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Applied Maths Chapter 4 Solutions
Mensuration
EXERCISE- 4.1
Q.1 Find the area of a triangle whose sides are 34 cm, 20 cm, and 42 cm. Hence, find the length of the altitude corresponding to the shortest side.
Ans. S = (42 + 20 + 34)/2
= 96/2 = 48 48
Area = √S(S – a) (S – b) (S – c)
= √48(48 – 42) (48 – 34) (48 – 12)
= √48 x 6 x 14 x 28
= √2 x 2 x 2 x 6 x 6 x 2 x 7 x 2 x 2 x 7
= 2 x 2 x 2 x 6 x 7
= 336 cm^2
Area of triangle = 1/2 x b x h
336 = 1/2 x 20 x h
336/10 = h
h = 33.6 cm
Q.2 The sides of a triangular plot are in the ratio 3:5:7 and its perimeter is 300 m. Find its area. Take √3 = 1.732.
Ans. a = 3x, b = 5x, c = 7x, Perimeter = 300m
3x + 5x + 7x = 300
15x = 300
x = 20
a = 3 x 20 = 60m
b = 5 x 20 = 100 m
c = 7 x 20 = 140 m
S = Perimeter/2
= 300/2 = 150 150
Area = √S(S – a) (S – b) (S – c)
= √150(150 – 60) (150 – 100) (150 – 140)
= √150 x 90 x 50 x 10
= 5 x 10 x 3 √3
= 150 x 1.732
= 2598.00 m^2
Q.3 If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.
Ans. The perimeter of an equilateral triangle = 3a
3a = 36
a = 12
Area of an equilateral triangle = √3/4 a^2
= √3 /4(12)^2
= √3 /4 x 144
= √3 x 36
= 1.732 x 36
= 62.4 cm^2
By using Pythagoras’ theorem
(12)^2 = (h)^2 + (6)^2
144 = h^2 + 36
108 = h^2
h = √108
h = √2 x 2 x 3 x 3 x 3
= 2 x 3 √3
= 6 x 1.732
=10.4cm
Q.4 Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.
Ans. b + 12 + 12 = 30
b = 30 – 24
b = 6
Area of triangle = 1/4 x b√4a^2-b^2
= 1/4 x 6 √4(12)^2 + (6)^2
= 3/2 √576 – 36
= 3/2 √540
= 3/2 √3 x 3 x 2 x 2 x 3 x 5
= 3/2 x 2 x 3 √15
= 6 √15
= 6 x 3.87
= 34.83 cm^2
Q.5 If the area of an isosceles triangle is 60 cm^2 and the length of each of its equal sides is 13 cm, find its base.
Ans. Area = 60 cm^2
Sides: a =13
b = 13
c = x
S = (a + b + c)/2
= (13 + 13 + x)/2
= (26 + x)/2
= 13 + x
Area = √S(S – a) (S – b) (S – c)
60 = √[13+x/2][13+x/2-13][13+x/2-13][13+x/2-x]
60^2 = [13 + x/2][x/2][x/2][13 – x/2]
3600 = x^2/4[13 + x/2][13 – x/2]
3600 = x^2/4 (13^2 – [x/2]^2)
3600 = x^2/4 [169 – x^2/4]
3600 x 4 = 169x^2 – x^4/4
3600 x 16 = 169 x 4 x x^2 + x^4
57600 = 676 x^2 – x^4
x^4 – 676x^2 + 57600 =0
Let, x^2 be a
So, a^2 – 676a + 57400 = 0
a^2 – 576a – 100a 57600 = 0
a(a – 576) – 100(a – 576) = 0
a – 576 = 0 , a – 100 = 0
a = 576 , a = 100
Therefore,
x^2 = 576 , x^2 – 100 = 0
x = 24 , x = 10
Q.6 If the perimeter of a right-angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.
Ans. Perimeter = 60 cm
Hypotenuse = 25 cm
Now,
a + b + c = 60
a + b + 25 = 60
a + b = 60 – 25
a + b = 35
Squaring both sides,
(a + b)^2 = 35^2
a^2 + b^2 + 2ab = 1225
we know that,
in a right-angle triangle a^2 + b^2 = c^2
Therefore, c^2 + 2ab = 1225
25^2 + 2ab = 1225
325 + 2ab = 1225
2ab = 1225 – 625
ab = 300
Now, the Area of the triangle = 1/2 x b x h
Here, area = 1/2 x a x b
Area = 1/2 x 300
= 150 cm^2
Q.7 The sides of a right-angled triangle containing the right angle are 5x cm and (3x-1)cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm^2.
