Area of a Triangle - Pre-Calculus

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Question

Find the area of this triangle:

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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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Question

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

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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 ?

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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 ?

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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 ?

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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 ?

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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 ?

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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 ?

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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 ?

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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 ?

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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.

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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?

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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:

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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:

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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:

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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:

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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?

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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:

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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:

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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:

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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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