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2 years ago

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The lengths of two sides of a triangle are 15 inches each. The third side measures 10 inches. What type of triangle is this? Exp

lain your answers using geometric terms.

1 answer:

zloy xaker [14]2 years ago

5 0

Answer: Isosceles

<u>Step-by-step explanation:</u>

When two sides of a triangle are congruent, it is an isosceles triangle.

When three sides of a triangle are congruent, it is an equilateral triangle.

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stellarik [79]

Answer:

Sheridan's Work is correct

Step-by-step explanation:

we know that

The lengths side of a right triangle must satisfy the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

where

a and b are the legs

c is the hypotenuse (the greater side)

In this problem

Let

$a=7\ cm\\c=13\ cm$

substitute

$13^{2}=7^{2}+b^{2}$

Solve for b

$169=49+b^{2}$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

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we have that

<em>Jayden's Work</em>

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\b=13\ cm$

substitute and solve for c

$7^{2}+13^{2}=c^{2}$

$49+169=c^{2}$

$218=c^{2}$

$c=\sqrt{218}\ cm$

$c=14.76\ cm$

Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle

<em>Sheridan's Work</em>

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\c=13\ cm$

substitute

$7^{2}+b^{2}=13^{2}$

Solve for b

$49+b^{2}=169$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

therefore

Sheridan's Work is correct

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