Respuesta :
Answer:
The base of triangle is [tex]\frac{8}{\sqrt{3}} \ m[/tex] and the height of triangle is [tex]\frac{12}{\sqrt{3}} \ m.[/tex]
Step-by-step explanation:
Given:
A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.
Now, to find the base and height of the triangle.
The base of triangle = [tex]x\times\frac{1}{2} =\frac{x}{2} \ m.[/tex]
The height of triangle = [tex]x\times \frac{3}{4} =\frac{3x}{4}\ m.[/tex]
The area of triangle = [tex]16\ m^2.[/tex]
Now, we put the formula of area to solve:
[tex]Area=\frac{1}{2} \times base\times height[/tex]
[tex]16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}[/tex]
[tex]16=\frac{3x^2}{16}[/tex]
Multiplying both sides by 16 we get:
[tex]256=3x^2[/tex]
Dividing both sides by 3 we get:
[tex]\frac{256}{3} =x^2[/tex]
Using square root on both sides we get:
[tex]\frac{16}{\sqrt{3}}=x[/tex]
[tex]x=\frac{16}{\sqrt{3}}[/tex]
Now, by substituting the value of [tex]x[/tex] to get the base and height:
[tex]Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.[/tex]
So, the base of triangle = [tex]\frac{8}{\sqrt{3}} \ m.[/tex]
[tex]Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.[/tex]
Thus, the height of triangle = [tex]\frac{12}{\sqrt{3}} \ m.[/tex]
Therefore, the base of triangle is [tex]\frac{8}{\sqrt{3}} \ m[/tex] and the height of triangle is [tex]\frac{12}{\sqrt{3}} \ m.[/tex]
