Quantitative Aptitude - SPLessons
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Area Problems

shape Introduction

Area Problems deals with parameters like sides, length, area, breadth, centroid, median, perimeter, height etc. of all the geometrical shapes like square, rectangle, triangle, circle and so on.


shape Methods

Area: It is defined as the surface enclosed by its sides.
Square: It is defined as a four sided shape that is made up of four straight sides that are the same lengths and that has four right angles.
Triangle: The plane figure formed by connecting three points not in a straight line by straight line segments like a three sided polygon.
    (i) Sum of angles of a triangle is 180 degrees.
    (ii) The sum of any two sides of a triangle is greater than the third side.
    (iii) The line joining the mid - point of a side of a triangle to the opposite vertex is called the meridian.
    (iv) The point where three medians of a triangle meet, is called centroid. The centroid divides each of the medians in the ratio 2 : 1.
    (v) In an isosceles triangle, the altitude from the vertex bisects the base.
    (vi) The median of a triangle divides it into two triangles of the same area.
    (vii) The area of a triangle formed by joining the mid - points of the sides of a given triangle is one - fourth of the area of the given triangle.

Rectangle: The plane figure formed with four straight sides and four right angles, especially one with unequal adjacent sides, in contrast to a square.
Circle: A line that is curved so that its ends meet and every point on the line is the same distance from the centre.
Quadrilateral: Quadrilateral means four sided flat shape.
    (i) The diagonals of a parallelogram bisect each other.
    (ii) Each diagonal of a parallelogram divides it into two triangles of the same area.
    (iii) The diagonals of a rectangle are equal and bisect each other.
    (iv) The diagonals of a square are equal and bisect each other at right angles.
    (v) The diagonals of a rhombus are unequal and bisect each other at right angles.
    (vi) A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
    (vii) Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.

Example: Given a square where the length of each side (edge) is 5cm. Find the area of this square.
    [latex]A = L^{2}[/latex]
    = [latex]5^{2}[/latex]
    = 5 x 5
    = [latex]25 \ cm^{2}[/latex]

Example: Given a rectangle with the length of 4ft and the width of 3ft. Find its area.
    [latex]A = lw[/latex]
    = (4)(3)
    = [latex]12ft^{2}[/latex]

Example: Given a circle with the radius 3cm. Find its area. Take [latex]\pi[/latex] = 3.14.
    [latex]A = \pi r^{2}[/latex]
    = (3.14)[latex](3)^{2}[/latex]
    = (3.14)(9)
    = 28.26 [latex]cm^{2}[/latex]

Example: Given a parallelogram with the base 5 in and the height 3 in. Find the area of this parallelogram.
    [latex]A = bh[/latex]
    = (5)(3)
    = [latex]15 in^{2}[/latex]

Example: Given a triangle with the base 4cm and the height 2cm. Find its area.
    [latex]A = \frac{1}{2}bh[/latex]
    = [latex]\frac{1}{2}(4)(2)[/latex]
    = [latex]\frac{1}{2}(8)[/latex]
    = [latex]4cm^{2}[/latex]

Example: Given a trapezoid with the height of 4 in, and two parallel sides of 2 in and 3 in respectively. Calculate its area.
    [latex]A = (\frac{a + b}{2})h[/latex]
    = [latex](\frac{2 + 3}{2})4[/latex]
    = [latex](\frac{5}{2})4[/latex]
    = [latex]10in^{2}[/latex]

shape Formulae

1 Pythagoras theorem: In a right-angled triangle, [latex](hypotenuse)^2[/latex] = [latex](base)^2[/latex] + [latex](height)^2[/latex].
2: Area of a triangle = (length x breadth) Therefore, length = [latex]\frac{area}{breadth}[/latex] and breadth = [latex]\frac{area}{length}[/latex].
3: Perimeter of a rectangle = 2(length x breadth).
4: Area of a square = [latex](side)^2 [/latex] = [latex]\frac{1}{2}(diagonal)^2[/latex].
5: Area of four walls of a room = 2(length x breadth) x height.
6: Area of a triangle = [latex]\frac{1}{2}[/latex] x base x height.
7: Area of a triangle = [latex]\sqrt{s(s - a)(s - b)(s - c)}[/latex], where [latex]a, b, c[/latex] are sides of the triangle and [latex]s[/latex] = [latex]\frac{1}{2}(a + b + c)[/latex].
8: Area of an equilateral triangle = [latex]\frac{\sqrt{3}}{4} * (side)^2[/latex].
9: Radius of incircle of an equilateral triangle of side [latex]a[/latex] is [latex]\frac{a}{2\sqrt{3}}[/latex].
10: Radius of circumference of an equilateral triangle of side [latex]a[/latex] is [latex]\frac{a}{\sqrt{3}}[/latex].
11: Radius of incircle of a triangle of area [latex]\bigtriangleup[/latex] and semi-perimeter [latex]s[/latex] = [latex]\frac{\bigtriangleup}{s}[/latex].
12: Area of a parallelogram = base x height.
13: Area of a trapezium = [latex]\frac{1}{2}[/latex] x (sum of parallel sides) x distance between them.
14: Area of a circle = [latex]\pi R^2[/latex], where R is the radius.
15: Circumference of a circle = 2[latex]\pi[/latex]R.
16: Length of an arc = [latex]\frac{2\pi R\theta}{360}[/latex], where [latex]\theta[/latex] is the central angle.
17: Area of a sector = [latex]\frac{1}{2}[/latex](arc x R) = [latex]\frac{\pi R^2\theta}{360}[/latex].
18: Area of a semi circle = [latex]\frac{\pi R^2}{2}[/latex].
19: Area of isosceles triangle = [latex]\frac{a}{4}\sqrt{4b^2 - a^2}[/latex]square units, where [latex]a, b[/latex] are linear units.
20: Circumference of a semi-circle = [latex]\pi R[/latex].
21: Side of a rhombus = [latex]\frac{1}{2}\sqrt{(d_{1})^2 + (d_{2})^2}[/latex] linear units, where [latex]d_{1}[/latex] and [latex]d_{2}[/latex] are the lengths of diagonals.

