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Class X - Maths: Mensuration
One Word Answer Questions:
Q) Definition of cone?
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Q) Definition of sphere?
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Short Answer Questions:
Q) Definition of a cube?
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Q) Definition of cuboid?
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Q) Definition of cylinder?
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Q) The point A, B, C, D have respective co-ordinates (-4, -3), (10, -15) , (20, 9) and (0, 2). Find the area of quadrilateral ABC?
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Q) The polar co-ordinates of the vertices of a triangle are (-a, π/8), (a, 3π/7) and (-2a, - 3π/5) find the area of the triangle?
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Q) If the co-ordinates of the vertices of a △ ABC be (3, 5), (5, 8) and (8, 9) and if D and E divide AB and AC, respectively internally in the ratio 1 : 2, then show that the area of △ ABC = 9 . the area of △ADE?
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Q) Find the altitude of the rhombus whose area is 345 cm2 and its perimeter is 210 cm?
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Q) Find the area of the rhombus having each side equal to 20 cm and one of its diagonals equal to 18 cm?
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Long Answer Questions:
Q) The length of the parallel sides of a trapezium are in the rat: 5 : 2 and the distance between them is 20 cm. If the area of trapezium is 345 cm2, find the length of the parallel sides?
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Q) Find the area of a trapezium whose parallel sides are AB = 12 cm, CD = 36 cm and the non-parallel sides are BC = 15 cm and AG = 15 cm?
    
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Q) Construct a quadrilateral ABCD in which AB = 3.6 cm, △ABC = 80°, BC = 4 cm, △BAD = 120° and AD = 5 cm.
    
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Q) The co-ordinates of the points A, B, C, D are (10, -7), (-7, 2), (15, 13) and (24, -15) respectively. Find the ratio in which AC divides BD?
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Content
  • Mensuration
  • Prism
  • Tetrahedron
  • Area of Trapezium
  • Area of a General Quadrilateral
  • Quadrilateral form a Parallelogram
  • Construction of Quadrilaterals
  • Worked-out problems to find the area of a triangle given 3 points
  • Alternative Method
  • Area of special quadrilaterals
  • Area of a Polygon
  • Area of Rhombus
  • Worked-out examples on area of rhombus
  • Common Solid Figures
  • Definition of a cube
  • Definition of cuboid
  • Definition of cylinder
  • Definition of cone
  • Definition of sphere
  • Volume and Capacity
  • Area of the lateral faces of a right prism = perimeter of the base × height.
  • Area of the whole surface of a right prism = Area of the lateral faces + 2 × area of the base.
  • Volume of a right prism = area of the base × height.
  • Area of the lateral faces of a right pyramid = `(1/2)` × perimeter of the base × slant height.
  • Area of whole surface of a right pyramid = area of the lateral faces + area of the base.
  • Volume of a right pyramid = `(1/3)` × area of the base × height.
  • The area of its slant sides = `((3sqrt3)/4)a^2` sq. units
  • The area of its whole surface = `((sqrt3)a^2)` sq. units
  • The volume of the tetrahedron = `((sqrt3)/12 a^3)` cu. units.
  • Area of trapezium ABCD = Area of Δ ABD + Area of Δ CBD
    `(1/2)` × h × (a + b)
  • `(1/2)` (sum of parallel sides) × (perpendicular distance between them)
  • The area of a triangle formed by joining the points (x1, y1), (x2, y2) and (x3, y3) is
    =1/2 |y1 (x2 - x3) + y2 (x3 - x1) + y3 (x1 - x2)| sq. units
Mensuration
    Prism:

    (i) Area of the lateral faces of a right prism = perimeter of the base × height.
    (ii) Area of the whole surface of a right prism = Area of the lateral faces + 2 × area of the base.
    (iii) Volume of a right prism = area of the base × height.

    Pyramid:

    (i) Area of the lateral faces of a right pyramid = `(1/2)` × perimeter of the base × slant height.
    (ii) Area of whole surface of a right pyramid = area of the lateral faces + area of the base.
    (iii) Volume of a right pyramid =`(1/3)` × area of the base × height.

    Tetrahedron:

    If a be the length of an edge of a regular tetrahedron then

    (i) The area of its slant sides = `(3sqrt3)/4 a^2`sq. units.
    (ii) The area of its whole surface = `sqrt3a^2` sq. units.
    (iii) The volume of the tetrahedron =`sqrt2/(12)a^2` cu. units.

    Area of Trapezium

    Area of trapezium ABCD = Area of Δ ABD + Area of Δ CBD

    = `1/2` × a × h + `1/2` × b × h
    = `1/2` × h × (a + b)
    = `1/2` (sum of parallel sides) × (perpendicular distance between them)

    mensuration1

    Worked-out examples on area of trapezium

    1. The length of the parallel sides of a trapezium are in the rat: 3 : 2 and the distance between them is 10 cm. If the area of trapezium is 325 cm2, find the length of the parallel sides.

