Question

The area of a right triangle is sixty square centimetres. Its base is one centimetre less than twice its height. If the base and height of the right triangle are both decreased by 2 centimetres, how much will the hypotenuse of the right triangle decrease?
1. 2 centimetres

2. 17 centimetres

3. sqrt(205) centimetres

4. 4 centimetres

5. None of the above

Answer 1

Answer:

5. None of the above

Step-by-step explanation:

We have this information:

A = 60 cm2

b = 2*h - 1

Let's find out b and h:

The equation of the are of a triangle is A = (b*h)/2, if we replace it with our information, we have

$60 = ((2*h - 1) * h)/2\\ 120 = 2*h^2 - h\\ 0 = 2*h^2 - h - 120$

Let's find h with the quadratic formula:

$\fbox {Quadratic formula:} \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$h = \frac{-(-1)\pm\sqrt{(-1)^2-4*2*(-120)}}{2*2} = \frac{1\pm\sqrt{961}}{4} = \frac{1\pm\ 31}{4}$

h = 8 or h = -7.5

But h represents the height of the triangle, so it has to be a positive number, that's h = 8.

If we replace this in the equation we had for b, we have that b = 2*8 - 1 = 15.

Now we can calculate the hypotenuse with the Pythagorean equation  

$\fbox {Pythagorean equation:}$ The square of the length of the hypotenuse (the side opposite the right angle) of a right triangle is equal to the sum of the squares of the two legs (the two sides that meet at a right angle).

The base and the height are our legs. We will use "H" for the hypotenuse

$H^2 = b^2 + h^2 = 15^2 + 8^2 = 289\\ H = \sqrt {289} = 17$

H = 17

If we decrease the base and the height by 2 centimeters, we have  

b' = 15 - 2 = 13 and h' = 8 - 2 = 6

With this, let's calculate the new hypotenuse:

$H'^2 = b'^2 + h'^2 = 13^2 + 6^2 = 205\\ H' = \sqrt {205} \approx 14.3$

$H' \approx 14.3$

So, the hypotenuse decreases $H - H' \approx 17 - 14.3 = 2.7$