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Find the area of the shape below

Find The Area Of The Shape Below class=

Sagot :

Answer:

Q1) 104m²

Q2) 121 in.²

Step-by-step 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 = [tex]\frac{1}{2} \times b \times h[/tex].
  • For a rectangle with length 'l' & width 'w' , its area = [tex]l \times w[/tex]

SOLUTION :-

Area of the triangle = [tex]\frac{1}{2} \times 8 \times 6 = 24m^2[/tex]

Area of the rectangle = [tex]10 \times 8 = 80m^2[/tex]

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 = [tex]\frac{1}{2} \times b \times h[/tex].
  • For a trapezium with parallel sides whose lengths are 'a' & 'b' and height 'h' , its area = [tex]\frac{1}{2} \times (a +b) \times h[/tex]

SOLUTION :-

Area of the triangle = [tex]\frac{1}{2} \times 3 \times 7 = \frac{21}{2} = 10.5 \; in.^2[/tex]

Area of the trapezium = [tex]\frac{1}{2} \times (6 +7) \times 17 = \frac{221}{2} = 110.5\; in^2[/tex]

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

                              = 10.5 in.² + 110.5 in.²

                              = 121 in.²

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