    Solution:

    Let the common ration be x,
    Then the two parallel sides are 3x, 2x
    Distance between them = 10 cm
    Area of trapezium = 325 cm2
    Area of trapezium = `(1/2)` (p1 + p2) h
    325 =`( 1/2)` (3x + 2x) 10
    ⇒ 325 = 5x × 5
    ⇒ 325 = 25x
    ⇒ x = `(325/25)`
    Therefore, 3x = 3 × 13 = 39 and 2x = 2 × 13 = 26

    Therefore, the length of parallel sides area are 26 cm and 39 cm.

    2. ABCD is a trapezium in which AB ∥ CD, AD `_|_` DC, AB = 20 cm, BC = 13 cm and DC = 25 cm. Find the area of the trapezium.

    mensuration2

    Solution:

    From B draw BP perpendicular DC
    Therefore, AB = DP = 20 cm
    So, PC = DC - DP
    = (25 - 20) cm
    = 5 cm

    Now, area of trapezium ABCD = Area of rectangle ABPD + Area of Δ BPC
    ΔBPC is right angled at ∠BPC
    Therefore, using Pythagoras theorem,
    BC2 = BP2 + PC2
    132 = BP2 + 52
    ⇒ 169 = BP2 + 25
    ⇒ 169 - 25 = BP2
    ⇒ 144 = BP2
    ⇒ BP = 12

    Now, area of trapezium ABCD = Area of rectangle ABPD + Area of ΔBPC
    = AB × BP + `(1/2)` × PC × BP
    = 20 × 12 + `(1/2)` × 5 × 12
    = 240 + 30
    = 270 cm2

    3. Find the area of a trapezium whose parallel sides are AB = 12 cm, CD = 36 cm and the non-parallel sides are BC = 15 cm and AG = 15 cm.

    mensuration3

    Solution:

    In trapezium ABCD,
    draw CE ∥ DA.
    Now CE = 15 cm
    Since, DC = 12 cm so, AE = 12 cm
    Also, EB = AB - AE = 36 - 12 = 24 cm
    Now, in Δ EBC
    S = `((15 + 15 + 24)/2)`
    = `(54/2 )`
    = 27
    = `sqrt(27 × 12 × 12 × 3) ` = `sqrt(3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 3 × 3)` = 3 × 3 × 3 × 2 × 2 = 108 cm2

    Draw CP ⊥ EB.
    Area of ΔEBC = `(1/2)` × EB × CP
    108 = `(1/2)` × 24 × CP
    `(108/12)` = CP
    ⇒ CP = 9 cmTherefore, h = 9 cm

    Now, area of triangle = `sqrt(s(s - a) (s - b) (s - c))`
    = `sqrt(27 (27 - 15) (27 - 15 ) (27 - 24)) `

    Now, area of trapezium = `(1/2)`(p1 + p2) × h
    = `1/2 × 48 × 9`
    = 216 cm2

    4. The area of a trapezium is 165 cm2 and its height is 10 cm. If one of the parallel sides is double of the other, find the two parallel sides.

    Solution:

    Let one side of trapezium is x, then other side parallel to it = 2x
    Area of trapezium = 165 cm2
    Height of trapezium = 10 cm
    Now, area of trapezium = `1/2` (p1 + p2) `× h`
    ⇒ 165 = `1/2(x + 2x) × 10`
    ⇒ 165 = 3x × 5
    ⇒ 165 = 15x
    ⇒ x = `(165)/(15)`
    ⇒ x = 11
    Therefore, 2x = 2 × 11 = 22
    Therefore, the two parallel sides are of length 11 cm and 22 cm.

Area of a General Quadrilateral
    Quadrilateral form a Parallelogram:

    Statement of the Theorem: Prove that the lines joining the middle points of the adjacent sides of a quadrilateral form a parallelogram.

    Proof: Let ABCD be a quadrilateral and length of its side AB is 2a.

    mensuration4

    Let us choose origin of rectangular Cartesian co-ordinates at the vertex A and x-axis along the sideAB and AY as the y-axis. Then, the co-ordinates of A and B are (0, 0) and (2a, 0) respectively. Referred to the chosen axes, let (2b, 2c) and (2d, 2e) be the co-ordinates of the vertices C and D respectively. If J, K, L, M be the mid-points of the sides AB, BC, CD, and, DA, respectively, then the co-ordinates of J, K, L and M are (a, 0 ), (a + b, c), (b + d, c + e) and (d, e) respectively.

    Now, the co-ordinates of the mid-point of the diagonal JL of the quadrilateral JKLM are `((a + b + d)/2, (c + e)/2)`

    Again, the co-ordinates of the mid-point of the diagonal MK of the same quadrilateral are `((a + b + d)/2, (c + e)/2)`.

    Clearly, the diagonals JL and MK of the quadrilateral JKLM bisect each other at `((a + b + d)/2, (c + e)/2)`. Hence, the quadrilateral JKLM is a parallelogram. Proved.

    Construction of Quadrilaterals:

    We consider the following five cases and explain the construction in each case by an example.
    We divide the required quadrilateral into two triangles which can be easily constructed.
    These two triangles together will form a quadrilateral.

