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2 years ago

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AB||CD . Find the value of x.A. 25B. 45C. 15D. 10

1 answer:

Tom [10]2 years ago

4 0

Answer:

The value of x is 10.

3x+15=5x-5

3x=5x-5-15

3x=5x-20

3x-5x=(-20)

(-2x)=(-20)

x=10

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What is the perimeter of the trapezoid with vertices Q(8, 8), R(14, 16), S(20, 16), and T(22, 8)? Round to the nearest hundredth

uysha [10]

Answer:

The perimeter of the trapezoid is $38.25\ units$

Step-by-step explanation:

we know that

The perimeter of the trapezoid is the sum of its four side lengths

so

In this problem

$P=QR+RS+ST+QT$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

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step 1

Find the distance QR

$Q(8, 8), R(14, 16)$

substitute the values in the formula

$d=\sqrt{(16-8)^{2}+(14-8)^{2}}$

$d=\sqrt{(8)^{2}+(6)^{2}}$

$d=\sqrt{100}$

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step 2

Find the distance RS

$R(14, 16), S(20, 16)$

substitute the values in the formula

$d=\sqrt{(16-16)^{2}+(20-14)^{2}}$

$d=\sqrt{(0)^{2}+(6)^{2}}$

$d=\sqrt{36}$

$RS=6\ units$

step 3

Find the distance ST

$S(20, 16),T(22, 8)$

substitute the values in the formula

$d=\sqrt{(8-16)^{2}+(22-20)^{2}}$

$d=\sqrt{(-8)^{2}+(2)^{2}}$

$d=\sqrt{68}$

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step 4

Find the distance QT

$Q(8, 8),T(22, 8)$

substitute the values in the formula

$d=\sqrt{(8-8)^{2}+(22-8)^{2}}$

$d=\sqrt{(0)^{2}+(14)^{2}}$

$d=\sqrt{196}$

$QT=14\ units$

step 5

Find the perimeter

$P=10+6+8.25+14=38.25\ units$

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