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Helga [31]
3 years ago
11

An isosceles trapezoid with a perimeter of 42 inches. Each of the congruent non parallel sides is 5 inches long, and the trapezo

id is 3 inches tall. How long are the two parallel sides?
A. 10 in, 22 in
B. 16 in, 16 in
C. 10 in, 16 in
D. 12 in, 20 in
Mathematics
1 answer:
Ratling [72]3 years ago
8 0

Answer:

Longer - 20 in and Shorter - 12 in

Step-by-step explanation:

We know that when we subtract the shorter side from longer we get two cuts named  x:

2 x = a - b  => a - b = 2 x

One of this section ( x ) with height ( h ) and lateral (congruent non parallel) side ( c ) make right triangle, from which we get:

x² = c² - h²  and c = 5 in, h = 3 in

x² = 5² - 3² = 25 - 9 = 16 => x = √16 = 4 => x = 4 in

Now we will replace x = 4  in the equation  a - b = 2 · 4 = 8 and get first equation of the system.

a - b = 8

We also know that the formula for calculating perimeter is:

P = a + b + 2 c  where P = 42 in and c = 5 in

a + b = 42 - 2 · 5 = 42 - 10 = 32

Now  we get the second equation of the system:

a + b = 32

a - b = 8  

When we add first equation to the second we get:

2 a = 40 => a = 40 / 2 = 20 => a = 20 in

When we replace a = 20 in the first equation we get:

20 + b = 32  => b = 32 - 20 = 12 => b = 12 in

God with you!!!

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Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral
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Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

l_{AB} = \sqrt{\left (6-4  \right )^{2}+\left (6-1  \right )^{2}} = \sqrt{29}

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

l_{BC} = \sqrt{\left (1-6  \right )^{2}+\left (4-6  \right )^{2}} = \sqrt{29}

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

l_{CD} = \sqrt{\left (-1-1  \right )^{2}+\left (-1-4  \right )^{2}} = \sqrt{29}

The length of side DA, with D = (-1, -1) and A = (1,4)   gives;

l_{DA} = \sqrt{\left (4-(-1)  \right )^{2}+\left (1-(-1)  \right )^{2}} = \sqrt{29}

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

Therefore;

Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

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