    I. Construction of quadrilaterals when four sides and one diagonal are given:

    1. Construct a quadrilateral ABCD in which AB = 4.8 cm, BC = 4.3 cm, CD = 3.6 cm, AD = 4.2 cm and diagonal AC = 6 cm.

    Solution:

      mensuration5

    Solution:

    First we draw a rough sketch of quadrilateral ABCD and write down its dimensions, as shown. We may divide it into two triangles, namely ΔABC and ΔACD.(Rough Sketch) →

    Steps of Construction:

    mensuration6

    Step 1: Draw AB = 4.8 cm.
    Step 2: With A as center and radius equal to 6 cm, draw an arc.
    Step 3: With B as center and radius equal to 4.3 cm, draw another arc, cutting the previous arc at C.
    Step4: Join BC.
    Step 5: With A as center and radius equal to 4.2 cm, draw an arc.
    Step 6: With C as center and radius equal to 3.6 cm, draw another arc, cutting the previous arc at D.
    Step 7: Join AD and CD.

    Then, ABCD is the required quadrilateral.

    II. Construction of quadrilaterals when three sides and two diagonals are given:

    2. Construct a quadrilateral ABCD in which AB = 4 cm BC = 3.8 cm, AD = 3 cm, diagonal AC = 5 cm and diagonal BD = 4.6 cm.

    mensuration7

    Solution:

    First we draw a rough sketch of quadrilateral ABCD and write down its dimensions, as shown. We may divide it into two triangles, namely ΔABC and ΔABD.(Rough Sketch) →

    Steps of Construction:

    mensuration8

    Step 1: Draw AB = 4 cm.
    Step 2: With A as center and radius equal to 5 cm, draw an arc.
    Step 3: With B as center and radius equal to 3.8 cm, draw another arc, cutting the previous arc at C.
    Step 4: Join BC.
    Step 5: With A as center and radius equal to 3 cm, draw an arc.
    Step 6: With B as center and radius equal to 4.6 cm draw another arc, cutting the previous arc at D.
    Step 7: Join AD and CD.

    Then, ABCD is the required quadrilateral.

    III. Construction of quadrilaterals when three sides and two included angles are given:

    3. Construct a quadrilateral ABCD in which AB = 3.6 cm, ∠ABC = 80°, BC = 4 cm, ∠BAD = 120° and AD = 5 cm.

    mensuration9

    Solution:

    First we draw a rough sketch of quadrilateral ABCD and write down its dimensions, as shown (Rough Sketch) →

    Steps of Construction:

    mensuration10

    Step 1: Draw AB = 3.6 cm.
    Step 2: Make ∠ABX = 80°.
    Step 3: With B as center and radius equal to 4 cm, draw an arc, cutting BX at C.
    Step 4: Make ∠BAY = 120°.
    Step 5: With A as center and 5 cm as radius, draw an arc, cutting AY at D. Step 6: Join CD.

    Then, ABCD is the required quadrilateral.

    IV. Construction of quadrilaterals when two adjacent sides and three angles are given:

    4. Construct a quadrilateral PQRS in which PQ = 4.5 cm ∠PQR = 120°, QR = 3.8 cm, ∠QRS = 100° and ∠QPS = 60°.

    mensuration11

    Solution:

    First we draw a rough sketch of quadrilateral PQRS and write down its dimensions, as shown. (Rough Sketch) →

    Steps of Construction:

    mensuration12

    Step 1: Draw PQ = 4.5 cm.
    Step 2: Make ∠PQX = 120°.
    Step 3: With Q as center and radius 3.8 cm, draw an arc, cutting QX at R. Join QR.
    Step 4: Make ∠QRY = 100°.
    Step 5: ∠QPZ = 60° so that PZ and RY intersect each other at the point S.

    Then, PQRS is the required quadrilateral.

    Area of a Triangle Given 3 Points:

    Solving the problems on area of a triangle given 3 points with the help of the formula, in the below examples use the formula to find the area of a triangle given 3 points

    The area of a triangle formed by joining the points (x1, y1), (x2, y2) and (x3, y3) is

    =`1/2` |y1 (x2 - x3) + y2 (x3 - x1) + y3 (x1 - x2)| sq. units

Worked-out problems to find the area of a triangle given 3 points:

    1. Find the value of x for which the area of the triangle with vertices at (-1, -4), (x, 1) and (x, -4) is `12 1/2` sq. units.

    Solution:

    The area of the triangle with vertices at (-1, -4), (x, 1) and (x, -4) is `1/2` |(- 1 - 4x - 4x) - (- 4x + x + 4)|
    = `1/2` |- 1 - 8x + 3x - 41 = `1/2` |- 5x - 5| sq. units.

    By problem, `1/2`|-1 - 5x - 5| = `12 1/2` = `(25)/2`
    Therefore, 5x + 5 = ± 25
    or, x + 1 = ± 5

    Therefore, x = 4 or, - 6.

    2. The point A, B, C have respective co-ordinates (3, 4), (-4, 3) and (8, -6). Find the area of Δ ABC and the length of the perpendicular from A on BC.

    Solution:

    The required area of the triangle ABC.
    = `1/2` |(9 + 24 + 32) - (- 16 + 24 - 18)| sq. unites.
    = `1/2` |65 + 10| sq. units = `(75)/2 `sq. units.

