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3 years ago

5

Duc has a trapezoid shaped section of his yard fenced off for his pets. The lengths of the parallel sides of the trapezoid are 8

ft and 14 ft. The height of the trapezoid section is 4 ft.
What is the area of the trapezoid section of the yard?

2 answers:

posledela3 years ago

7 0

The area is equal to 44 ft.

Hope this answer helps!

notka56 [123]3 years ago

6 0

Answer:

LOOK THE RIGHT ANWER IS DOWN BELLOW

The answer is 44 ft.

I hope this helps

dont forget to hit that thanks button

Step-by-step explanation:

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$\bold{\huge{\pink{\underline{ Solution }}}}$

<u>We </u><u>have </u><u>given </u><u>in </u><u>the </u><u>question </u><u>that</u><u>, </u>

<u>The </u><u>sum </u><u>of </u><u>2</u><u> </u><u>numbers </u><u>is </u><u>equal </u><u>to </u><u>1</u><u>1</u><u> </u>
<u>The </u><u>difference </u><u>between </u><u>two </u><u>numbers </u><u>is </u><u>1</u><u>9</u><u> </u><u>.</u>

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<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>x </u><u>and </u><u>y</u><u>. </u>

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Let the two numbers be x and y

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

$\sf{ x + y = 11......eq(1) }$

$\sf{ x - y = 19......eq(2) }$

<u>Solving </u><u>eq(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>get </u><u>:</u><u>-</u><u> </u>

$\sf{ x + y = 11 }$

$\sf{ x = 11 - y ......eq(3 ) }$

<u>Subsituting </u><u>eq(</u><u>3</u><u> </u><u>)</u><u> </u><u>in </u><u>eq</u><u>(</u><u>2</u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{ x - y = 19 }$

$\sf{ ( 11 - y) - y = 19 }$

$\sf{ 11 - y - y = 19 }$

$\sf{ 11 - 2y = 19 }$

$\sf{ - 2y = 19 - 11 }$

$\sf{ - 2y = 8}$

$\sf{ y = 8/(-2) }$

$\sf{ y = - 4 }$

$\sf{\red{Thus ,\: the\: value\: of \: y = -4}}$

<u>Now</u><u>, </u><u> </u><u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>y </u><u>in </u><u>eq(</u><u> </u><u>3</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{ x = 11 - y }$

$\sf{ x = 11 - (-4 ) }$

$\sf{ x = 11 + 4 }$

$\sf{ x = 15 }$

Hence, The value of x and y are 15 and (-4) .

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