In triangle ABC, we have AB=3 and AC=4. Side BC and the median from A to BC have the same length. What is BC?
Not making sense to me, I think the answer must be simple, but I don't know how to solve I applied the law of sines but to no avail. Help is appreciated, thanks.
median length x = BC length
median hits BC at D
triangle ABC
4^2 = 3^2 + x^2 - 6 x cos B
triangle ABD
x^2 = 3^2 + (x/2)^2 - 3 x cos B
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16 = 9 + x^2 - 6x cos B
x^2 = 9 +x^2/4 -3 x cos B
x^2 -6 x cos B -7 = 0
3x^2/4 +3 x cos B - 9 = 0
x^2 - 6 x cos B - 7 = 0
(3/2)x^2 + 6 x cos B -18 = 0
(5/2)x^2 - 25 = 0
5 x^2 = 50
x^2 = 10
x = sqrt(10)
thanks a lot man!
To find the length of side BC, let's use the properties of a median in a triangle.
In triangle ABC, let D be the midpoint of BC, and let M be the foot of the perpendicular from A to BC.
Since the median from A to BC has the same length as side BC, we can say that AM = MD.
Now let's use the Pythagorean theorem in triangle ABM.
According to the given information, AB = 3 and AC = 4. We need to find BC.
We have AM = MD, so using the Pythagorean theorem:
AB^2 = AM^2 + BM^2
3^2 = (4/2)^2 + BM^2
9 = 2^2 + BM^2
9 = 4 + BM^2
BM^2 = 9 - 4
BM^2 = 5
Now, let's use the Pythagorean theorem in triangle ACM.
AC^2 = AM^2 + CM^2
4^2 = (4/2)^2 + CM^2
16 = 2^2 + CM^2
16 = 4 + CM^2
CM^2 = 16 - 4
CM^2 = 12
Since we know that CM = BM (medians divide the side they connect in a 2:1 ratio), we can equate BM^2 and CM^2:
BM^2 = CM^2
5 = 12
BM^2 = 5
This implies that the equation 5 = 12 is false, which means there is no such triangle that satisfies the conditions. Therefore, we cannot determine the length of side BC.
To solve this problem, let's start by drawing a diagram to visualize the information given.
We have triangle ABC, where AB = 3 units and AC = 4 units. Let's label the midpoint of side BC as M and the length of BC as x.
Since the median from A to BC has the same length as BC, we can label the length of AM as x units as well.
Now, we can apply the Law of Cosines to triangle ABC to find the value of x. The Law of Cosines states that for any triangle with sides a, b, and c, and angle C opposite side c, the following equation holds:
c^2 = a^2 + b^2 - 2ab*cos(C)
In our case, we have side a = 3, side b = 4, and angle C opposite side c = angle BAC = 90 degrees (since the median from A to BC forms a right angle with BC).
Plugging in the values into the equation, we have:
x^2 = 3^2 + 4^2 - 2*3*4*cos(90)
Simplifying further:
x^2 = 9 + 16 - 24*cos(90)
Since cos(90) = 0, we can simplify the equation to:
x^2 = 9 + 16
x^2 = 25
Taking the square root of both sides:
x = √25
x = 5
Therefore, the length of side BC is 5 units.