8. Find the slope. Make sure to reduce.

1. The diagonal of a quadrilateral

zalisa [80]

Consider the length of diagonal is 8.5 cm instead of 8.5 m because length of perpendiculars are in cm.

Given:

Length of the diagonal of a quadrilateral = 8.5 cm

Lengths of the perpendiculars dropped on it from the remaining opposite vertices are 3.5 cm and 4.5 cm.

To find:

The area of the quadrilateral.

Solution:

Diagonal divides the quadrilateral in 2 triangles. If diagonal is the base of both triangles then the lengths of the perpendiculars dropped on it from the remaining opposite vertices are heights of those triangles.

According to the question,

Triangle 1 : Base = 8.5 cm and Height = 3.5 cm

Triangle 2 : Base = 8.5 cm and Height = 4.5 cm

Area of a triangle is

$Area=\dfrac{1}{2}\times base \times height$

Using this formula, we get

$Area(\Delta 1)=\dfrac{1}{2}\times 8.5\times 3.5$

$Area(\Delta 1)=14.875$

and

$Area(\Delta 2)=\dfrac{1}{2}\times 8.5\times 4.5$

$Area(\Delta 2)=19.125$

Now, area of the quadrilateral is

$Area=Area(\Delta 1)+Area(\Delta 2)$

$Area=14.875+19.125$

$Area=34$

Therefore, the area of the quadrilateral is 34 cm².

6 0

3 years ago

Suppose the initial height of a Pumpkin is 12 feet and the pumpkin is being launched with a velocity of 61 feet per second. Use

iogann1982 [59]

<h2>Hello!</h2>

The answer is:

The maximum height before landing will be 69.7804 feet.

Since there is no information about the angle of the launch, we can safely assume that it's launched vertically.

So, we can calculate the maximum height of the pumpkin using the following formulas:

$y=y_o+v_{o}*t-\frac{1}{2}gt^{2}$

$v=vo-gt$

Where,

y, is the final height

$y_o$ , is the initial height

g, is the acceleration of gravity , and it's equal to:

$g=32.2\frac{ft}{s^{2} }$

t, is the time.

Now, we are given the following information:

$y_{o}=12ft\\\\v=61\frac{ft}{s}$

Then, to calculate the maximum height, we must remember that at the maximum height, the speed tends to 0, so, calculating we have:

Time calculation,

We need to use the following equation,

$v=vo-gt$

So, substituting we have:

$v=61\frac{ft}{s}-32.2\frac{ft}{s^{2}}*t\\\\-61\frac{ft}{s}=-32.2\frac{ft}{s^{2}}*t\\\\t=\frac{-61{ft}{s}}{-32.2\frac{ft}{s^{2}}}=1.8944s$

We know that it will take 1.8944 seconds to the pumpkin to reach its maximum height.

Maximum height calculation,

Now, calculating the maximum height, we need to use the following equation:

$y=y_o+v_{o}*t-\frac{1}{2}gt^{2}$

Substituting and calculating, we have:

$y=y_o+v_{o}*t-\frac{1}{2}gt^{2}$

$y=12ft+61\frac{ft}{s}*1.8944s-\frac{1}{2}32.2\frac{ft}{s^{2}}*(1.8944s)^{2}$

$y=12ft+61\frac{ft}{s}*1.8944s-\frac{1}{2}32.2\frac{ft}{s^{2}}*(1.8944s)^{2}\\\\y_{max}=12ft+115.5584ft-16.1\frac{ft}{s^{2}}*(3.5887s^{2})\\\\y_{max}=127.5584ft-57.7780ft=69.7804ft$