Find the area and perimeter. The area is ... and the perimeter is ...

Mathematics

1 answer:

Schach [20]3 years ago

7 0

Answer:

Area: 6a^2

Perimeter: 12a^2

Step-by-step explanation:

To find the area of a triangle, use the formula A=1/2 b*h. The base is 3a^2 and the height is 4a^2.

A= 1/2 (3a^2)(4a^2)

A = 1/2 (12a^4)

A = 6a^4

To find the perimeter of the triangle, add each of the sides of the triangle. The third side, the hypotenuse, can be found using Pythagorean theorem.

$a^2+b^2=c^2\\(3a^2)^2+(4a^2)^2 = c^2\\(9a^4)+(16a^4)=c^2\\25a^4=c^2\\5a^2=c$

Now add each of the sides: 3a^2 + 4a^2 +5a^2 = 12a^2

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Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$