Question

Find the area of the shape below

Answer 1

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