MATH SOLVE

4 months ago

Q:
# the longest side of an obtuse triangle measures 20 cm. the two shorter sides meausre x cm and 3x cm. rounded to the nearest tenth , what is the greatest posible of x?

Accepted Solution

A:

Answer:

the greatest possible value for xΒ is 6

Explanation:

Assume that the sides of the triangle are:

a = x

b = 3x

c = 20

We are given that c is the longest side.

For the triangle to be obtuse:

c^2 > a^2 + b^2

Substitute with the values of a, b and c and solve for x as follows:

c^2 > a^2 + b^2

(20)^2 > (x)^2 + (3x)^2

400 > x^2 + 9x^2

400 > 10 x^2

40 > x^2

Now, we will get the zeros of the function. This means that we will solve for x^2 = 40 as follows:

x^2 = 40

x = + or -Β β40

either x = 6.32

Or x = -6.32

Since we want x^2 < 40, therefore, the greatest possible value for x to satisfy this condition is 6

Hope this helps :)

the greatest possible value for xΒ is 6

Explanation:

Assume that the sides of the triangle are:

a = x

b = 3x

c = 20

We are given that c is the longest side.

For the triangle to be obtuse:

c^2 > a^2 + b^2

Substitute with the values of a, b and c and solve for x as follows:

c^2 > a^2 + b^2

(20)^2 > (x)^2 + (3x)^2

400 > x^2 + 9x^2

400 > 10 x^2

40 > x^2

Now, we will get the zeros of the function. This means that we will solve for x^2 = 40 as follows:

x^2 = 40

x = + or -Β β40

either x = 6.32

Or x = -6.32

Since we want x^2 < 40, therefore, the greatest possible value for x to satisfy this condition is 6

Hope this helps :)