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Mashutka [201]
1 year ago
9

Find the perimeter and the area of the figure.

Mathematics
2 answers:
Gwar [14]1 year ago
6 0

\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 :)

joja [24]1 year ago
5 0

Step-by-step explanation:

perimeter = 7 + 6 + 9.5 + 6.5

= 29

area of rectangle = 6 X 7 = 42

area of triangle = ½ X 2.5. X 6 = 7.5

total area = 42 + 7.5 = 49.5

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To solve for the A or the principal amount plus interest you can use two formulas:

A = P + I

Where: P = Principal
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or you can use 

A = P (1+ rt)

Where: P = principal
             r  = rate in decimal
             t  = time in years

With your given you can use the second one, without having to use the first. 
Given that the Principal amount is $222 and the rate is 12% and time is 10 years, we first need to convert your rate into decimal by dividing the value in percent by 100 which will yield 0.12. 

Then now we can just input the data that you know into the formula:
A = P(1+ rt)
   = $222(1 + (0.12)(10))
   = $222(2.2) 
   = $488.40

Your A is then equal to $488.40

If you need to get the simple interest all you need to use is the first formula given:

A = P + I
for the interest you transpose the P to the side of the A and you will get:
I = A - P
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$266.40 is the added interest to the principal amount. 
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sp2606 [1]

Answer:

y=\frac{1}{6} (x+5)^2-4.5

y=\frac{-1}{20} (x-10)^2+1

Step-by-step explanation:

focus at (-5, -3), and directrix y = -6

Directrix y=-6 so its  a vertical parabola

so equation is

(x-h)^2 = 4p(y-k)

(h,k) is the center

P is the distance between focus and vertex

distance between focus and directrix = 2p

distance between -3  and y=-6 is 3

2p = 3

p = 3/2 or p = 1.5

Focus is (h, k+p)

given focus is (-5, -3) so h= -5  and k+p = -3

k+p=-3, plug in 1.5 for p

k + 1.5 = -3

subtract 1.5 on both sides

k = -4.5

(x-h)^2 = 4p(y-k)

(x+5)^2= 4(1.5) (y+4.5)

(x+5)^2= 6(y+4.5)

divide by 6 on both sides

then subtract 4.5 on both sides

y=\frac{1}{6} (x+5)^2-4.5

focus at (10, -4), and directrix y = 6.

Directrix y=6 so its  a vertical parabola

so equation is

(x-h)^2 = 4p(y-k)

distance between focus and directrix = 2p

distance between -4  and y=6 is -4-6=-10

2p = -10

p = -5

Focus is (h, k+p)

given focus is (10, -4) so h= 10  and k+p = -4

k+p=-4, plug in 5 for p

k - 5 = -4

add 5 on both sides

k = 1

(x-h)^2 = 4p(y-k)

(x-10)^2= 4(-5) (y-1)

(x-10)^2= -20(y-1)

divide by -20 on both sides and add 1 on both sides

y=\frac{-1}{20} (x-10)^2+1


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