Question

Find the area of the shape below

Answer 1

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 = $\frac{1}{2} \times b \times h$.
• For a rectangle with length 'l' & width 'w' , its area = $l \times w$

SOLUTION :-

Area of the triangle = $\frac{1}{2} \times 8 \times 6 = 24m^2$

Area of the rectangle = $10 \times 8 = 80m^2$

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

SOLUTION :-

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

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

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

                              = 10.5 in.² + 110.5 in.²

                              = 121 in.²