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scoundrel [369]
3 years ago
8

Geometry help! Please prove the following is a square with full explanation, I'm really lost.

Mathematics
1 answer:
Hoochie [10]3 years ago
5 0

We have to show that this is a square, meaning all the sides are perpendicular and the same distance.  

A(3,4), B(2,-2), C(-4,-1), D(-3,5)

For each side we calculate the squared distance and the slope.  We don't have to bother to take the square root.  We'll subtract the points first because that difference, called the direction vector, goes into both the calculation of the slope and of the distance.

AB=B-A=(2 - 3, -2 - 4) = (-1, -6).   slope=-6/-1=6  AB²=(-1)²+(-6)²=37

BC=C-B=(-6, 1).  slope=1/-6=-1/6.   BC²=(-6)²+1²

CD=D-C=(1, 6).   slope=6/1=6    CD²=1²+6²=37

DA=A-D=(6,-1)   slope=-1/6     DA²=37

We see the negative reciprocal slopes, alternating between 6 and -1/6, indicating perpendicular sides.  We see they all have squared distance 37.  Equal perpendicular sides proves its a square.

If we do these problems enough we learn:  There's no point taking the square root.  In fact, there's really no point computing the slopes and the squared distances; we can see it's a square from the pattern of the direction vectors of the sides:  (-1, -6) (-6, 1) (1, 6) (6,-1)

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Answer:

Volume of the cone in ascending order.

V_{2}=270\pi\ units^{3}

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let V_{2}. V_{3}. and\  V_{4}. be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

d_{1} = 20\ and\ h_{1}=12

d_{2} = 18\ and\ h_{2}=10

r_{3} = 10\ and\ h_{3}=9

r_{4} = 11\ and\ h_{14}=9

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Solution:

The volume of the cone is given below.

V=\pi r^{2} \frac{h}{3}----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for d_{1} = 20\ and\ h_{1}=12

r_{1} = \frac{d_{1}}{2}

r_{1} = \frac{20}{2}=10\ units

V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}

V_{1}=\pi (10)^{2} \frac{12}{3}

V_{1}=\pi\times 100\times 4

V_{1}=400\pi\ units^{3}

Similarly, for volume of the cone for d_{2} = 18\ and\ h_{2}=10

r_{2} = \frac{d_{2}}{2}

r_{2} = \frac{18}{2}=9\ units

V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}

V_{2}=\pi (9)^{2} \frac{10}{3}

V_{2}=\pi\times 81\times \frac{10}{3}

V_{2}=\pi\times 27\times 10

V_{2}=270\pi\ units^{3}

Similarly, for volume of the cone for r_{3} = 10\ and\ h_{3}=9

V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}

V_{3}=\pi (10)^{2} \frac{9}{3}

V_{3}=\pi\times 100\times 3

V_{3}=\pi\times 300

V_{3}=300\pi\ units^{3}

Similarly, for volume of the cone for r_{4} = 11\ and\ h_{4}=9

V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}

V_{4}=\pi (11)^{2} \frac{9}{3}

V_{4}=\pi\times 121\times 3

V_{4}=\pi\times 363

V_{4}=363\pi\ units^{3}

So, the volume of the cone in ascending order.

V_{2}=270\pi\ units^{3}

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