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 :-
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 :-
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
