greatwhiteshark
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The dimensions of a triangular lots are 100 feet by 50 feet by 75 feet. If the price of such a land is $3 per square foot, how much does the lot cost?
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Cost of Triangular Lot
- Thread starter greatwhiteshark
- Start date
The easiest way is Heron's formula. If you look at
http://www.csgnetwork.com/herontriangleareacalc.html
It is covered.
If you need some other way of finding the area, give us a hint.
http://www.csgnetwork.com/herontriangleareacalc.html
It is covered.
If you need some other way of finding the area, give us a hint.
greatwhiteshark
Full Member
- Joined
- May 8, 2005
- Messages
- 279
Gene
This question was taken from a chapter titled: Area of Triangle Using Trigonometry. I multiplied 100ft by 50ft by 75 feet and then divided by the measure of a square foot. Afterward, I multiplied the quotient by $3 but got the wrong answer.
HELP.
Janet
This question was taken from a chapter titled: Area of Triangle Using Trigonometry. I multiplied 100ft by 50ft by 75 feet and then divided by the measure of a square foot. Afterward, I multiplied the quotient by $3 but got the wrong answer.
HELP.
Janet
Okay, you can still use Heron to check your answer. I don't know where you gat that method for area. If you multiply three lengths you get a volume, not an area. ft*ft*ft = ft^3 HAS to be volume. Area has to end up with ft^2 (or inches or miles)^2. Some length squared.
The first area of a triangle is (1/2)*base * height (note base * height. Two lengths). You can check first to see if you are lucky and have a right triangle.
75²+50² = 90²
Nope that isn't 100² so no right triangle. Further 100 > 90 so it is an obtuse triangle. That will help in your sketch. Draw a horizontal line and call it 50. Sketch a circle around the right end with a radius half again the length of that line and another around the left end twice the length. Where they meet is the third point of the mystery triangle. It helps to have a sketch of what you are realy working with.
Now lets call the left corner A, the other bottom corner B and the intersection corner C.
Extend AB to the right, below C at D. Now you have two right triangles.
BC² = BD²+DC²
75² = x² + y²
and AC² = AD² + DC²
100² = (50 + x)² + y²
You want to solve for y which is the height of the triangle.
Then
area = (1/2)*base*height = (1/2) * 50 * y
The first area of a triangle is (1/2)*base * height (note base * height. Two lengths). You can check first to see if you are lucky and have a right triangle.
75²+50² = 90²
Nope that isn't 100² so no right triangle. Further 100 > 90 so it is an obtuse triangle. That will help in your sketch. Draw a horizontal line and call it 50. Sketch a circle around the right end with a radius half again the length of that line and another around the left end twice the length. Where they meet is the third point of the mystery triangle. It helps to have a sketch of what you are realy working with.
Now lets call the left corner A, the other bottom corner B and the intersection corner C.
Extend AB to the right, below C at D. Now you have two right triangles.
BC² = BD²+DC²
75² = x² + y²
and AC² = AD² + DC²
100² = (50 + x)² + y²
You want to solve for y which is the height of the triangle.
Then
area = (1/2)*base*height = (1/2) * 50 * y
Heron's Formula
Heron's formula is just a nifty formula that's handy if you have three sides of a triangle and don't want to apply the law of cosines/sines to find the angles.
To use it, you take half the perimeter, = "s", and...
sqrt[s(s-a)(s-b)(s-c)]
...with a, b, c being equal to any of the three sides (it doesn't matter which, just be consistent.)
I figured it'd be important to point out that these values would give you four distances multiplied together, not three. When you take the square root of these multiplied distances, you get (some distance)^2.
for example, if you were to use inches...
inches*inches*inches*inches = sqrt(in^4) = in^2
Both methods are useful in the right situation
Heron's formula is just a nifty formula that's handy if you have three sides of a triangle and don't want to apply the law of cosines/sines to find the angles.
To use it, you take half the perimeter, = "s", and...
sqrt[s(s-a)(s-b)(s-c)]
...with a, b, c being equal to any of the three sides (it doesn't matter which, just be consistent.)
I figured it'd be important to point out that these values would give you four distances multiplied together, not three. When you take the square root of these multiplied distances, you get (some distance)^2.
for example, if you were to use inches...
inches*inches*inches*inches = sqrt(in^4) = in^2
Both methods are useful in the right situation