Ans. According to the question,
a = 5x , b = 3x – 1
a & b are not hypotenuse
Now, Area = 1/2 x b x h
60 = 1/2 x (5x) x (3x – 1)
120 = 15x^2 – 5x
24 = 3x^2 – x
3x^2 – x -24 = 0
3x^2 – 9x + 8x – 24 = 0
3x(x – 2) + 8(x – 3) = 0
(x – 8) (3x + 8) = 0
x = 8 , x = -8/3
Therefore, a = 15
b = 8
Using Pythagoras theorem,
a^2 + b^2 = c^2
15^2 + 8^2 = c^2
225 + 64 = c^2
289 = c^2
So, c = √289
= 17 cm
Q.8 In the adjoining figure, ABC is an isosceles right-angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.
Ans. In ΔABC,
AB – AD + DB = 3 + 4 = 7
AC = AE + EC = 3 + 4 = 7
∠A = 90°
Area = 1/2 x b x h
= 1/2 x 7 x 7
= 49/2 cm^2
= 24.5 cm^2
In ΔADE,
∠A = 90°
Therefore,
AD^2 + AE^2 = DE^2 (By Pythagoras theorem)
3^2 + 3^2 = DE^2
9 + 9 = DE^2
18 = DE^2
√18 = DE
3√2 = DE
In ΔABC,
AB = AC (Isosceles triangle)
∠A = ∠C (angle opposite to equal sides)
So, ∠B = 45° (Angle sum property)
In ΔDGB,
∠G = 90° (Angle of a rectangle)
∠B = 45°
So, ∠D = 45° (Angle sum property)
Hence, ΔDBG is an isosceles right triangle.
In ΔDGB, BG = DG = x
Also, BG^2 + DG^2 = BD^2
x^2 + x^2 = 4^2
2x^2 = 16
x^2 = 8
x = 2√2
Therefore, the Area of the rectangle = DE x DG
= 3√2 x 2√2
= 12cm^2
Required area = area of a triangle – and area of a rectangle
= 24.5 – 12
= 12.5 cm^2
Q.9 The diagram given alongside shows two paths drawn inside a rectangular field 50m long and 35 m wide. The width of each path is 5 meters. Find the area of the shaded portion.
Ans. Area of rectangle ABCD = 35 x 5 = 175 cm^2
Area of rectangle PQRS = 50 x 5 = 250 cm^2
Area of square WXYZ = 5 x 5
= 25 m^2
Required area = 175 + 250 – 25
= 400 m^2
Q.10 A footpath of uniform width runs all around the inside of a rectangular field 50 m long and 38 m wide. If the area of the path is 492 m^2, find its width.
Ans. Area of path = Area of outer rectangle – Area of inner rectangle
492 = 50 x 38 -[50 – 2x][38 – 2x]
492 =1900 – [1900 – 76x – 100x + 4x^2]
492 = 1900 – 1900 + 176x – 4x^2
4x^2 – 176x + 492 = 0
x^2 – 44x + 123 = 0
x^2 – 41x – 3x + 123 = 0
x(x – 41) – 3(x – 41) = 0
(x – 41) (x – 3) = 0
x – 41 = 0 , x – 3 = 0
Therefore, the width of the park = 3m
Q.11 A wire when bent in the form of an equilateral triangle encloses an area of 36 √3 cm^2, Find the area enclosed by the same wire when bent to form:
(i) a square and
(ii) a rectangle whose length is 2 cm more than its width.
Ans. Area of equilateral triangle = √3/4a^2
36√3 = √3/4a^2
36 x 4 = 144 a^2
√36 x 4 = a
6 x 2 = a
12 = a
Therefore, the length of a wire = 12 x 3 = 36 cm
(i) For Square:
Perimeter = 36 cm
4a = 36
a = 9 cm
Therefore, area of square = a^2
Area = 9^2
Area = 81 cm^2
(ii) For Rectangle:
l = b + 2
Perimeter = 36 cm
2(l + b) = 36
l + b = 18
b + 2 + b = 18
2b = 18 – 2
2b = 16
b = 8
Therefore, b = 8 cm
l = 10 cm
Area of rectangle = l x b
= 10 x 8 = 80 cm^2
Q.12 The area of a trapezium is 540 cm^2. If the ratio of parallel sides is 7:5 and the distance between them is 18 cm, find the length of parallel sides.