shape Samples

1. The area of four walls of a room is 600[latex]m^2[/latex] and its length is twice its breadth. If the height of the room is 11 m, then find the area of the floor in [latex]m^2[/latex]?
    Let, breadth of a room = [latex]x[/latex] m
    Length of the room = 2[latex]x[/latex] m
    Height of the room = 11 m
    Total area of 4 waits of room = 2(2[latex]x[/latex] + [latex]x[/latex]) x 11 [latex]m^2[/latex] = 66[latex]m^2[/latex]
    By hypothesis, 66[latex]x[/latex] = 660
    ⇒ [latex]x[/latex] = 10 m
    Hence, area of floor = 2[latex]x[/latex] + [latex]x[/latex] = 2[latex]x^2[/latex] = 2[latex](10)^2[/latex] = 2 x 100 = 200[latex]m^2[/latex]

2. The perimeter of two squares are 60 cm and 44 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of two squares?
    Side of first square = [latex]\frac{60}{4}[/latex] = 15 cm
    Side of second square = [latex]\frac{44}{4}[/latex] = 11 cm
    Area of third square = [latex](15)^2 - (11)^2[/latex] [latex](cm)^2[/latex]
    = (225 - 121)[latex](cm)^2[/latex]
    = 104 [latex](cm)^2[/latex]
    Side of third square = [latex]\sqrt{104}[/latex] cm = 10.19 cm
    Therefore, required perimeter = (10.19 x 4) cm = 40.76 cm.

3. Find the area of the square, one of whose diagonals is 6 m long?
    Diagonal = 6 m
    Now, consider
    Area of square = [latex]\frac{1}{2}(diagonal)^2[/latex]
    ⇒ Area of square = [latex]\frac{1}{2}(6)^2[/latex]
    ⇒ Area of square = [latex]\frac{1}{2}36[/latex]
    ⇒ Area of square = 18 [latex]m^2[/latex]
    Therefore, Area of square = 18 [latex]m^2[/latex]

4. If the length of a certain rectangle is decreased by 5 cm and the width is increased by 4 cm, a square with the same area as the original rectangle would result . Find the perimeter of the original rectangle?
    [latex]x[/latex] and [latex]y[/latex] be the length and breadth of the rectangle respectively.
    Then, [latex]x[/latex] - 5 = [latex]y[/latex] + 4
    [latex]x[/latex] - [latex]y[/latex] = 9 -------- (i)
    Now, Area of rectangle = length x breadth and
    Area of square = ([latex]x[/latex] - 5)([latex]y[/latex] + 4)
    Therefore, ([latex]x[/latex] - 5)([latex]y[/latex] + 4) = [latex]xy[/latex]
    ⇒ 4[latex]x[/latex] - 5[latex]y[/latex] = 20 -------- (ii)
    By soling (i) and (ii),
    [latex]x[/latex] = 25 and [latex]y[/latex] = 16.
    Therefore, Perimeter of the rectangle = 2([latex]x[/latex] + [latex]y[/latex]) = 2(16 + 25) = 82 cm.

5. Find the area of the triangle whose base sides measure 10 cm, 13 cm and 15 cm?
    Given that,
    Let base sides [latex]a[/latex] = 10 cm, [latex]b[/latex] = 13 cm and [latex]c[/latex] = 15 cm
    Now, consider [latex]s[/latex] = [latex]\frac{1}{2}(a + b + c)[/latex]
    ⇒ [latex]s[/latex] = [latex]\frac{1}{2}(10 + 13 + 15)[/latex]
    ⇒ [latex]s[/latex] = 19
    (s - a) = 19 - 10 = 9
    (s - b) = 19 - 13 = 6
    (s - c) = 19 - 15 = 4
    Therefore, area = [latex]\sqrt{s(s - a)(s - b)(s - c)}[/latex] = [latex]\sqrt{19 * 9 * 6 * 3}[/latex] = [latex]\sqrt{3078}[/latex] = 55.47 [latex](cm)^2[/latex]
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