    Again, BC = distance between the points B and C
    = `sqrt[(8 + 4)^2 + (- 6 - 3)^2] = sqrt[44 + 81] = sqrt225 = 15` units.

    Let p be the required length of the perpendicular from A on BC then,
    `1/2 BC xx p` = area of the triangle ABC
    or, `1/2 xx 15 xx p = (75)/2`
    or, p = 5

    Therefore, the required length of the perpendicular from A on BC is 5 units.

    3. The point A, B, C, D have respective co-ordinates (-2, -3), (6, -5) , (18, 9) and (0, 12). Find the area of quadrilateral ABC.

    Solution:

      We have, the area of the triangle ABC
      = `1/2 |(10 + 54 - 54) - (- 18 - 90 - 18)|` sq. units
      = `1/2 (10 + 126)` sq. units
      = 68 sq. units.

      Again, area of the triangle ACD
      = `1/2 |(- 18 + 216 + 0) - (- 54 + 0 - 24)|`sq. units
      = `1/2 (198 + 78)` sq. units
      = 138 sq. units.

      Therefore, the required area of the quadrilateral ABCD
      = area of the Δ ABC + area of the ΔACD
      = (68 + 138) sq. units
      = 206 sq. units.

    Alternative Method

    This method is analogous with the short-cut method of getting the area of a triangle. Suppose, we want to find the area of the quadrilateral whose vertices have co-ordinates (x1, y1), (x2, y2), (x3, y3) and (x4, y4). For this, we write the co-ordinates of the vertices in four rows repeating the first written co-ordinates in the fifth row. Now take the sum of the products of digits shown by (↘) and from this sum subtract the sum of the products of digits shown by (↗). The required area of the quadrilateral will be equal to half of the difference obtained. Thus, the area of the quadrilateral

    `1/2` |(x1y2 + x2y3 + x3y4 + x4y1) - (x2y1 + x3y2 + x4y3 + x1y4)| sq. units.

    The above method can be used to find the area of a polygon of any number of sides when the co-ordinates of its vertices are given.]

    Solution:

      The required area of the quadrilateral ABCD

      = `1/2 |(10 + 54 + 216 + 0) - (- 18 - 90 + 0 - 24)|` sq. units.
      = `1/2 (280 + 132)` sq. units.
      = `1/2 xx 412 `sq. units.
      = 206 sq. units.

    4. The co-ordinates of the points A, B, C, D are (0, -1), (-1, 2), (15, 2) and (4, -5) respectively. Find the ratio in which AC divides BD.

    Solution:

      Let us assume that the line-segment AC divides the line -segment BD in the ratio m : n at P.
      Therefore, P divides the line-segment BD in the ratio m : n. Hence, the co-ordinates of P are.

      `[(m xx 4 + n xx (-1))/(m + n), (m xx (-5) + n xx 2)/(m + n)] + [(4m - n)/(m + n), (5m + 2n)/(m + n)]`.

      Clearly, the points A, C and P are collinear. Therefore, the area of the triangle formed by the point A, C and P must be zero.

      Therefore, `1/2 [( ((15) (- 5m + 2n))/(m + n) - (4m - n)/(m + n) ) - ((- 15 + 2 xx (4m - n))/((m + n) + 0))]` = 0
      or, `[(15 (-5m + 2n))/(m + n) - (4m - n)/(m + n) + ((15 - 2) (4m - n))/(m + n)]=0`
      or, - 75m + 30n - 4m + n + 15m + 15n - 8m + 2n = 0.
      or, - 72m + 48n = 0
      or, 72m = 48n
      or, `m/n = 2/3`.

      Therefore, the line-segment AC divides the line-segment BD internally in the ratio 2 : 3.

    5. The polar co-ordinates of the vertices of a triangle are `(-a, pi/6), (a, pi/2) and (-2a, - (2pi)/3)` find the area of the triangle.

    Solution:

      The area of the triangle formed by joining the given points

      = `1/2 |a xx (-2a) sin(- (2pi)/3 - pi/2) + (-2a) (-a) sin(pi/6 + (2pi)/3) - (-a) xx a sin(pi/6 + pi/2)|` sq. units. [ using above formula]
      = `1/2 |2a^2 sin(pi + pi/6) + 2a^2 sin(pi - pi/6) -2a^2 sin(pi/2 - pi/6)|`sq. units.
      = `1/2 |-2a^2 sin(pi/6) + 2a^2 sin(pi/6) - a^2 cos(pi/6)|` sq. units
      . = `1/2xx a^2xx (sqrt3/2)` sq. units
      = `(sqrt3/4) a^2` sq. units.

    6. The centre of a circle is at (2, 6) and a chord of this circle of length 24 units is bisected at (- 1, 2). Find the radius of the circle.

    Solution:

      Let C (2, 6) be the centre of the circle and its chord AB of length 24 units is bisected at D (- 1, 2).

