Question

A night triangle has an area (A = bh) of 9 km? and a hypotonuse of 3/5 km Find the dimensions of the triangle

Answer 1

The dimensions of the triangle (x, y) = (6, 3) or (3, 6)

That is if x = 6, y = 3

if x = 3, y = 6

Step-by-step explanation:

Area = 1/2 bh

Area 9km²

hypotenuse = 3√5 km

opposite = x

adjacent = y

a² + b² = c²

using pythagoras' theorem:

Hypotenuse² = opposite² + adjacent²

(3√5)² = x² + y²

9(5) = x² + y²

45 = x² + y² .....equation 1

Area = 1/2 × y × x

9 = xy/2

2(9) = xy

18 = xy ... equation 2

y = 18/x ... equation 3

Sum of two squares:

(x+y)² = x² +2xy + y²

x² + y² = (x+y)² - 2xy

45 = (x+y)² - 2(18)

45 + 2(18) = (x+y)²

45 + 36 = (x+y)²

81 = (x+y)²

√81 = √(x+y)²

`\begin{gathered} \pm\sqrt[]{81\text{ }}\text{ = (x+y) } \\ \pm9\text{ = x + y } \\ x+y\text{ = -9 or x+y = 9} \end{gathered}`

inserting y = 18/x in the (x+y) equation above:

when x+y = 9

x + (18/x) = 9

(x² + 18)/x = 9

x² + 18 = 9x

x² - 9x + 18 = 0

x² - 3x - 6x + 18 = 0

x(x -3) -6(x -3) = 0

x -6 = 0 or x -3 = 0

x = 6 or x = 3

when x+y = -9

x + (18/x) = -9

(x² + 18)/x = -9

x² + 18 = -9x

x² + 9x + 18 = 0

x² + 3x + 6x + 18 = 0

x(x +3) +6(x +3) = 0

x +6 = 0 or x +3 = 0

x = -6 or x = -3

Since x cannot be a negative number because we dealing with dimension of the triangle, x+y = -9 is wrong

So we go after x+y = 9

x = 6 or 3

when x = 6

Insert into x+y = 9

6+y = 9

y = 9-6

y = 3

When x = 3

x+y = 9

3+y = 9

y = 9-3

y = 6

Hence, the dimensions of the triangle (x, y) = (6, 3) or (3, 6)

That is if x = 6, y = 3

if x = 3, y = 6