Question

the hypotenuse of a right angled triangle is 1 cm more than twice its altitude. If the base is 7 cm more than the altitude then the sides of the triangle

Answer 1

Let the altitude of the right angled triangle be denoted by the variable x cm.

Then, the hypotenuse will be:

2x + 1 cm.

And the base of the right angled triangle is:

x + 7 cm.

Using Pythagoras Theorem to find the value of x:

(2x+1)² = x² + (x+7)².

4x² + 4x + 1 = x² + x² + 14x + 49.

4x² + 4x + 1 = 2x² + 14x + 49.

2x² - 10x - 48 = 0.

Dividing the equation by 2:

x² - 5x - 24 = 0.

The above equation is of the form ax² + bx + c = 0, where a = 1, b = - 5 and c = - 24, which is a quadratic equation.

Using quadratic formula to find the possible value of x:

$x = \frac{ -( - 5) \pm \sqrt{25 - 4(1)( - 24)} }{2} \\ \\ x = \frac{5 \pm \sqrt{25 + 96} }{2} \\ \\ x = \frac{5 \pm \sqrt{121} }{2} \\ \\ x = \frac{5 \pm11}{2}$

Since x is the length of the altitude of the right angled triangle it should be positive.

So, we consider only the addition of the numbers given in the numerator since the 2nd number is larger than the fit and subtraction will give a negative number.

x = 16/ 2.

x = 8 cm.

2x + 1 = 2(8) + 1 = 16 + 1 = 17 cm

x + 7 = 8 + 7 = 15 cm.

Therefore:

Altitude = 8 cm.

Base = 15 cm.

Hypotenuse = 17 cm.

Answer 2

Answer

$(2x + 1 cm) \\ </p><p>(x + 7 cm)$

Pythagoras theorem

(2x+1)² = x² + (x+7)²

4x² + 4x + 1 = x² + x² + 14x + 49

4x² + 4x + 1 = 2x² + 14x + 49

2x² - 10x - 48 = 0

x² - 5x - 24 = 0

ax² + bx + c = 0

a = 1, b = - 5

c = - 24

$\begin{lgathered}x = \frac{ -( - 5) \pm \sqrt{25 - 4(1)( - 24)} }{2} \\ \\ x = \frac{5 \pm \sqrt{25 + 96} }{2} \\ \\ x = \frac{5 \pm \sqrt{121} }{2} \\ \\ x = \frac{5 \pm11}{2}\end{lgathered}$

$(x = 16/ 2) \\ </p><p>(x = 8 cm) \\ </p><p>(2x + 1 = 2) \\ </p><p>(8 + 1 = 16 + 1 = 17 cm) \\ </p><p>(x + 7 = 8 + 7 = 15 cm)$

So, Altitude = 8 cm,Base = 15 cm and Hypotenuse = 17 cm.