How would you find the dimensions and interior angles to the nearest degree of a right angle triangle when its height is greater than its base by 1.85 cm and has an area of 20.535 square cm?

Answer 1

First, you need to find the lengths of the sides of the triangle.

let the base = x

so the height = x + 1.85cm

area = 20.535 cm2

The area of a right triangle is 1/2*b*h, so using the base & height above we have:

(1/2)*(x)*(x+1.85) = 20.535

x2 + 1.85x = 41.07

x2+1.85x-41.07 = 0

From algebra you can use the quadratic equation to solve for x: x = [-b +- SQRT (b2-4ac)]/2a where a=1, b=1.85, c=-41.07. You will get two answers one of which will be negative & the other positive. You only need to use the positive root, which in this case is 5.55.

So the base is 5.55 cm & the height of the triangle is 7.40 cm. Now pick one of the angles and use the tangent function to solve for the angle measurement. I used the angle adjacent to the base:

tan (X) = opposite/adjacent or height/base

tan (X) = 7.40 / 5.55

(x) = tan-1(1.33333)

(X) = 53.1 or 53o

the other interior angle is simply 180o - (90o+53o) = 37o

So the interior angles are 53o & 37o

Additional Information:-

The dimensions of the triangle must include its hypotenuse so use Pythagoras' theorem:-

5.552+7.42 = 85.5625 and the square root of this is 9.25

Therefore the dimensions are: hypotenuse 9.25 cm, height 7.4 cm and base 5.55 cm