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Rama09 [41]
2 years ago
12

How many square feet of outdoor carpet will we need for this hole

Mathematics
1 answer:
Goshia [24]2 years ago
6 0

Answer:

30

Step-by-step explanation:

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3 years ago
Find the perimeter and the area of the figure.
Gwar [14]

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

<u>the </u><u>given </u><u>figure </u><u>is </u><u>a </u><u>composition</u><u> </u><u>of </u><u>a </u><u>rectangle</u><u> </u><u>as </u><u>well </u><u>as </u><u>a </u><u>right </u><u>angled </u><u>triangle </u><u>!</u>

<u>we've</u><u> </u><u>been </u><u>given </u><u>the </u><u>two </u><u>sides </u><u>of </u><u>the </u><u>rectangle </u><u>and </u><u>we're</u><u> </u><u>required</u><u> </u><u>to </u><u>find </u><u>out </u><u>the </u><u>height </u><u>of </u><u>the </u><u>triangle </u><u>,</u><u> </u><u>so </u><u>as </u><u>to </u><u>find </u><u>it's</u><u> </u><u>area </u><u>~</u>

<u>we </u><u>know </u><u>the </u><u>the </u><u>opposite</u><u> </u><u>sides </u><u>of </u><u>a </u><u>rectangle </u><u>are </u><u>equal</u><u> </u><u>,</u><u> </u><u>therefore </u><u>we </u><u>can </u><u>break </u><u>the </u><u>longest </u><u>side </u><u>(</u><u> </u><u>length </u><u>=</u><u> </u><u>9</u><u>.</u><u>5</u><u> </u><u>cm </u><u>)</u><u> </u><u>into </u><u>two </u><u>parts </u><u>!</u><u> </u><u>the </u><u>first </u><u>part </u><u>of </u><u>length </u><u>=</u><u> </u><u>7</u><u> </u><u>cm </u><u>which </u><u>is </u><u>the </u><u>length </u><u>of </u><u>the </u><u>rectangle </u><u>and </u><u>the </u><u>rest </u><u>2</u><u>.</u><u>5</u><u> </u><u>cm </u><u>(</u><u> </u><u>9</u><u>.</u><u>5</u><u> </u><u>-</u><u> </u><u>7</u><u> </u><u>=</u><u> </u><u>2</u><u>.</u><u>5</u><u> </u><u>)</u><u> </u><u>will </u><u>become </u><u>the </u><u>height </u><u>of </u><u>the </u><u>triangle </u><u>!</u>

<h3><u>For </u><u>perimeter</u><u> </u><u>of </u><u>the </u><u>figure </u><u>-</u></h3>

perimeter \: of \: figure = perimeter \: of \: rectangle + perimeter \: of \: triangle \\  \\

now ,

<u>perimeter</u><u> </u><u>of </u><u>rectangle </u><u>=</u><u> </u><u>2</u><u> </u><u>(</u><u> </u><u>l </u><u>+</u><u> </u><u>b </u><u>)</u>

where ,

<u>l </u><u>=</u><u> </u><u>length </u>

<u>b </u><u>=</u><u> </u><u>breadth </u>

\longrightarrow \: perimeter = 2(7 + 6) \\ \longrightarrow \: 2(13) \\ \longrightarrow \: 26 \: cm

and ,

perimeter \: of \: \triangle = 6.5 + 2.5 + 6 \\ \longrightarrow \: 15 \: cm

<u>Perimeter</u><u> </u><u>of </u><u>figure </u><u>in </u><u>total </u><u>=</u><u> </u><u>2</u><u>6</u><u> </u><u>cm </u><u>+</u><u> </u><u>1</u><u>5</u><u> </u><u>cm</u>

thus ,

\qquad\quad\bold\red{perimeter \: = \: 41 \: cm}

<h3><u>For </u><u>area </u><u>of </u><u>the </u><u>figure </u><u>-</u></h3>

area \: of \: figure = area \: of \: rectangle  + area \: of \: rectangle \\

now ,

<u>area </u><u>of </u><u>rectangle</u><u> </u><u>=</u><u> </u><u>l </u><u>×</u><u> </u><u>b</u>

where ,

<u>l </u><u>=</u><u> </u><u>length </u>

<u>b </u><u>=</u><u> </u><u>breadth</u>

area \: of \: rectangle = 7 \times 6 \\ \longrightarrow \: 42 \: cm {}^{2}

and ,

area \: of\triangle =  \frac{1}{2}  \times base \times height \\  \\ \longrightarrow \:   \frac{1}{\cancel2}  \times \cancel6 \times 2.5 \\  \\ \longrightarrow \: 3 \times 2.5 \\  \\ \longrightarrow \: 7.5 \: cm {}^{2}

<u>Area </u><u>of </u><u>figure</u><u> </u><u>in </u><u>total </u><u>=</u><u> </u><u>4</u><u>2</u><u> </u><u>cm²</u><u> </u><u>+</u><u> </u><u>7</u><u>.</u><u>5</u><u> </u><u>cm²</u>

thus ,

\qquad\quad\bold\red{Area \: = \: 49.5 \: cm^{2}}

hope helpful :)

6 0
1 year ago
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