      Therefore, CD2 = (2 + 1)2 + (6 - 2)2
      = 9 + 16 = 25 and DB = `1/2 xx AB =1/2 xx 24 = 12 `

      Join CB. Now, D is the mid-point of the chord AB; hence, CD is perpendicular to AB. Therefore, from the triangle BCD we get,
      BC2 = CD2 + BD2 = 25 + 122 = 25 + 144 = 169
      or, BC = 13

      Therefore, the required radius of the circle = 13 units.

    7. If the co-ordinates of the vertices of a Δ ABC be (3, 0), (0, 6) and (6, 9) and if D and E divide AB and AC, respectively internally in the ratio 1 : 2, then show that the area of Δ ABC = 9 . the area of ΔADE.

    Solution:

      By question D divides AB internally in the ratio 1 : 2; hence, the co-ordinates of D are
      `((1 xx 0 + 2 xx 3)/(1 + 2), (1 xx 6 + 2 xx 0)/(1 + 2)) = (6/3, 6/3) = (2, 2)`.

      Again, E divides AC internally in the ratio 1 : 2; hence, the co-ordinates of E are
      `((1 xx 6 + 2 xx 3)/(1 + 2), (1 xx 9 + 2 xx 0)/(1 + 2)) = ((12)/3, 9/3) = (4, 3)`.

      Now, the area of the triangle ABC
      = `1/2 |(18 + 0 + 0) - (0 + 36 + 27)|` sq. units.
      = `1/2 |18 - 63|` sq. units.
      = `(45)/2` sq. units.

      And the area of the triangle ADE
      = `1/2 |( 6 + 6 + 0) - (0 + 8 + 9)|` sq. units.
      = `1/2 |12 - 17|` sq. units.
      = `5/2` sq. units.

      therefore, area of the Δ ABC
      = `(45)/2` sq. units = `9 xx 5/2` sq. units.
      = 9. area of the Δ ADE. Proved.

    The above worked-out problems on area of a triangle given 3 points are explained step-by-step with the help of the formula.

    Area of special quadrilaterals :

      We can use the same method of splitting into triangles (which we called "triangulation")
      To find a formula for the area of a rhombus.

      mensuration13
      Fig 11.13

      In Fig 11.13 ABCD is a rhombus.
      Therefore, its diagonals are perpendicular bisectors of each other.
      Area of rhombus ABCD = (area of Δ ACD) + (area of Δ ABC) Fig 11.11
      Fig 11.10 Fig 11.12 MENSURATION 175 TRY THESE Fig 11.14
      THINK, DISCUSS AND WRITE
      = ( `1/2` × AC × OD) + ( `1/2` × AC × OB)
      = `1/2` AC × (OD + OB)
      = `1/2` AC × BD
      = `1/2` d1 × d2
      where AC = d1 and BD = d2 In other words,
      area of a rhombus is half the product of its diagonals.

    Example 2: Find the area of a rhombus whose diagonals are of lengths 10 cm and 8.2 cm.

    Solution:

      Area of the rhombus = `1/ 2` d1 d2
      where d1 , d2 are lengths of diagonals.
      = 1 2 × 10 × 8.2 cm2 = 41 cm2

Area of a Polygon

    Polygon: A figure bounded by four or more straight lines is called a polygon.

    Regular Polygon: A polygon is said to be regular when all its sides are equal and all its angles are equal.

    A polygon is named according to the number of sides it contains.

Given below are the names of some polygons and the number of sides contained by them.

  • Quadrilateral - 4
  • Pentagon - 5
  • Hexagon - 6
  • Heptagon - 7
  • Octagon - 8
  • Nonagon - 9
  • Decagon - 10
  • Undecagon - 11
  • Dodecagon - 12
  • Quindecagon - 15

    Central Point of a Polygon:

    The inscribed and the circumscribed circles of a polygon have the same centre, called the central point of the polygon.

    Radius of the Inscribed Circle of a Polygon:

    The length of perpendicular from the central point of a polygon upon any one of its sides, is the radius of the inscribed circle of the polygon.
    The radius of the inscribed circle of a polygon is denoted by r.

    Radius of the Circumscribed Circle of a Polygon:

    The line segment joining the central point of a polygon to any vertex is the radius of the circumscribed circle of the polygon. The radius of the circumscribed circle of a polygon is denoted by R.
    In the figure given below, ABCDEF is a polygon having central point O and one of its sides a unit. OL ⊥ AB.
    Then, OL = r and OB = R

    Area of a polygon of n sides

    = n × (area ΔOAB) = n × `1/2` × AB × OL
    = (`n/2` × a × r)

    mensuration14

    Now, A = `1/2` nar ⇔ a = `(2A)/(nr)` ⇔ na = `(2A)/r` ⇔ Perimeter = `(2A)/r`
    From right ΔOLB, we have:

    OL2 = OB2 - LB2 ⇔ r2 = `{R^2 - (a/2)^2}`
    `<=> r = sqrt(R^2 - (a^2/4)`

    Therefore, area of the polygon =` {n/2 xx a xx sqrt(R^2 - a^2/4)}` square units.

    n area of a polygon some of the particular cases such as;

    (i) Hexagon:

    mensuration15

    OL2 = (OB2 - LB2)
    = `{a^2 - (a/2)^2} = (a^2 - a^2/4) = (3a^2)/4`
    `=> OL = {sqrt3/2 xx a}`

    ⇒ Area ΔOAB =`1/2 `× AB × OL
    = `{1/2 xx a xx sqrt3/2 xx a}`
    = `((sqrt3)a^2)4`

    ⇔ area of hexagon ABCDEF = `{6 xx (sqrt3)a^2/4}` square units
    = `{3(sqrt3)a^2/2}` square units.

