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2 years ago

Find the area a of the triangle whose sides have the given lengths. a = 20, b = 15, c = 25 a =?

Mathematics

1 answer:

PIT_PIT [208]2 years ago

3 0

The area of a triangle with sides a = 20, b = 15, and c = 25 is 150.

The sides of the triangle are given as a = 20, b = 15, and c = 25.

We will use Hero's formula to find the area of this triangle.

<h3>What is Heron's formula?</h3>

It is a three-face polygon that consists of three edges and three vertices.

We use Heron's formula to find the area of a triangle with 3 sides:

Herons formula:

Area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\$

Where a, b, and c are sides of a triangle.

And s = semi perimeter of a triangle.

s = $\frac{a+b+c}{2}$

If the sum of two sides of a triangle is greater than the third side of a triangle then the sides of a triangle are true.

Let the given sides be:

a = 20, b = 15 and c = 25.

(20 + 15) > 25

(20 + 25) > 15

(15 + 25) > 20 so the given sides are true.

Now,

Semi perimeter of the triangle:

s = (a+b+c) / 2

s = (20+15+25) / 2

s = 60 / 2

s = 30

Putting s = 30 in the area of the triangle.

we get,

Area of the triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\$

Area of the triangle = $\sqrt{30(30-20)(30-15)(30-25)}\\\\\sqrt{30\times10\times15\times5}\\\\\sqrt{22500}\\\\150$

Thus, the area of a triangle is 150.

Learn more about the Area of triangles here:

brainly.com/question/11952845

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Which of the following is not one of the 8th roots of unity?

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1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$