*Videos, solutions, examples, lessons, worksheets, and activities to help Algebra students learn how to solve word problems that involve area of triangles.*

A good first step, after reading the entire exercise, is to draw a right triangle and try to figure out how to label it.

Once you've got a helpful diagram, the math is usually pretty straightforward.

Note: Unless you are told to give your answer in decimal form, or to round, or in some other way not to give an "exact" answer, you should probably assume that the "exact" form is what they're wanting.

For instance, if they hadn't told me to round in the exercise above, my value for the height should have been the value with the radical. That width will be twice the base of one of the right triangles.

Simplify this equation step by step and solve it: 2x 2(x - 6) (x (x-6)) = 60 (after multiplying both sides of the previous equation by 2), 2x 2x x x - 12 - 6 = 60, 6x - 18 = 60, 6x = 60 18, 6x = 78, x = 13. Hence, the second side is of 13 cm - 6 cm = 7 cm long, the third side is (13 7)/2 = 20/2 = 10 cm long in accordance with the problem condition.

You can check that the sum of the side measures is equal to the given value of the perimeter: 13 cm 7 cm 10 cm = 30 cm. The triangle side measures are 13 cm, 7 cm and 10 cm.

Problem 5 Find the measures of the triangle sides, if the second side is in 6 cm shorter than the first one, the third side measure is half of the sum of the measures of the first and the second sides, and the perimeter of the triangle is 30 cm.

Solution Let x be the length of the first side (in centimeters).

For navigation over the lessons on Properties of Triangles use this file/link Properties of Trianles.

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link GEOMETRY - YOUR ONLINE TEXTBOOK.

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