Use trigonometric functions to calculate the area of a triangle

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Pre-Calculus › Use trigonometric functions to calculate the area of a triangle

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1

What is the area of a triangle with side lengths , , and ?

$area=4\sqrt{6}$

$area=\sqrt{92}$

$area=16\sqrt{6}$

Explanation

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(4+5+7)$

$s=\frac{1}{2}(16)$

So the area is

$area=\sqrt{8(8-4)(8-5)(8-7)}$

$area=\sqrt{96}$

$area=4\sqrt{6}$

2

Find the area of a quarter circle with radius 9cm.

$\frac{81\pi }{4}cm^2$

$\frac{9\pi }{4}cm^2$

$\frac{81}{4}cm^2$

$\frac{91\pi }{4}cm^2$

Explanation

To find the area of a quarter circle, first remember what the area of a circle is. The formula is $A=\pi r^2$ . Since we're only interested in a quarter of the circle, let's use the formula $A=\frac{\pi r^2}{4}$ . Then, plug in the radius so that your answer is: $\frac{81\pi }{4}cm^2$ .

3

Find the area of a triangle with sides 25 meters, 42 meters, and 23 meters.

None of these.

Explanation

We use Heron's formula for finding the area of oblique triangles or triangles without right angles.

Heron's formula is

$\sqrt{s(s-a)(s-b)(s-c)}$ .

We find s by

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

Find s.

$s=\frac{a+b+c}{2}=\frac{25+42+23}{2}=45$

Input into Heron's formula:

$\sqrt{45(45-25)(45-42)(45-23)}=244m^2$

4

Find the area of this triangle:

Tri area d

Explanation

Use the area formula to find area that is associated with the side angle side theorem for triangles.

$area= \frac{ 1}{2 } a b \sin C$

where and are side lengths and is the included angle.

Plugging these values into the formula above, we arrive at our final answer.

$area = \frac{ 1}{2} (11) (23 ) \sin (41^o ) \approx 82.99$

5

Find the area of this triangle:

Tri area h

Explanation

Use Heron's Formula

$area = \sqrt{s (s - a)(s-b)(s-c)}$

where $s = \frac{ 1}{2} (a + b + c )$

and

we can find to be:

$s = \frac{ 1}{2} ( 5 + 5 + 5 ) = 7.5$ .

From here, plug in all our known values and solve.

$\area = \sqrt{ 7.5 ( 7.5 - 5 ) (7.5 - 5 ) ( 7.5 - 5 )} \ \= \sqrt{ 7.5 (2.5 )^3 } \approx 10.83$

6

What is the area of a triangle with side lengths , , and ?

Explanation

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(7+24+25)$

$s=\frac{1}{2}(56)$

So the area is

$area=\sqrt{28(28-7)(28-24)(28-25)}$

$area=\sqrt{7056}$

7

Find the area of this triangle:

Tri area a

Explanation

Find the area using the formula associated the side angle side theorem of a triangle,

$area = \frac{ 1}{2} ab \sin C$

where and are side lengths and is the included angle.

In this particular case,

$a=15, b=19, C=20^\circ$

therefore the area is found to be,

$area = \frac {1}{2} (15)(19) \sin (20^o ) \approx 48.74$ .

8

What is the area of a triangle if the sides of a triangle are , , and ?

$area=\sqrt{15}$

$area=4\sqrt{5}$

$area=\sqrt{5}$

$area=4\sqrt{5}$

$area=16\sqrt{5}$

Explanation

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(4+2+4)$

$s=\frac{1}{2}(10)$

So the area is

$area=\sqrt{5(5-4)(5-2)(5-4)}$

$area=\sqrt{15}$

9

What is the area of a triangle with sides , , and ?

$area=10\sqrt{2}$

$area=100\sqrt{2}$

$area=20\sqrt{2}$

Explanation

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(6+5+9)$

$s=\frac{1}{2}(20)$

So the area is

$area=\sqrt{10(10-6)(10-5)(10-9)}$

$area=\sqrt{200}$

$area=10\sqrt{2}$

10

Find the area of this triangle:

Tri area g

Explanation

To find the area, use Heron's Formula,

$area = \sqrt{s(s-a) (s - b )(s - c )}$

where

$s = \frac{ 1}{ 2} (a + b + c )$

and

.

Here,

$s= \frac{ 1}{2} ( 6 + 7 + 9 ) = \frac{ 1}{2} \cdot 22 = 11$ .

Now plug in all known values and solve.

$\ area = \sqrt{ 11 (11- 6 ) ( 11 - 7 ) ( 11 - 9 )} \ \= \sqrt{ 11 \cdot 5 \cdot 4 \cdot 2 } \approx 20.98$

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