Area of a Triangle - Pre-Calculus

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Find the area of this triangle:

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To solve, use Heron's Formula,

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

where , , and are the side lengths and

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

In this particular case,

thus,

$s = $\frac{ 1}{2}$ ( 13 + 11 + 7 ) = $\frac{ 1}{2}$ \cdot 31 = 15.5$ .

Plugging these values into the Heron's Formula we arrive at our answer.

$\ area = $\sqrt{ 15.5 ( 15.5 - 11 ) (15.5 - 13 ) ( 15.5 - 7 )}$ \ \= $\sqrt{ 15.5 \cdot 4.5 \cdot 2.5 \cdot 8. 5 }$ \approx 38.50$

Find the exact area of a triangle with side lengths of , , and .

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Use the Heron's Formula:

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

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

Solve for .

$s=\frac{a+b+c}{2}$=\frac{2+3+4}{2}$=\frac{9}{2}$$

Solve for the area.

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

$A=$\sqrt{\frac{9}{2}$\left($\frac{9}{2}$-2\right)\left($\frac{9}{2}$-3\right)\left($\frac{9}{2}$-4\right)}$

$A=$\sqrt{\frac{9}{2}$\left($\frac{5}{2}\right)\left($\frac{3}{2}\right)\left($\frac{1}{2}\right)}=$\sqrt{\frac{135}{16}$}=\frac{\sqrt{135}$}{4}$

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

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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}$$

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

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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}$$

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

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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}$$

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

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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

$s=\frac{1}{2}$(3+4+5)$

$s=\frac{1}{2}$(12)$

So the area is

$area=$\sqrt{6(6-3)(6-4)(6-5)}$$

$area=$\sqrt{36}$$

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

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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

$s=\frac{1}{2}$(5+12+13)$

$s=\frac{1}{2}$(30)$

So the area is

$area=$\sqrt{15(15-5)(15-12)(15-13)}$$

$area=$\sqrt{900}$$

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

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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}$$

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

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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

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

$s=\frac{1}{2}$(14)$

So the area is

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

$area=$\sqrt{84}$$

$area=2$\sqrt{21}$$

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

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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

$s=\frac{1}{2}$(3+3+2)$

$s=\frac{1}{2}$(8)$

So the area is

$area=$\sqrt{4(4-3)(4-3)(4-2)}$$

$area=$\sqrt{8}$$

$area=2$\sqrt{2}$$

In triangle , $m\angle C = 30^\circ$ , , and . Find the area of the triangle.

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When given the lengths of two sides and the measure of the angle included by the two sides, the area formula is:

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

Plugging in the given values we are able to calculate the area.

$A = \frac {1}{2}\cdot 40\cdot 20 \cdot \sin 30^\circ= 200$

On a Cartesian plane plot a circle centered at the origin of radius . Now draw a segment from to . Shade the area between the segment and the boundary of the circle, above the segment.

What is the area of the shaded region?

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The shaded area is one quarter of the area of the circle, minus the area of an isosceles right triangle with legs along the radius.

The circle has radius , so the area of the circle is

$A_C=\pi $r^2$$

$=\pi $(6)^2$=36\pi$ .

The part of the circle in the first quadrant is $$\frac{36\pi}{4}$=9\pi$

If we subtract the area under the line connecting we get the correct answer.

To find this area, recognize the geometry as a triangle. The two points give you the base and height so its area is,

$A_T=\frac{1}{2}$bh=\frac{1}{2}$\cdot6\cdot 6=18$ .

Therefore, the area of the region of interest is $9\pi-18$ .

Find the area of the shaded segment of the given circle that has a radius of 8 cm:

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The area of a segment can be calculated with the following formula:

Finding the area of the sector is calculated as such:

$A_${sector}=\frac{120^{\circ}$$$}{360^{\circ}$}\cdot \pi $\cdot8^{2}$$

$A_${sector}=67cm^{2}$$

To find the area of the triangle, the 120 degree isosceles can be divided into two 30 60 90 triangles:

The area for the triangle can be calculated as such:

$A_${triangle}=27.7cm^{2}$$

Going back to the original formula for segment area:

$A_${segment}=67cm^{2}$-27.7cm2=39.3cm2$

Find the area of the shaded segment:

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The area of the segment can be calculated with the following equation:

The area of the sector can be found as such:

$A_${sector}=\frac{90^{\circ}$$$}{360^{\circ}$}\cdot \pi \cdot(8; $cm)^{2}$$

The area of the triangle can be determined easily because it is a right triangle, at 90 degrees:

Now that we have both variables needed for the formula, we can determine the area of the segment:

Find the area of the shaded segment BC:

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The area of a segment can be calculated as such:

We can find the area of the sector as such:

$A_${sector}=\frac{160^{\circ}$$$}{360^{\circ}$}\cdot \pi \cdot(6; $cm)^{2}$=50.27; $cm^{2}$$

To find the area of the triangle, we must divide the isosceles into two right triangles with a bisector:

To determine base, we calculate:

$0.9848=\frac{1/2base}{6 ; cm}$$

To determine height, we calculate:

$0.1736=\frac{height}{6; cm}$$

Now that we have all the needed values, we can calculate the area of the triangle:

$Area=\frac{1}{2}$\cdot11.82; cm\cdot1.04; cm=6.15; $cm^{2}$$

The total segment area for BC can now be calculated:

Find the area of the segment BC:

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To solve for the area of the segment, we must first find the area of the sector and the triangle, to fulfill the given formula:

To find the area of the sector:

To find the area of the triangle, we first must divide it into two right triangles:

We can find the height of the triangle as such:

$0.9659=\frac{height}{4: cm}$$

We can find the base employing a similar method:

$0.2588=\frac{1/2base}{4: cm}$$

The area of the triangle can then be calculated as such:

$Area=\frac{1}{2}$\cdot2.07: cm\cdot3.86: cm=4: $cm^2$$

To find the Area of the segment, we can now subtract the triangle area from the sector area:

The triangular fence of the T-rex in Jurassic park has sides a,b, and c measuring 100m, 110m, and 120m respectively. What is area of its enclosure?

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Use Heron's formula to calculate the area:

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

A, B, and C are side lengths and s is calculated by:

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

Calculate s:

$s=\frac{100+110+120}{2}$=165$

Plug side lengths and s into Heron's formula:

$\A=$\sqrt{165(165-100)(165-110)(165-120)}$\ \=$\sqrt{165(65)(55)(45)}$\ $\=5,152m^2$$

Find the area of this triangle:

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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$

Find the area of this triangle:

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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$

Find the area of this triangle:

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To find the area, use the formula associated with side, angle, side triangles which states,

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

where and are side lengths and is the included angle.

In our case,

$a=8, b=17, C=90^\circ$ .

Plug the values into the area formula and solve.

$area = $\frac{ 1}{2}$ (8 )( 17 ) \cdot \sin( 90 ^ o ) = 68$