    Therefore, area of a hexagon = `{3(sqrt3)a^2/2}` square units.

    (ii) Octagon:

    mensuration16

    BM is the side of a square whose diagonal is BC = a.
    Therefore, `BM = a/sqrt2`.

    Now, OL = ON + LN
    = `ON + BM = (a/2 + a/sqrt2)`

    ⇔ Area of given octagon
    = 8 × area of ΔOAB = 8 × `1/2` × AB × OL
    = `4 xx a xx (a/2 + a/sqrt2) = 2a^2 (1 + sqrt2)` square units.

    Therefore, area of an octagon = `2a^2 (1 + sqrt2)` square units.

    We will solve the examples on different names of the area of a polygon.

    1. Find the area of a regular hexagon each of whose sides measures 6 cm.

    Solution:

    Side of the given hexagon = 6 cm.

    Area of the hexagon = `{3sqrt(3)a^2/2}` cm2
    = `(3 xx 1.732 xx 6 xx 6)/2 cm^2`
    = `93.528 cm^2`.

    2. Find the area of a regular octagon each of whose sides measures 5 cm.

    Solution:

    Side of the given octagon = 5 cm.

    Area of the octagon = `[2a^2 (1 + sqrt2)]` square units
    = [2 × 5 × 5 × (1 + 1.414)] cm2
    = (50 × 2.414) cm2
    = 120.7 cm2.

    3. Find the area of a regular pentagon each of whose sides measures 5 cm and the radius of the inscribed circle is 3.5 cm.

    Solution:

    Here a = 5 cm, r = 3.5 cm and n = 5.

    Area of the pentagon = `(n/2× a × r)` square units
    = `(5/2 xx 5 xx 7/2) cm^2`
    = `43.75 cm^2`.

    4. Each side of a regular pentagon measures 8 cm and the radius of its circumscribed circle is 7 cm. Find the area of the pentagon.

    Solution:

    Area of the pentagon = `(n/2 xx a xx sqrt(R^2 - a^2/4))` square units
    = `(5/2 xx 8 xx sqrt(72 - 64/4)) cm^2`
    = `(20 xx sqrt(49 - 16)) cm^2`
    = `(20 xx sqrt33) cm^2`
    = `(20 xx 5.74) cm^2`
    = `(114.8) cm^2`.

Area of Rhombus

    ABCD is a rhombus whose base AB = b, DB ⊥ AC DB = d1 AC = d2 and the altitude from C on AB is CE, i.e., h.

    mensuration17

    Area of rhombus ABCD = 2 Area of Δ ABC
    = 2 × `1/2` AB × CD sq units.
    = 2 × `1/2` b × h sq. units
    = base x height sq. units

    Also, area of rhombus = 4 × area of Δ AOB
    = 4 × `1/2` × AO × OB sq. units
    = 4 × `1/2`× `1/2` d2 × `1/2` d1 sq. units
    = 4 × `1/8` d1 × d2 square units
    = `1/2` × d1 × d2; where d1 and d2 are diagonals.
    Therefore, area of rhombus = `1/2` (product of diagonals) square units

    Perimeter of rhombus = 4 × side

    Worked-out examples on area of rhombus:

    1. Find the area of the rhombus having each side equal to 17 cm and one of its diagonals equal to 16 cm.

    Solution:

    mensuration18

    ABCD is a rhombus in which AB = BC = CD = DA = 17 cm
    AC = 16 cm
    Therefore, AO = 8 cm
    In Δ AOD,
    AD2 = AO2 + OD2
    ⇒ 172 = 82 + OD2
    ⇒ 289 = 64 + OD2
    ⇒ 225 = OD2
    ⇒ OD = 15

    Therefore, BD = 2 OD
    = 2 × 15
    = 30 cm

    Now, area of rhombus
    = `1/2` × d1 × d2
    = `1/2` × 16 × 30
    = 240 `cm^2`

    2. Find the altitude of the rhombus whose area is 315 cm2 and its perimeter is 180 cm.

    Solution:

    Since, the perimeter of rhombus = 180 cm
    Therefore, side of rhombus = `P/4 = (180)/4`= 45 cm
    Now, area of rhombus = b × h
    ⇒ 315 = 45 × h
    ⇒ `h = (315)/(45) `
    ⇒ h =7 cm

    Therefore, altitude of the rhombus is 7 cm.

    3. The floor of building consists of 2000 tiles which are rhombus shaped and each of its diagonals are 40 cm and 25 cm in length. Find the total cost of polishing the floor, if the cost per m2 is $5.

