Area of a Triangle - Pre-Calculus

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Question

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

Answer

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.

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

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

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

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Question

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

Answer

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

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

$area=\sqrt{96}$

$area=4\sqrt{6}$

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Question

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

Answer

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

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

$area=\sqrt{15}$

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Question

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

Answer

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

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

$area=\sqrt{200}$

$area=10\sqrt{2}$

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Question

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

Answer

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

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

$area=\sqrt{36}$

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Question

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

Answer

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

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

$area=\sqrt{900}$

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Question

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

Answer

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

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

$area=\sqrt{7056}$

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Question

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

Answer

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

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

$area=\sqrt{84}$

$area=2\sqrt{21}$

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Question

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

Answer

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

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

$area=\sqrt{8}$

$area=2\sqrt{2}$

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Question

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

Answer

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.

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

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Question

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?

Answer

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 $9\pi-18$ .

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Question

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

Answer

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

$A_{segment}=A_{sector}-A_{triangle}$

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:

$A_{triangle}=\frac{1}{2}\cdot8\sqrt{3}\cdot4=16\sqrt{3}$

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

Going back to the original formula for segment area:

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

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Question

Find the area of the shaded segment:

Answer

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

$A_{segment}=A_{sector}-A_{triangle}$

The area of the sector can be found as such:

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

$A_{sector}=16\pi=50.3; cm^{2}$

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

$A_{triangle}=\frac{1}{2}\cdot8; cm\cdot8; cm=32; cm^{2}$

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

$A_{segment}=50.3; cm^{2}-32; cm^{2}=18.3; cm^{2}$

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Question

Find the area of the shaded segment BC:

Answer

The area of a segment can be calculated as such:

$A_{segment}=A_{sector}-A_{triangle}$

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:

$sin(80^{\circ})=\frac{1/2base}{6; cm}$

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

To determine height, we calculate:

$sin(10^{\circ})=\frac{height}{6; cm}$

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

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:

$A_{segment}=50.27; cm^{2}-6.15; cm^{2}=44.12; cm^{2}$

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Question

Find the area of the segment BC:

Answer

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:

$A_{segment}=A_{sector}-A_{triangle}$

To find the area of the sector:

$Area=\frac{30^{\circ}}{360^{\circ}}\cdot \pi \cdot(4cm)^{2}=4.2cm^{2}$

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:

$sin(75^{\circ})=\frac{height}{4: cm}$

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

We can find the base employing a similar method:

$sin(15^{\circ})=\frac{1/2base}{4: cm}$

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

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:

$A_{segment}=4.2: cm^{2}-4: cm^2=0.2: cm^{2}$

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Question

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?

Answer

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:

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

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Question

Find the area of this triangle:

Answer

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.

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

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Question

Find the area of this triangle:

Answer

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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Question

Find the area of this triangle:

Answer

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.

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

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Question

Find the area of this triangle:

Answer

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.

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

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