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3 years ago

The sides of a quadrilateral are x,x+3,x+4,x+5. it’s perimeter is 52, find the lengths of the sides

Mathematics

1 answer:

wariber [46]3 years ago

4 0

Answer: 10 , 13 , 14 , 15

Step-by-step explanation:

To find the perimeter of a quadrilateral , we will add all its side , that is

Perimeter = x + x + 3 + x + 4 + x + 5

That is

52 = x + x + 3 + x + 4 + x + 5

52 = 4x + 12

subtract 12 from both sides

40 = 4x

divide through by 4

x = 10

therefore , the lengths of the sides are

10 , 10+3 , 10 + 4 , 10 + 5 , that is

10 , 13 , 14 , 15

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Find the equation of the line perpendicular to y = 3x + 6 and containing the point (-9,-5).

neonofarm [45]

The equation of the line perpendicular to y = 3x + 6 and containing the point (-9,-5) is $y = \frac{-1}{3}x - 8$

Solution:

Given that line perpendicular to y = 3x + 6 and containing the point (-9, -5)

We have to find the equation of line

The slope intercept form is given as:

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

Let us first find the slope of line

The given equation of line is y = 3x + 6

On comparing the given equation of line y = 3x + 6 with eqn 1, we get,

m = 3

Thus the slope of given equation of line is 3

We know that product of slopes of given line and slope of line perpendicular to given line is equal to -1

Slope of given line $\times$ slope of line perpendicular to given line = -1

$3 \times \text{ slope of line perpendicular to given line }= -1$

$\text{ slope of line perpendicular to given line } = \frac{-1}{3}$

Let us now find the equation of line with slope $m = \frac{-1}{3}$ and containing the point (-9, -5)

Substitute $m = \frac{-1}{3}$ and (x, y) = (-9, -5) in eqn 1

$-5 = \frac{-1}{3}(-9) + c\\\\-5 = 3 + c\\\\c = -8$

Thus the required equation of line is:

Substitute $m = \frac{-1}{3}$ and c = -8 in eqn 1

$y = \frac{-1}{3}x - 8$

Thus the required equation of line perpendicular to given line is found

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