Mathematics, 06.08.2021 05:30 ricksterv5000

Find the length of the third side. If necessary, round to the nearest tenth. 5 10

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Answer from: kgflol5692

$\boxed {\boxed {\sf 8.7}}$

Step-by-step explanation:

We are asked to find the length of the third side in a triangle, given the other 2 sides.

Since this is a right triangle (note the small square in the corner of the triangle representing a 90 degree /right angle), we can use the Pythagorean Theorem.

a^2 + b^2 =c^2

In this theorem, a and b are the legs of the triangle and c is the hypotenuse.

We know that the unknown side (we can say it is a) and the side measuring 5 are the legs because they form the right angle. The side measuring 10 is the hypotenuse because it is opposite the right angle.

b= 5 c= 10

Substitute the values into the formula.

a^2 + (5)^2 = (10)^2

Solve the exponents.

(5)²= 5*5 = 25 (10)²= 10*10= 100

a^2 + 25=100

We are solving for a, so we must isolate the variable. 25 is being added to a. The inverse operation of addition is subtraction, so we subtract 25 from both sides.

a^2 +25-25=100-25

a^2=100-25

a^2 = 75

a is being squared. The inverse of a square is the square root, so we take the square root of both sides.

$\sqrt {a^2}= \sqrt{75}$

$a= \sqrt{75}$

a= 8.660254038

Round to the nearest tenth. The 6 in the hundredth place tells us to round the 6 up to a 7 in the tenth place.

$a \approx 8.7$

The length of the third side is approximately 8.7

Answer from: queenhall934

Perpendicular=P=5Hypontenuse=H=10Base=B=?

Using Pythagorean theorem

$\boxed{\sf B^2=H^2-P^2}$

Putting values

$\\ \sf \longmapsto B^2=10^2-5^2$

$\\ \sf \longmapsto B^2=100-25$

$\\ \sf \longmapsto B^2=75$

$\\ \sf \longmapsto B=\sqrt{75}$

$\\ \sf \longmapsto B=\sqrt{25\times 3}$

$\\ \sf \longmapsto B=5\sqrt{3}$

$\\ \sf \longmapsto B=5\times 1.732$

$\\ \sf \longmapsto B=8.66$

$\\ \sf \longmapsto B\approx 8.7$

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