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pishuonlain [190]
3 years ago
15

The area of a triangle is 24sq inches. If the base of the triangle is 6 inches, what is the height?

Mathematics
2 answers:
Helen [10]3 years ago
4 0

Answer:

8 inches

Step-by-step explanation:

To find the area of a triangle, we use the formula

A = 1/2 bh where b is the length of the base and h is the height

A = 24 and b = 6

24 = 1/2 (6) * h

24 =3h

Divide each side by 3

24/3 = 3h/3

8 =h

The height is 8 inches

eimsori [14]3 years ago
3 0
<h2>Hello!</h2>

The answer is:

The height of the triangle is equal to 8 inches.

height=8in

<h2>Why?</h2>

To find the height of the triangle, we need to use the formula to calculate the area of a triangle, it's given by the following expression:

Area=\frac{base*height}{2}

So, since we already know the area and the base, we need to isolate the height from the formula, so, isolating we have:

Area=\frac{base*height}{2}\\\\\frac{Area*2}{base}=height

We know that:

Area=24in^{2} \\base=6in

Now, substituting we have:

height=\frac{Area*2}{base}

height=\frac{24in^{2} *2}{6in}=8in

Hence, we have that the height of the triangle is equal to 8 inches.

height=8in

Have a nice day!

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Read 2 more answers
two similar triangles, the perimeter of the first is 74cm. and the side lengths of the other are 4.5cm, 6cm and 8cm find the len
Amiraneli [1.4K]

\huge\mathfrak{\underline{Answer:}}

❒ The length of longest side of first Trianlge :

‎ㅤ‎ㅤ‎ㅤ‎ㅤ‎ㅤ\large\bf{32\:cm}

__________________________________________

\large\bf{\underline{given:}}

  • Two similar triangles
  • Perimeter of first Trianlge is 74 cm
  • Side length of second triangle are 4.5cm , 6cm and 8cm

\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){$\bf A$}\put(0.5,-0.3){$\bf C$}\put(5.2,-0.3){$\bf B$}\end{picture}

\setlength{\unitlength}{0.75 cm}\begin{picture}(0,1)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){$\bf P$}\put(0.5,-0.3){$\bf R$}\put(5.2,-0.3){$\bf Q$}\end{picture}

__________________________________________

\large\bf{\underline{To \:find :}}

  • The length of longest side of first Trianlge

__________________________________________

\large\bf{\underline{Perimeter\:of\: second\: triangle:}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=sum\:of\:all\:sides}

‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=4.5+6+8}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=18.5}

❒ The ratio between the perimeter of two triangles :

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=\frac{perimeter\:of\:1st\: triangle}{perimeter\:of\:2nd\: triangle}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=\frac{74}{18.5}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{=4:1}

❒ The two triangles are similar , therefore the ratio of their sides will also be similar , i.e 4:1

➽ Let the sides of first Trianlge be x , y and z

\large\bf{\underline{Therefore:}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{x}{4.5}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{y}{6}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{z}{8}=\frac{4}{1}}

\large\bf{\underline{Hence:}}

❒ Length of first side :

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{x}{4.5}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹x=4\times 4.5}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹x=18}

❒ Length of second side :

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{y}{6}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹y=4\times 6}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹y=24}

❒ Length of third side :

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹\frac{z}{8}=\frac{4}{1}}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹z=4\times 8}

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎\large\bf{⟹z=32}

__________________________________________

\large\mathfrak{Length\:of\: largest\:side:}

\huge\bf{=32cm}

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