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Angelica used four congruent equilateral triangles. Each side of the triangles was 14 inches. The height of each triangle was 12.1 inches. Which figure did she construct? What is the surface area of her figure? 7 Check your guesses from Exercise 3. Were you correct?​

Sagot :

Answer:

in simplified version   it would be x ≤-4

Step-by-step explanation:

multiply each term in x ≤-4 by -1

Angelica constructs a regular tetrahedron

The surface area of the regular tetrahedron is 339.08 [tex]in^{2}[/tex]

What is equilateral triangle?

"It is a triangle that has all its sides equal in length and all angles are 60° in measure."

What are congruent triangles?

"Two or more triangles are congruent if all three corresponding sides are equal and all the three corresponding angles are equal in measure."

What is a regular tetrahedron?

"It is a triangular pyramid having congruent equilateral triangles for each of its faces."

Formula of area of regular tetrahedron:

[tex]A=\sqrt{3}\times a^{2}[/tex], where 'a' is the edge length of tetrahedron

For given question,

Angelica used four congruent equilateral triangles.

She constructs a regular tetrahedron using four congruent equilateral triangles.

The length of each edge of tetrahedron = side length of equilateral triangle

So, each edge of tetrahedron measures 14 inches.

Using the formula of area of the regular tetrahedron,

[tex]A=\sqrt{3}\times a^{2} \\\\A=\sqrt{3}\times (14)^{2}\\\\A=1.73\times 196\\\\A=339.08~~in^2[/tex]

Therefore, the surface area of the regular tetrahedron is 339.08 [tex]in^{2}[/tex]

Learn more about regular tetrahedron here:

https://brainly.com/question/17132878

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