Lesson HOW TO solve problems on right triangles

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<H2>How to solve problems on right triangles</H2>

In this lesson typical problems on right triangles are presented. These problems relate to the side measures of right triangles. The base for the solution of these problems is the Pythagorean Theorem. This theorem states that if a and b are measures of the right triangle legs and c is its hypotenuse, then

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

For the proof (the proofs) of the Theorem see the lessons <A HREF=http://www.algebra.com/algebra/homework/Pythagorean-theorem/The-Pythagorean-Theorem.lesson>The Pythagorean Theorem</A> and <A HREF=http://www.algebra.com/algebra/homework/Pythagorean-theorem/More-proofs-of-the-Pythagorean-Theorem.lesson>More proofs of the Pythagorean Theorem</A> under the topic Pythagorean Theorem of the section Geometry in this site.

The Pythagorean Theorem allows to calculate the length of the hypotenuse of the right triangle, if the leg measures are known, and/or to calculate the length of the leg 
of the right triangle, if the another leg measure and the hypotenuse measure are known.

The examples below show how to use the Pythagorean Theorem in different circumstances.

Problem 1

Find the perimeter of the right triangle, if one leg is 12 cm long and the hypotenuse is in 1 cm longer than this leg.

Solution

First, the hypotenuse is 12 cm + 1 cm = 13 cm long in accordance with the condition.
Now, apply the Pythagorean Theorem to find the second leg:

{{{sqrt(13^2 -12^2) = sqrt(169-144) = sqrt(25) = 5}}}.

Now calculate the perimeter of the triangle as the sum of measures of its three sides:
5 + 12 + 13 = 30 cm.

Answer. The perimeter of the triangle is 30 cm.

Problem 2

Find the sides of the right triangle, if one leg is in 1 cm longer than another, and the hypotenuse is in 1 cm longer than the longer leg.

Solution

The way to solve the problem is to reduce it to the quadratic equation by applying the Pythagorean Theorem and then to solve the quadratic equation.

Let x be the shorter leg of the triangle. 
Then another leg is (x+1) cm long, and the hypotenuse is ((x+1)+1) cm = (x+2) cm long.
By applying the Pythagorean theorem you get the equation

{{{x^2 + (x+1)^2 = (x+2)^2}}}.

Open the brackets, move all the terms with the unknown x from the right side to the left (do not forget to change the sign), cancel the like terms step by step.

{{{x^2 + x^2 + 2x + 1 = x^2 + 4x + 4}}},

{{{x^2 - 2x - 3 = 0}}}.

You got the quadratic equation in the standard form. Find its roots by applying the quadratic formula

{{{x = (2 +- sqrt(2^2 - 4*1*(-3)))/2 = (2 +- sqrt(16))/2 = (2 +- 4)/2}}}.

So, the two roots are {{{x[1] = 3}}} and {{{x[2] = -1}}}.
Only first root suits to the problem condition, because we are looking for positive roots for the leg measure.

Thus the shorter leg is 3 cm long, the longer leg is 3 cm + 1 cm = 4 cm long and the hypotenuse is 4 cm + 1 cm = 5 cm long.

Answer. The right triangle sides are 3 cm, 4 cm and 5 cm long.

Problem 3

Find the sides of the right triangle, if one leg is 8 cm long and the perimeter of the triangle is 24 cm.

Solution

Let x be the measure of the other leg of the right triangle.
Then the hypotenuse measure is {{{sqrt(x^2+ 8^2) = sqrt(x^2 + 64)}}} in accordance with the Pythagorean Theorem.
So, the perimeter of this triangle is equal to

{{{8 + x + sqrt(x^2 + 64)}}}.

Since the perimeter is equal to 24 cm, we have the equation

{{{8 + x + sqrt(x^2 + 64) = 24}}}.

To solve this equation, keep the square root term in the left side and move other terms from the left side to the right changing the sign. Then combine the like terms.

{{{sqrt(x^2 + 64) = 24 - 8 - x}}},

{{{sqrt(x^2 + 64) = 16 -x}}}.

Now, square both sides. You get
{{{x^2 + 64 = (16 - x)^2}}},

{{{x^2 + 64 = 16^2 - 2*16*x + x^2}}},

{{{x^2 + 64 = 256 - 32*x + x^2}}}.

Cancel the terms {{{x^2}}} in both sides and simplify the equation, step by step. You get

32x = 256 - 64,
32x = 192,
x = 6.

So, we found the unknown measure of the leg. It is 6 cm.
Thus the legs are 6 cm and 8 cm long.
Hence, the hypotenuse measure is equal to 
{{{sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10}}} cm.

Answer. The legs of the right triangle are 6 cm and 8 cm long. The hypotenuse is 10 cm long.

For navigation over the lessons on Properties of Triangles use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/Compendium-of-properties-of-triangles.lesson>Properties of Trianles</A>.

For navigation over the lessons on Properties of right-angled triangles use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Pythagorean-theorem/Properties-of-right-triangles.lesson>PROPERTIES OF RIGHT TRIANGLES</A>.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Triangles/GEOMETRY-your-online-textbook.lesson>GEOMETRY - YOUR ONLINE TEXTBOOK</A>.