Answer:
[tex]\boxed{\boxed{ \sf \: Area \: of \: Shape \: C \: = 9 \: sq.in}}[/tex]
Step-by-step explanation:
Given:
Data obtained from diagram
length of rectangle X (l) = 2 in
width of rectangle X (w) = 1 in
length of rectangle Y (L) = 3 in
width of rectangle Y (W) = 2 in
Base of triangle b = 1 in
height of triangle = 2 in
To find:
Area of shape is C which is a combination of rectangle X, Y & triangle = ?
solution:
To find the overall area of shape C we will have to find each shapes area individually.
let's find out the area of Rectangle X (Ax),
[tex] \sf Area \: of \: rectangle \: = length \times width[/tex]
Substituting the value of length & width of Rectangle X in above formula,
[tex]A_x = l \times w \\ A_x =2 \times 1 \\ A_x =2 \: {in}^{2} [/tex]
[tex] \boxed{ \sf Area \: of \: Rectangle \: X = 2 \: sq. in}[/tex]
Similarly for area of rectangle Y (Ay),
[tex]A_y \: = L \times W \\ A_y =3 \times 2 \\ A_y =6 \: {in}^{2} [/tex]
[tex] \boxed{ \sf Area \: of \: Rectangle \:Y = 6 \: sq. in}[/tex]
Now, let's find out the area of triangle
[tex] \sf \: Area \: of \: triangle = \frac{1}{2} \times base \times height[/tex]
Substituting the value of base & height of triangle in above formula.
[tex]A_t = \frac{1}{2} \times1 \times 2 \\ A_t = \frac{1}{\cancel2} \times1 \times \cancel2 \\ A_t =1 \: {in}^{2} [/tex]
[tex] \boxed{ \sf Area \: of \: Triangle = 1\: sq. in}[/tex]
Now, let's find out the area of required shape C,
Area of shape C = Area of Rectangle X + Area of Rectangle Y + Area of Triangle
[tex]A_c = A_x + A_y + A_t[/tex]
[tex]A_c = A_x + A_y + A_t \\ A_c = 2 + 6 + 1 \\ A_c = 9 \:{in}^{2} [/tex]
[tex] \sf \: Answer \rightarrow The \: area \: of \: shape \: C \: is \: 9 \: {in}^{2} [/tex]
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