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A figure is broken into a rectangle and a triangle. The triangle has a base of 2 and two-thirds feet and height of 3 feet. The rectangle has a base of 5 feet and height of 1 and one-third feet.
The irregular figure can be broken into a triangle and a rectangle as shown with the dashed line.

The length of b, the base of the triangle, is
ft.
The area of the triangle is

The area of the rectangle is

The area of the irregular figure is

A figure is broken into a rectangle and a triangle The triangle has a base of 2 and twothirds feet and height of 3 feet The rectangle has a base of 5 feet and h class=

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mhanifa

Answer:

  • See below

Step-by-step explanation:

The length of b, the base of the triangle, is

  • 5 - 2 1/3 = 3 - 1/3 = 2 2/3 ft

The area of the triangle is

  • A = 1/2*3*(2 2/3) = (3/2)(8/3) = 8/2 = 4 ft²

The area of the rectangle is

  • 1 1/3 * 5 = 4/3 * 5 = 20/3 = 6 2/3 ft²

The area of the irregular figure is

  • 4 + 6 2/3 = 10 2/3 ft²
semsee45

Answer:

[tex]b=2\frac23\:\sf ft[/tex]

Area of triangle = 4 ft²

Area of rectangle = [tex]6\frac23\: \sf ft^2[/tex]

Area of irregular figure = [tex]10\frac23\: \sf ft^2[/tex]

Step-by-step explanation:

[tex]\begin{aligned}\sf Base\:of\:triangle\:(b) & = 5-2\frac13\\\\& = \dfrac{15}{3}-\dfrac73\\\\ & = \dfrac83\\\\ & = 2\frac23\:\sf ft \end{aligned}[/tex]

[tex]\begin{aligned}\sf Area\:of\:a\:triangle & =\dfrac12 \sf \times base \times height\\\\& = \dfrac12 \times b \times 3\\\\ & = \dfrac12 \times \dfrac83 \times \dfrac31\\\\ & = \dfrac{24}{6}\\\\ & = 4\: \sf ft^2\end{aligned}[/tex]

[tex]\begin{aligned}\sf Area\:of\:rectangle& =\sf length \times width\\\\& = 5 \times 1\frac13\\\\ & = \dfrac51\times \dfrac43\\\\& = \dfrac{20}{3}\\\\ & = 6\frac23\: \sf ft^2\end{aligned}[/tex]

[tex]\begin{aligned} \sf Area\:of\:irregular\:figure & = \sf area\:of\:triangle+area\:of\:rectangle\\\\ & = 4 + 6\frac23\\\\ & = 10\frac23\: \sf ft^2\end{aligned}[/tex]

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