Ans. Area of trapezium = 1/2 x h x (a + b)
ATQ, Area = 540 cm^2
h = 18 cm
Ratio of parallel sides – 7 : 5
Therefore, a = 7x
b = 5x
Area = 1/2 x h x (a + b)
540 = 1/2 x 18 x (7x + 5x)
540 = 9 x 12x
540/9 = 12x
60 = 12x
x = 5
Therefore,
a = 7 x 5 = 35 cm
b = 5 x 5 = 25 cm
Q.13 A sheet is 11 cm long and 2 cm wide. Circular pieces 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.
Ans.
No. of circles in 1 row = 11/0.5 = 22
No. of circles in 1 column= 2/0.5 = 4
Total no. of circles = 22 x 4
= 88 circles
Q.14 A road 3.5 m wide surrounds a circular plot whose circumference is 44 m. Find the cost of paving the road at Rs.50 per m^2.
Ans. Circular plot = 44 m
2 лr = 44
лr = 22
22/7 x r = 22
r = 7 m
Area of road = Area of outer circle – Area of inner circle
= лR^2 – лr^2
= л(10.5^2 – 7^2)
= 22/7 x (10.5 + 7) (10.5 – 7)
= 22/7 x 17.5 x 3.5
= 22/7 x 17.5 x 35/10
= 11 x 17.5
= 192.5 m
Cost = 192.5 x 50
= Rs.9625
Q.15 A copper wire when bent in the form of a square encloses an area of 121 cm^2. If the same wire is bent into the form of a circle, find the area of the circle.
Ans. Area of square = 121 cm^2
a^2 = 121
a = 11 cm
Length of wire = Perimeter of square
= 11 x 4
= 44 cm
Length of wire = Circumference of a circle
44 = 2лr
22 = 22/7 x r
r = 7 cm
Therefore,
Area = лr^2
= 22/7 x 7 x 7
= 22 x 7 = 154 cm^2
Q.16 In the adjoining figure, ABCD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semicircles are drawn with AD and BC as diameters. Find the area of the shaded region. Take (Π = 22/7)
Ans. Area of 2 Δ = 1/2 x Area of square
= 1/2 x 21 x 21
= 441/2 cm^2
Area of 2 semicircles = 2 x 1/2 x 22/7 x 21/2 x 21/2
= (11 x 3 x 21)/2
= 693/2 cm^2
Therefore, required area = 441/2 + 693/2 = 1134/2
= 567 cm^2
Q.17 (a) In the figure (i) given below, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.
(b) In figure (ii) given below, ABC is an isosceles right-angled triangle with ∠ABC = 90°. A semicircle is drawn with AC as the diameter. If AB = BC = 7 cm, find the area of the shaded region. Take л =22/7
Ans. (a) Required area = Area of quadrant – Area of triangle
= 1/4лr^2 – 1/2 x b x h
= [1/4 x 22/7 x 35/10 x 35/10] – [1/2 x 35/10 x 2]
= 35/10 [11/4 – 1]
= 3.5 x 7/4
= 3.5 x 1.75
= 6.125 cm^2
(b) In ΔABC, ∠A = 90°
Therefore, AB^2 + BC^2 = AC^2 [By pythagoras theorem]
7^2 + 7^2 = AC^2
98 = AC^2
√98 = AC
7√2 = AC
Diameter = 7√2 , r = 7√2/2
Required Area = Area of semi-circle – Area of triangle
= 1/2лr^2 – 1/2 x b x h
= [1/2 x 22/7 x 49/2] – [1/2 x 7 x 7]
= 11/2 x 49/2 – 49/2
= 49/2 [11/7 – 1]
= 49/2 [11-7/7]
= 49/2 x 4/7
= 14 cm^2
Q.18 (a) The quadrants shown in Figure (i) below are each of a radius of 7 cm. Calculate the area of the shaded portion.
