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Find the area of the shape below

Find the area of the shape below-example-1
User Erowlin
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Answer:

Q1) 104m²

Q2) 121 in.²

Explanation:

Q1)

GIVEN :-

  • Figure given in the question comprises of a triangle & a rectangle.
  • Length of the rectangle = 10 m
  • Width of the rectangle = 8 m
  • Base of the triangle = 8 m
  • Height of the triangle = 6 m

TO FIND :-

  • Area of the figure

GENERAL FORMULAES TO BE USED IN THIS QUESTION :-

  • For a triangle with height 'h' & base 'b' , its area =
    (1)/(2) * b * h.
  • For a rectangle with length 'l' & width 'w' , its area =
    l * w

SOLUTION :-

Area of the triangle =
(1)/(2) * 8 * 6 = 24m^2

Area of the rectangle =
10 * 8 = 80m^2

Area of the figure = (Area of the triangle) + (Area of the rectangle)

= 24m²+ 80m²

= 104m²

Q2)

GIVEN :-

  • Figure given in the question comprises of a triangle & a trapezium.
  • Lengths of parallel sides of trapezium are 6 in. & 7 in.
  • Height of trapezium = 20 - 3 = 17 in.
  • Base of the triangle = 3 in.
  • Height of the triangle = 7 in.

TO FIND :-

  • Area of the figure

GENERAL FORMULAES TO BE USED IN THIS QUESTION :-

  • For a triangle with height 'h' & base 'b' , its area =
    (1)/(2) * b * h.
  • For a trapezium with parallel sides whose lengths are 'a' & 'b' and height 'h' , its area =
    (1)/(2) * (a +b) * h

SOLUTION :-

Area of the triangle =
(1)/(2) * 3 * 7 = (21)/(2) = 10.5 \; in.^2

Area of the trapezium =
(1)/(2) * (6 +7) * 17 = (221)/(2) = 110.5\; in^2

Area of the figure = (Area of the triangle) + (Area of the trapezium)

= 10.5 in.² + 110.5 in.²

= 121 in.²

User Raedawn
by
7.6k points

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