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3 years ago

If MN = 5, NO = 13, then LM = (Blank 1). (Round your answer to one decimal place, as necessary.)

Mathematics

1 answer:

Nana76 [90]3 years ago

3 0

Answer:

8.1 = LM

Step-by-step explanation:

(whole secant) x (external part) = (tangent)^2

13 * 5 = LM^2

65 = LM^2

sqrt(65) = LM

8.06225 = LM

To one decimal place

8.1 = LM

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What is the perimeter of rhombus WXYZ?

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<u>Answer</u>:

The perimeter of rhombus WXYZ is $4 \sqrt{13}$

<u>Step-by-step explanation:</u>

Step 1 :Finding length of XY

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 5

$x_2$ =3

$y_1$ = -1

$y_2$ =2

XY = $\sqrt{(3-5)^2 +(2 -(-1))^2}$

XY = $\sqrt{(3-5)^2 +(2 +1))^2}$

XY = $\sqrt{(-2)^2 +(3))^2}$

XY = $\sqrt{4 +9}$

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Step 2 :Finding length of YZ

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 3

$x_2$ =5

$y_1$ = 2

$y_2$ =5

YZ = $\sqrt{(5-3)^2 +(5-2)^2}$

YZ = $\sqrt{(2)^2 +(3)^2}$

YZ = $\sqrt{4 +9}$

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Step 3 : :Finding length of ZW

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 5

$x_2$ =7

$y_1$ = 5

$y_2$ =2

ZW = $\sqrt{(7-5)^2 +(5-2)^2}$

ZW = $\sqrt{(2)^2 +(3)^2}$

ZW = $\sqrt{4 +9}$

ZW = $\sqrt{(13)}$

Step 4 :Finding length of WX

Distance formula = $\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

here

$x_1$ = 7

$x_2$ =5

$y_1$ = 2

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Step 5: finding the perimeter of the rhombus

Perimeter= 4 X side

=> $4 \times \sqrt{13}$

=> $4 \sqrt{13}$

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