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Liono4ka [1.6K]
2 years ago
15

Suppose a triangle has sides a, b, and c and the angle opposite the side of length a is acute what must be true?

Mathematics
1 answer:
ANTONII [103]2 years ago
4 0

The true statement about the triangle is (a) b^2 + c^2 > a^2

<h3>How to determine the true inequality?</h3>

The sides are given as:

a, b and c

The angle opposite of side length a is an acute angle

The above means that:

The side a is the longest side of the triangle.

The Pythagoras theorem states that:

a^2 = b^2 + c^2

Since the triangle is not a right triangle, and the angle opposite a is acute.

Then it means that the square of a is less than the sum of squares of other sides.

This gives

a^2 < b^2 + c^2

Rewrite as:

b^2 + c^2 > a^2

Hence, the true statement about the triangle is (a) b^2 + c^2 > a^2

Read more about triangles at:

brainly.com/question/2515964

#SPJ1

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Karo-lina-s [1.5K]

The perimeter of the equatorial triangle is 24 units

<h3><u>Solution:</u></h3>

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The triangle is shown below

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Now, using Pythagoras Theorem in Triangle ABD,

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\mathrm{AB}^{2}=\mathrm{BD}^{2}+\mathrm{AD}^{2}

\begin{array}{l}{a^{2}=\left(\frac{a}{2}\right)^{2}+(4 \sqrt{3})^{2}} \\\\ {a^{2}-\left(\frac{a}{2}\right)^{2}=(4 \sqrt{3})^{2}}\end{array}

\frac{4 a^{2}-a^{2}}{4}=16 \times 3

\begin{array}{l}{\frac{3 a^{2}}{4}=16 \times 3} \\\\ {3 a^{2}=192} \\\\ {a^{2}=192 \div 3=64}\end{array}

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