If it's a right angle triangle then use Pythagoras' theorem to find the 3rd side
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A rectangle and a triangle have equal areas. The length of the rectangle is 12 inches, and its width is 8 inches. If the base of the triangle is 32 inches, what is the length, in inches, of the altitude drawn to the base? WRONG WRONG WRONG NO NO NO ::"::":""::":"""::"":":""::::::::::::: ::"::": ::":: :: ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::"
To find the length of a line drawn between 2 vertices which are not next to each other, first draw a right triangle such that the line is the hypotenuse and the other two lines are drawn parallel to the x-axis and y-axis. Since the length of the other two lines are known, you can then calculate the hypotenuse to find the length of the line between the two vertices.
The rectangle has an area of 12 x 8 = 96 square inches. The triangle has an area of (1/2 * 32 * x). So, 16x = 96, therefore x = 6. The height of the triangle is 6 inches.
Sure. If one of the base angles is more than 90 degrees, then the altitude (height) is outside the triangle. Yes. This only occurs with an obtuse triangle. Because an altitude is a line drawn from a vertex to the opposite side and is perpendicular with that opposite side, it can only occur if it is outside the triangle. Look at the triangle in related links. If you look at the vertex on the top, the only way to draw the altitude would be to draw outside the triangle.
The altitude will be 6 inches in length. The area of the rectangle is length (l) times width (w) or l x w = 12 x 8 = 96 The area of a triangle is 1/2 the base (b) times the height (h), or altitude, and, because the area of the triangle is equal to the area of the rectangle, both will have an area of 96. For the triangle, 1/2 b x h = 1/2 x 32 x h = 96, and 16 x h = 96, and h = 6