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3 years ago

Abdul drove 320 miles using 12 gallons of gas. At this rate, how many gallons of gas would he need to drive 296 miles?

2 answers:

8 0

He would need 11.1 gallons of gas if he was/would drive 296 miles.

12 divided by 320 = .0375

.0375 x 296 = 11.1 Gallons of Gas needed to drive the distance.

Gekata [30.6K]3 years ago

4 0

Well, to solve this, first of all, you will need to compare. So, Abdul drove 320 MILES using 12 GALLONS, how many gallons will he need to drive 296 MILES. This can be represented mathematically by;
320 miles=12 gallons
296 miles=x
I replaced the unknown value with the letter x to make it easier to identify. Anyway, you need to cross and multiply. So 296 will multiply 12 and 320 will multiply x.
296 multiplied by 12 is 3552 and 320 multiplied by x is 320x. So;
3552=320x
Divide the both sides by 320 to reveal the value of x. 320 divided by 320 is x. And 3552 divided by 320 is 11.1. So;
x=11.1 gallons of gas
Hope i helped

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

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

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CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

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Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

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