side of a square × √2
⇒ Side of a square = 24/√2 = (12×√2×√2)/√2 = 12√2
∴ Area of a square = (side)2 = (12√2)2 = 288 cm2
6. A steel wire, when bent in the form of a square, encloses an area of 121 cm2. The same wire is bent in the form of a circle. Find area the circle.
⇒ 121 = (Side)2
⇒ Side of a square = 11 cm
Now,
⇒ 4×side = Perimeter of a circle
⇒ Perimeter of a circle = 44 cm
⇒ 2Ï€r = 44 (r is radius of a circle)
⇒ r = 44/2Ï€ = 44/2(22/7) = 7cm
∴ Area of a circle = Ï€r2 = (22/7)×7×7 = 154 cm2
7. The perimeter of a semicircular plate is 108 cm. find its area.
Answer
Perimeter of a semi - circular plate = 108 cm
⇒ Ï€r + 2r = 108 (r is radius of a circle)
⇒ r(Ï€ + 2) = 108
⇒ r = 108/(Ï€ + 2) = 108/[(22/7) + 2] = (108×7)/36 = 21 cm
Now, area of a semi - circular plate = Ï€r2/2 = [(22/7)×21×21]/2 = 693 cm2
8. Two circles touch externally. The sum of their areas is 130Ï€ sq. cm and the distance between their centres is 14 cm. Find the radii of the circles.
Answer
Let the radii of two circles be r1 and r2 respectively.
Sum of the areas of two circles = 130Ï€ sq. cm
⇒ Ï€r12 + Ï€r22 = 130Ï€
⇒ 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.
Answer
Given the diameter of the front and rear wheels are 63cm = 0.63m and 1.54m respectively.
Radius of the rear wheel = 1.54/2 = 0.77 m
and radius of the front wheel = 0.63/2 = 0.315 m
Distance travelled by tractor in one revolution of rear wheel
= circumference of the rear wheel
= 2Ï€r
= 2×(22/7)×0.77 = 4.84 m
The rear wheel rotates at 24(6/11) revolutions per minute
= 270/11 revolutions per minute
Since in one revolution the distance travelled by the rear wheel = 4.84 m
So, in 270/11 revolutions, the tractor travels (270/11)×4.84 = 118.8 m
Let the number of revolutions made by the front wheel be x.
(i) Now, number of revolutions made by the front wheel in one minute x circumference of the wheel
= the distance travelled by the tractor in one minute
⇒ x × 2 ×(22/7)×0.315 = 118.8
⇒ x = (118.8×7)/(2×22×0.315) = 60
(ii) Distance travelled by the tractor in 40 minutes
= Number of revolutions made by the rear wheel in 40 minutes x circumference of the rear wheel
= 270/11 ×40×4.84 = 4752 m
10. Two circles touch each other externally. The sum of their areas is 74Ï€ cm2 and the distance between their centres is 12 cm. Find the diameters of the circle.
Answer
Let the radius of the circles be r1 and r2 .
So, r1 + r2 = 12 ⇒ r2 = 12 – r1
Sum of the areas of the circles = 74 π
⇒ Ï€r12 + Ï€r22 = 74Ï€
⇒ r12 + r22 = 74
⇒ r12 + (12-r1)2 = 74
⇒ r12 + 144 - 24 r1 + r12 = 74
⇒ 2 r12 - 24 r1 + 70 = 0
⇒ r12 - 12 r1 + 35 = 0
⇒ (r1 – 7)( r1 – 5) = 0
⇒ r1 = 7 or r1 = 5
If r1 = 7cm, then r2 = 5cm
If r1 = 5cm, then r2 = 7cm
So, the diameters of the circles will be 10 cm and 14cm.
11. If a square in inscribed in a circle, find the ratio of the areas of the circle and the square.
If AB = x , AC = x√2
Diameter of the circle = diagonal of the square
⇒ 2r = x√2
⇒ r = x√2/2
Area of the circle = πr2
= Ï€(x√2/2)2
= π(x22/4)
= πx2/2
Area of the square = x2
= π/2
= (22/7)×(1/2)
= 11/7
Hence, the required ratio is 11 : 7.