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Please help find the area of the figure above

Please Help Find The Area Of The Figure Above class=

Sagot :

Answer:

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View image ASKMEMYFRIEND

Answer:

Area of the given figure is [tex]{\boxed{\sf{\green{84}}}}[/tex] in²

Step-by-step explanation:

Firstly, finding the area of triangle by substituting the values in the formula

  • base = 6 in
  • height = 6 in

[tex]\begin{gathered} \qquad\longrightarrow{\sf{A_{(\triangle)} = \dfrac{1}{2} \times b \times h}}\\\\\qquad\longrightarrow{\sf{A_{(\triangle)} = \dfrac{1}{2} \times 6 \times 6}}\\\\ \qquad\longrightarrow{\sf{A_{(\triangle)} = \dfrac{1}{\cancel{2}} \times 6 \times \cancel{6}}}\\\\\qquad\longrightarrow{\sf{A_{(\triangle)} = 1 \times 6 \times 3}}\\\\\qquad\longrightarrow{\sf{A_{(\triangle)} = 6 \times 3}}\\\\\qquad\longrightarrow{\sf{A_{(\triangle)} = 18}}\\\\ \qquad{\star{\boxed{\sf{\purple{A_{(\triangle)} = 18 \: {in}^{2}}}}}}\end{gathered}[/tex]

Hence, the area of triangle is 18 in².

[tex]\rule{200}2[/tex]

Secondly, finding the area of rectangle by substituting the values in the formula :

  • length = 6 in
  • breadth = 11 in

[tex]\begin{gathered} \qquad{\twoheadrightarrow{\sf{A_{(Rectangle)} = l \times b}}} \\ \\ \qquad{\twoheadrightarrow{\sf{A_{(Rectangle)} = 6 \times 11}}} \\ \\ \qquad{\twoheadrightarrow{\sf{A_{(Rectangle)} = 66}}} \\ \\ \qquad{\star{\boxed{\sf{\pink{A_{(Rectangle)} = 66 \: {in}^{2}}}}}}\end{gathered}[/tex]

Hence, the area of rectangle is 66 in².

[tex]\rule{200}2[/tex]

Now, finding the total area of given polygon by substituting the values in the formula :

  • Area of triangle = 18 in²
  • Area of rectangle = 66 in²

[tex]\begin{gathered} \qquad{\rightarrow{\sf{Area = A_{( \triangle)} + A_{(Rectangle)}}}} \\ \\ \qquad{\rightarrow{\sf{Area = 18 + 66}}} \\ \\ \qquad{\rightarrow{\sf{Area = 84}}} \\ \\ \qquad{\star{\boxed{\sf{\red{Area = 84\: {in}^{2}}}}}}\end{gathered}[/tex]

Hence, the area of given figure is 84 in².

[tex]\rule{300}{2.5}[/tex]

View image AadhPrit
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