Question

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

Answer 1

• Perpendicular=P=5
• Hypontenuse=H=10
• Base=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=√(75)`

`\\ \sf \longmapsto B=√(25* 3)`

`\\ \sf \longmapsto B=5√(3)`

`\\ \sf \longmapsto B=5* 1.732`

`\\ \sf \longmapsto B=8.66`

`\\ \sf \longmapsto B\approx 8.7`

Answer 2

`\boxed {\boxed {\sf 8.7}}`

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}= √(75)`

`a= √(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