The area of a triangle is 1,440 cm2. The base of the triangle is 5 times the height. What is the height of the triangle?

Mathematics

2 answers:

Korvikt [17]3 years ago

4 0

Answer:

The height of the triangle = 24 cm

Step-by-step explanation:

∵ The area of the triangle = 1/2 × base × height

∵ The length of the base = 5 times the length of the height

Let the length of the height = x, then length of the base = 5x

∴ Area triangle = 1/2 × 5x × x = 2.5 x²

∵ Area of the triangle = 1440 cm²

∴ 2.5 x² = 1440

∴ x² = 1440 ÷ 2.5 = 576

∴ x = √576 = 24 cm

∴ The height of the triangle = 24 cm

madam [21]3 years ago

3 0

Answer: 24 cm

Step-by-step explanation:

The formula for calculate the area of a triangle is:

$A=bh/2$

Where b is the base and h is the height.

Then if the base of the triangle is 5 times the height:

$b=5h$

Substitute the above into the formula and solve for the height. Therefore, you obtain that the height is:

$1,440=\frac{5h*h}{2}\\1,440=\frac{5h^{2}}{2}$

$1,440cm^{2}=5h^{2}/2\\h=\sqrt{\frac{2(1,440)}{5}}\\h=24cm$

You might be interested in

Find the exact values of sin2 θ for cos θ = 3/18 on the interval 0° ≤ θ ≤ 90°

mote1985 [20]

Answer:

$sin(2\theta)=\frac{\sqrt{35} }{18}$

Step-by-step explanation:

Recall the formula for the sine of the double angle:

$sin(2\theta)=2*sin(\theta)*cos(\theta)$

we know that $cos(\theta)=\frac{3}{18}$ , and that $\theta$ is in the interval between 0 and 90 degrees, where both the functions sine and cosine are non-negative numbers. Based on such, we can find using the Pythagorean trigonometric property that relates sine and cosine of the same angle, what $sin(\theta)$ is:

$cos^2(\theta)+sin^2(\theta)=1\\sin^2(\theta)=1-cos^2(\theta)\\sin(\theta)=\sqrt{1-cos^2(\theta)} \\sin(\theta)=\sqrt{1-(\frac{3}{18} )^2}\\sin(\theta)=\sqrt{1-\frac{9}{324} }\\sin(\theta)=\sqrt{\frac{324-9}{324} }\\sin(\theta)=\sqrt{\frac{315}{324} }\\\\sin(\theta)=\frac{3}{18}\sqrt{35 }$

With this information, we can now complete the value of the sine of the double angle requested:

$sin(2\theta)=2*sin(\theta)*cos(\theta)\\sin(2\theta)=2*\frac{3}{18} \,\sqrt{35} \,\frac{3}{18}\\sin(2\theta)=\frac{2*3*3}{18*18}\,\sqrt{35} \\sin(2\theta)=\frac{\sqrt{35} }{18}$