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Mamont248 [21]
3 years ago
12

A cross section is made by the intersection of a plane and a square pyramid at an angle either parallel or perpendicular to the

base.
The cross section can be which of these shapes? Check all that apply.

-square
-triangle
-trapezoid
-circle
-non-square rectangle
Mathematics
2 answers:
alukav5142 [94]3 years ago
8 0
A cross section is the two dimensional shape that is created when a slice is made through a solid figure by an intersection of a plane and the solid body.

A square pyramid is a pyramid with a square base.

Case 1: When the plane intersects the square pyramid at an angle perpendicular to the base but not through the vertex. In this case a trapezoid is formed.

Case 2: When the plane intersects the square pyramid at an angle perpendicular to the base and through the vertex. In this case a triangle is formed.

Case 3: When the plane intersects the square pyramid at an angle parallel to the base. In this case a square is formed.

<span>Therefore, a cross section made by the intersection of a plane and a square pyramid at an angle either parallel or perpendicular to the base can be of shapes:

-square
-triangle
-trapezoid</span>
pentagon [3]3 years ago
6 0
I just did the unit test I got square, trapezoid and triangle 
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One factor in flood safety along a levee is the area that will absorb water should the levee break. The coordinates that make up
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Let

A(0,0)\\ B (2.8, 2.1)\\C(4.3, 4.4)

using a graph tool

see the attached figure

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we know that

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Hero's Formula is equal to

Area=\sqrt{p*(p-a)*(p-b)*(p-c)}

where

p is  is half the perimeter of the triangle

 a,b,c are the lengths of the sides of a triangle

so

Step 1

<u>Find the length sides of the triangle</u>

a) <u>Find the distance AB</u>

d =\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}

Substitute

d =\sqrt{(2.1-0)^{2} +(2.8-0)^{2}}

d =3.5\ mi}

b) <u>Find the distance AC</u>

d =\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}

Substitute

d =\sqrt{(4.4-0)^{2} +(4.3-0)^{2}}

d =6.15\ mi}

c) <u>Find the distance BC</u>

d =\sqrt{(y2-y1)^{2} +(x2-x1)^{2}}

Substitute

d =\sqrt{(4.4-2.1)^{2} +(4.3-2.8)^{2}}

d =2.75\ mi}

Step 2

<u>Find the perimeter of the triangle</u>

Perimeter= 3.5+6.15+2.75 =12.4\ mi

<u>Find the half of the perimeter </u>

p=\frac{12.4}{2} = 6.2\ mi

Step 3

<u>Find the area of the triangle</u>

Area=\sqrt{6.2*(6.2-3.5)*(6.2-6.15)*(6.2-2.75)}

Area=\sqrt{6.2*(2.7)*(0.05)*(3.45)}

Area= 1.69\ mi^{2}

therefore

<u>the answer is the option </u>

a) 1.65 mi^2.

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Answer:

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