    Solution:

    In each rhombus, the tile length of the diagonals = 40 cm and 25 cm
    Therefore, area of each tile = `1/2` × 40 × 25 = 500 cm2
    Therefore, area of 2000 tiles = 2000 × 500 cm2
    = 1000000 cm2
    = `(1000000)/(10000) `cm2
    = 100 m2

    For 1 m2 cost of polishing = $5 = $5 × 100 = $500.

Common Solid Figures

    In third grade geometry we will discuss in brief about some of the common solid figures named below:

    (i) Cube: Definition of cube, parts of a cube, properties of a cube.
    (ii) Cuboid: Definition of cuboid
    (iii) Cylinder: Definition of cylinder
    (iv) Cone: Definition of cone
    (v) Sphere: Definition of sphere

    Definition of a cube:

    An object which looks like solid box-shaped that has six identical square faces.
    A cube has 6 equal and plane surfaces. All the faces of a cube are square in shape.
    In a cube there are 6 plane surfaces. There are 8 vertices and 12 edges.

    mensuration19

    Two adjoining plane - surfaces meet at an edge. There are 12 edges in a cube and all the 12 edges are equal in length. These edges are straight edges.

    The meeting point of two edges is called a vertex. In a cube there are 8 such vertices.

    Parts of a cube:

    (i) Face: Face is also known as sides. A cube has six faces and all the faces of a cube are square in shapes. Each face has four equal sides.

    (ii) Edge: When two edges meet each other a line segment formed. There are 12 edges in a cube. All the 12 edges are equal in length because all faces are squares. These edges are straight edges.

    (iii) Vertex: When three edges meet each other a point formed. There are 8 vertices in a cube.

    (iv) Face Diagonals: Face Diagonals of a cube is the line segment that joins the opposite

    vertices of a face. There are 2 diagonals in each face so altogether there are 12 diagonals in the cube.

    (v) Space Diagonals: Space diagonals of a cube are the line segment that joins the opposite vertices of a cube, cutting through its interior. There are 4 space diagonals in a cube.

    Properties of a cube:

    Volume: The volume of a cube is s3 where s is the length of one edge.

    Surface Area: The surface area of a cube is 6s2, where s is the length of one edge.

    Definition of cuboid:

    The cuboid has 6 rectangular faces. The opposite rectangular plane surfaces are identical (equal in all respects). It has 8 vertices and 12 edges.
    In a cuboid there are 6 rectangular plane surfaces. There are 8 vertices and 12 edges.

    mensuration20

    A cube is also a cuboid having all its 6 faces equal and square. Thus, a cube has all the six faces identical, whereas a cuboid has the opposite faces identical.

    Properties of a cuboid:

    Formulas for the above rectangular box:

    Volume: The volume of a cuboid is lwh, where l is the length, w is the width and h is the height.

    Lateral Surface Area: The lateral surface area of a cuboid is 2lh + 2wh, where l is the length, w is the width and h is the height.

    Surface Area:The surface area of a cuboid is 2lw + 2lh + 2wh, where l is the length, w is the width and h is the height.

    Definition of cylinder:

    A cylinder stands on a circular plane surface having circular plane surfaces on its top and bottom. Thus a cylinder has two circular plane surfaces, one at its base and another at its top. It has a curved surface in the middle.
    It has two edges, at which the two plane surfaces meet with the curved surface. These edges are curved edges.
    In a cylinder there are 2 plane surfaces and 1 curved surface. There are 2 edges and no vertices.

    mensuration21

    The base and top of a cylinder are of the same shape (circular) and size. Thus, both are identical.

    Definition of cone:

    A cone has one plane circular surface, i.e. its base and only one curved surface. In a cone there is 1 plane surface and 1 curved surface. There are 1 edge and 1 vertex.

    mensuration22

    It has one edge which is formed by the circular plane surface meeting with the curved surface. The edge of a cone is a curved edge.

    Definition of sphere:

    The ball-like shape is called a sphere. In sphere there is curve surface, no edge and no vertex.

    mensuration23

    Some of the common solid figures are explained above in brief with the labelled diagram to get the basic ideas of the solid shapes.

    Note:

      (i) The surface of solid is called face.
      (ii) Edge is formed when two faces meet.
      (iii) The point where 3 faces meet is called the corner.

    Volume and Capacity:

    There is not much difference between these two words.

    (a) Volume refers to the amount of space occupied by an object.
    (b) Capacity refers to the quantity that a container holds.

    Note: If a water tin holds 100 cm3 of water then the capacity of the water tin is 100 cm3 . Capacity is also measured in terms of litres. The relation between litre and cm3 is, 1 mL = 1 cm3 ,1 L = 1000 cm3 . Thus, 1 m3 = 1000000 cm3 = 1000 L.

    Example 1: Find the height of a cuboid whose volume is 275 cm3 and base area is 25 cm2 .

    Solution:

    Volume of a cuboid = Base area × Height
    Hence height of the cuboid = (Volume of cuboid) /(Base area)
    =`(275)/(25)`
    = 11 cm Height of the cuboid is 11 cm.

    Example 2: A godown is in the form of a cuboid of measures 60 m × 40 m × 30 m.
    How many cuboidal boxes can be stored in it if the volume of one box is 0.8 m3 ?