(b) In figure (ii) given below, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (Use л = 3.14)
Ans. (a) Required Area = Area of square – Area of 4 quadrant
= 14 x 14 – 4 x 1/4 x 22/7 x 7 x 7
= 196 – 154
= 42 cm^2
(b) In ΔOAB, ∠A = 90°
Therefore, OA^2 + AB^2 = OB^2 [By pythagoras theorem]
20^2 + 20^2 = OB^2
800 = OB^2
20√2 = OB
So, radius of quadrant = 20√2 cm
Required area = Area of quadrant – Area of square
= 1/4лr^2 – 5^2
= 1/4 x 3.14 x 400 x 2 – 20^2
= 3.14 x 200 – 400
= 628 – 400
= 228 cm^2
Q.19 (a) In figure (i) given below, the points A, B, and C are centres of arcs of circles of radii 5 cm, 3 cm, and 2 cm respectively. Find the perimeter and the area of the shaded region. (Take л = 3.14)
(b) figure (ii) given below shows a kite, in which BCD is in the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and ACEF is an isosceles right-angled triangle whose equal sides are 6cm long. Find the area of the shaded region.
Ans. (a) Required Area = [Area of semi-circle 1 – Area of semicircle 2] + Area of semi-circle 3
= 1/2л5^2 – 1/2л3^2 + 1/2л2^2
= 1/2 x 3.15 [5^2 – 3^2 + 2^2]
= 3.14/2 [25 – 9 + 4]
= 3.14/2 x 20
= 3.14 x 10
= 31.4 cm^2
(b) Required area = Area of quadrant + Area of triangle
= 1/4лr^2 + 1/0 x b x h
= [1/4 x 22/4 x 42 x 42] + [1/2 x 6 x 6]
= 11 x 126 + 18
= 1386 + 18 = 1404 cm^2
Q.20 (a) In figure (i) given below, a piece of cardboard, in the shape of a trapezium ABCD, and AB|| DC and ∠BCD = 90°, quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE =2 cm. Calculate the area of the remaining piece of the cardboard.
(b) In figure (ii) given below, ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm, and BC = 21 cm. With AC as diameter, a semicircle is drawn and with BC as radius, a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.
Ans. (a) Required area = Area of trapezium – Area of Quadrant
= 1/2 x h x (a + b) – 1/4лr^2
= 1/2 x 35/10 x(3.5 + 5.5) – 1/4 x 22/7 x 35/10 x 35/10
= 1/2 x 35/10 x 9 – 1/4 x 11 x 35/10
= 63/4 – 38.5/4
= 24.5/4 = 6.125 cm^2
(b) In ΔABC, ∠B = 90°
Therefore, AB^2 + BC^2 = AC^2 [By pythagoras theorem]
28^2 + 21^2 = AC^2
784 + 441 = AC^2
1225 = AC^2
√1225 = AC
35 = AC
Required area = [Area of triangle + Area of semi-circle] – Area of quadrant
= [1/2 x b x h + 1/2лr^2] – 1/4лr^2
= [1/2 x 21 x 28 + 1/2 x 22/7 x 35/2 x 35/2] – 1/4 x 2/7 x 21 x 21
= (21 x 14) + (11 x 5 x 35)/4 – (11 x 3 x 21)/2
= 294 + (11 x 175)/4 – (11 x 63)/2
= 294 + (1925 – 1386)/4
= 294 + 134.75
= 428.75 cm^2
Q.21 In the adjoining figure, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and the major segment. Take π = 3.14.
Ans. Area of sector = θ/360лr^2
= 90/360 x 3.14 x 10^2
= 1/4 x 314
= 78.5 cm^2
Area of minor segment = Area of sector – Area of triangle
= 78.5 – 1/2 x 10 x 10
= 78.5 – 50 = 28.5 cm^2
Area of major segment = Area of circle – Area of minor segment
= лr^2 – 28.5
= 3.14 x 10^2 – 28.5
= 314 – 28.5
= 285.5 cm^2
FAQ’s related to Applied Maths Chapter 4 on Mensuration:
Q.1 What is mensuration?
Ans. Mensuration is the branch of mathematics that deals with the measurement of geometric figures, such as length, area, volume, and other related quantities.
Q.2 What are the basic formulas in mensuration?
Ans. Basic formulas in mensuration include formulas for calculating the perimeter and area of geometric shapes like squares, rectangles, triangles, circles, and the volume of solids like cubes, cylinders, and spheres.
Q.3 How is mensuration used in real life?
Ans. Mensuration is used in various real-life applications such as calculating the area of fields, determining the amount of material needed for construction projects, estimating the volume of containers, and designing objects in engineering and architecture.
These are few Frequently Asked Questions relating to Applied Maths Chapter 4
In Applied Maths chapter 4, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Applied Maths Chapter 4 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 4, we delve deep into advanced mathematical concepts that are crucial for understanding.
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