Triangle problem

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Triangle problem

Here is a geometric problem for advanced K-12 students:

The area of a triangle ABC is 48 m^2, with two sides AB = 10m, and AC = 12 m. Find the third side BC. We know that the third side is a whole number.

Posted 169 day ago

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matherapist	Best Answer There is actually a second answer to this problem. Using the area formula A=[s(s-a)(s-b)(s-c)]^.5 where s is the semi perimeter s=(a+b+c)/2 where a,b,c are the sides of the triangle. let a=10,b=12, then after substituting and a bit of algebra we have 48 x 4 = ([22^2-c^2][c^2-4])^.5 This reduces to: c^4-488c^2+38800 FACTORING we have [c^2-388][c^2-100] Thus c=+-SQRT(388) and +-SQRT(100) The negative values are thrown out and c=10 or 19.6977 both values are OK sinc e10>12-10 and 19.7 < 12+10 Math fanatic' s method is fine for an acute angle and the short diagonal of a paralellogram. Posted 168 day ago ( 1 ) ( 0 ) Report Abuse

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Answers (6)

atta

This ABC triangle is an Isosceles triangle with at most two congruent sides. By dividing the ABC Isosceles triangle into its right-triangular counterparts yields the height h^2 = AB^2 - AC^2/2 = sqrt(AB^2 - AC^2/2) = sqrt(10^2 - 6^2) = sqrt(64) = 8 = h. Given the height (h = 8) for the first right-triangle, the hypothenuse BC of the second right triangle can be derived via the Pythagorean Theorem, which is: BC^2 = h^2 + AC^2/2 = sqrt(h^2 + AC^2/2) = sqrt(8^2 + 6^2) = sqrt(100) = 10 = BC. By substituting the height (h = 8 && base AC = 12) into the original formula yields the following equality:
A = 0.5 * height * base , where A = 48 thus (48) = 0.5 * (8) * (12) = 48 thereby making the third side true for (BC = 10).

J.C

Posted 163 day ago

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Our triangle problem if solved by Heron's formula, gives both answers at once, but the method used by math_fanatic can be used twice to find the second solution as well.
It is given here so that the high school students would learn both methods used by math_fanatic and matherapist.

If Angle A > 90 degrees, then point D, the foot of height h, will fall on the extention of 12m side, and h will still be equal to 8m as shown by math_fanatic.
.
Now, in the right angle triangle ADB , length of AD will be (10^2 – 8^2)^0.5 = 6m, so the length of extended side, DC, will be 12+6 = 18m, which is the long leg of right angle triangle BDC.
So the length of BC, which is the third side of the original triangle, becomes: SQRT( 18^2 + 8^2) = 19.6977, as calculated by Heron’s formula.

Comment by: Professor rajbaba78

Posted 167 day ago

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I chose the matherapist’s solution as the Best Solution, but it lacks some details so will be somewhat difficult for high school students to follow exactly his solution.
Below is my version that has slightly more details, and since my idea is to teach more mathematics to high school students, I am submitting it for students benefit:

More than 2000 years ago, an Alexandrian Greek mathematician by the name of Heron proved that knowing the three sides of any triangle, a, b, and c, its area can be calculated by the following equation:

Area = [( s.(s-a).(s-b).(s-c)]^0.5, where s = half of the triangle’s perimeter = (a + b+ c)/2.
You don’t have to prove this formula at this time, although the solution is available.

In our problem, the area and two sides of the triangle are given.
Let the third side be equal to 2X. Therefor we can write:

s = (10 + 12 + 2X)/2 = (11 + x), it’s area becomes:

48 = [(11 + X).( 1 + X).(X – 1).( 11-X)]^0.5, or
using the formula (a + b).(a – b) = a^2 – b^2 , we get:

48 = [( 11^2 – X^2).( X^2 - 1^2)]^0.5 or
48 = [(121 – X^2).( X^2 – 1)]^0.5,
squaring both sides:
2304 = 122.X^2 – 121 – X^4
Collecting all terms on the left side, we get:

X^4 – 122 X^2 + 2425 = 0, which is a quadratic equation in X^2.

Therefore X^2 = [ + 122  (122^2 – 4.(1). (2425)^0.5]/2
Or X^2 = (122  72)/2, so X^2 = 97 or 25. Which results in:
X  9.8488 or X = 5. Now, per problem statement, we ignore the first answer and find the length of third side equal to 2X = 10m. <------Answer.

Posted 168 day ago

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I liked matherapist's answer. In the problem statement, I mentioned that the third side is a whole number,
because I new that the second solution will have a fractional number as matherapist has shown. I chose his answer as the Best Solution, although solution number 1 is also correct and acceptable.

matherapist has used Heron's formula which was developed by Alexandrian Greek mathematcian some 2110 ago, in fact 100 years BC.

CONGRATULATION TO matherapist. Your solution deserved to be marked as the best solution.
Commented by: Professor rajbaba78.

Posted 168 day ago

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matherapist

There is actually a second answer to this problem.
Using the area formula A=[s(s-a)(s-b)(s-c)]^.5 where s is the semi perimeter s=(a+b+c)/2 where a,b,c are the sides of the triangle.
let a=10,b=12, then after substituting and a bit of algebra we have
48 x 4 = ([22^2-c^2][c^2-4])^.5
This reduces to:
c^4-488c^2+38800
FACTORING we have
[c^2-388][c^2-100]
Thus c=+-SQRT(388) and +-SQRT(100)
The negative values are thrown out and c=10 or 19.6977
both values are OK sinc e10>12-10 and 19.7 < 12+10

Math fanatic' s method is fine for an acute angle and the short diagonal of a paralellogram.

Posted 168 day ago

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math_fanatic

Why do you keep posting these questions? The only ones who're going to answer are the tutors and the kids are never going to even look at our answers. It seems pointless but I'll keep answering in an attempt to keep my math skills alive and well.

Formula for the area of a triangle:
a = 1/2bh

let AC be the base. b = 12
48 = 1/2(12)h h = 8

Let D be the point where the height intersects the base.
Now we have a new sub triangle that's a right triangle, ABD. AB = 10 and BD = 8. AB is the hypotenuse so BD^2 + AD^2 = AB^2 => 64 + AD^2 = 100
AD^2 = 36 so AD = 6

Since AC is 12 and AD is 6, then CD is also 6 which means that triangle ABD is congruent to CBD so BC is 10, therefore ABC is an isosceles triangle with sides 10, 10, and 12.

Posted 169 day ago

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