    Solution:

    Volume of one box = 0.8 m3
    Volume of godown = 60 × 40 × 30 = 72000 m3
    Number of boxes that can be stored in the godown = Volume of the godown / Volume of one box
    = 60 × 40 × 30 0.8
    = 90,000

    Hence the number of cuboidal boxes that can be stored in the godown is 90,000

Surface Area of Cube, Cuboid and Cylinder
Surface Area of Cube

Imran, Monica and Jaspal are painting a cuboidal, cubical and a cylindrical box respectively of same height

They try to determine who has painted more area. Hari suggested that finding the surface area of each box would help them find it out.

To find the total surface area, find the area of each face and then add. The surface area of a solid is the sum of the areas of its faces. To clarify further, we take each shape one by one

Cuboid

Suppose you cut open a cuboidal box and lay it flat Fig . We can see a net as shown below (Fig ).

Write the dimension of each side. You know that a cuboid has three pairs of identical faces. What expression can you use to find the area of each face?

Find the total area of all the faces of the box. We see that the total surface area of a cuboid is area I + area II + area III + area IV +area V + area VI
=` (h xx l + b xx l + b xx h + l xx h + b xx h + l × b)`
So total surface area =` 2 (h xx l + b xx h + b xx l)
= 2(lb + bh + hl)`

where h, l and b are the height, length and width of the cuboid respectively.

Suppose the height, length and width of the box shown above are 20 cm, 15 cm and 10 cm respectively.
Then the total surface area = `2 (20 ×x 15 + 20 × 10 + 10 × 15)` =` 2 ( 300 + 200 + 150)
= 1300 m^2`.

Cube

Draw the pattern shown on a squared paper and cut it out [Fig ]. (You know that this pattern is a net of a cube. Fold it along the lines [Fig ] and tape the edges to form a cube [Fig].

  1. What is the length, width and height of the cube? Observe that all the faces of a cube are square in shape. This makes length, height and width of a cube equal
  2. Write the area of each of the faces. Are they equal?
  3. Write the total surface area of this cube.
  4. If each side of the cube is l, what will be the area of each face?.

Can we say that the total surface area of a cube of side l is 6l2 ?

Cylinders

Most of the cylinders we observe are right circular cylinders. For example, a tin, round pillars, tube lights, water pipes etc.

Example 1: An aquarium is in the form of a cuboid whose external measures are 80 cm × 30 cm × 40 cm. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed?

Solution:
The length of the aquarium = l = 80 cm
Width of the aquarium = b = 30 cm
Height of the aquarium = h = 40 cm
Area of the base =` l xx b = 80 xx 30 = 2400 cm^2`
Area of the side face = `b xx h = 30 xx 40 = 1200 cm^2`
Area of the back face = `l xx h = 80 xx 40 = 3200 cm^2`
Required area = Area of the base + area of the back face + (2 × area of a side face)
= `2400 + 3200 + (2 xx 1200) = 8000 cm^2`

Hence the area of the coloured paper required is `8000 cm^2`.

Volume of Cube, Cuboid and Cylinder

Amount of space occupied by a three dimensional object is called its volume. Try to compare the volume of objects surrounding you. For example, volume of a room is greater than the volume of an almirah kept inside it. Similarly, volume of your pencil box is greater than the volume of the pen and the eraser kept inside it. Can you measure volume of either of these objects?

Remember, we use square units to find the area of a region. Here we will use cubic units to find the volume of a solid, as cube is the most convenient solid shape (just as square is the most convenient shape to measure area of a region).

For finding the area we divide the region into square units, similarly, to find the volume of a solid we need to divide it into cubical units.
Observe that the volume of each of the adjoining solids is 8 cubic units (Fig).
We can say that the volume of a solid is measured by counting the number of unit cubes it contains. Cubic units which we generally use to measure volume are

1 cubic `cm = 1 cm × 1 cm × 1 cm = 1 cm^3`
=> = `10 mm × 10 mm × 10 mm = ............... mm^3`
1 cubic `m = 1 m × 1 m × 1 m = 1 m^3`
=> = ............................... `cm^3`
1 cubic `mm = 1 mm × 1 mm × 1 mm = 1 mm^3`
=> = `0.1 cm × 0.1 cm × 0.1 cm = ...................... cm^3`

We now find some expressions to find volume of a cuboid, cube and cylinder. Let us take each solid one by one.

Cuboid

Take 36 cubes of equal size (i.e., length of each cube is same). Arrange them to form a cuboid. You can arrange them in many ways. Observe the following table and fill in the blanks.

Cube

The cube is a special case of a cuboid, where l = b = h. Hence, volume of cube = `l × l × l = l ^3`

Cylinder

We know that volume of a cuboid can be found by finding the product of area of base and its height. Can we find the volume of a cylinder in the same way?

Just like cuboid, cylinder has got a top and a base which are congruent and parallel to each other. Its lateral surface is also perpendicular to the base, just like cuboid.
So the Volume of a cuboid = area of base × height
= l × b × h = lbh
Volume of cylinder = area of base × height
=` pir^2 × h = pir^2h`

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