Fig. 1,309.—In a right angle triangle the square on the hypothenuse is equal to the sum of the squares on the other two sides. That is: hypothenuse2 = base2 + altitude2. Draw AB, 4 inches long, and BC, 3 inches long and at right angles to AB. Join AC, which will be found to be 5 inches long. From the diagram, it must be clear that the square on AC = sum of squares on AB and BC; that is, 52 = 42 + 32. Further, 42 = 52 - 32; 32 = 52 - 42; 5 = √(42 + 32); 4 = √(52 - 32); 3 = √(52 - 42).
Ques. Why is the square of the hypothenuse of a right angle triangle equal to the sum of the squares of the other two sides?
Ans. This may be explained with the aid of fig. 1,309. Draw a line AB, 4 inches in length and erect a perpendicular BC, 3 inches in height; connect A and C, giving the right angle triangle ABC. It will be found that AC the hypothenuse of this triangle is 5 inches long. If squares be constructed on all three sides of the triangle, the square on the hypothenuse will have an area of 25 sq. ins.; the square on the base, 16 sq. ins., and the square on the altitude, 9 sq. ins. Then from the figure 52 = 42 + 32, that is 25 = 16 + 9.
Repeating equation (1), it is evident from the figure that
| hypothenuse2 | ⎫ | ⎧ | base2 + altitude2 | ⎫ | |
| ⎬ | = | ⎨ | ⎬ | ||
| 52 | ⎭ | ⎩ | 42 + 32 | ⎭ |
that is,
25 = 16 + 9.
In the right angle triangle, the following relations also hold:
| base2 | = | hypothenuse2 | - | altitude2 | (2) |
| (42 | = | 52 | - | 32) | |
| altitude2 | = | hypothenuse2 | - | base2 | (3) |
| (32 | = | 52 | - | 42) |
In working impedance problems, it is not the square of any of the quantities which the sides of the triangle are used to represent that is required, but the quantities themselves, that is, the sides. Hence extracting the square root in equations (1), (2) and (3), the following are obtained: