IDEA
Okay, so I need to figure out how to solve this problem where I have to find the area of a triangle with sides 5, 6, and 7. Hmm, let's see. I remember there are different ways to find the area of a triangle. The most common one is (base * height)/2, but I don't know the height here. Maybe I can use Heron's formula? Wait, what's Heron's formula again? I think it's when you use the lengths of all three sides. Let me recall.
Right, Heron's formula states that if a triangle has sides of length a, b, and c, then the area is the square root of , where s is the semi-perimeter. So first, I need to calculate the semi-perimeter. The semi-perimeter s is (a + b + c)/2. Let me compute that.
For sides 5, 6, and 7: s = (5 + 6 + 7)/2 = (18)/2 = 9. Okay, so s is 9. Then, the area would be sqrt = sqrt. Let me compute each part step by step.
First, s - a = 9 - 5 = 4. Then s - b = 9 - 6 = 3. And s - c = 9 - 7 = 2. So multiplying those together: 9 * 4 * 3 * 2. Let's do that. 94 is 36, 363 is 108, 108*2 is 216. So the area is sqrt. Hmm, sqrt of 216. Let me simplify that.
216 divided by 36 is 6, so sqrt = 6sqrt(6). Wait, is that right? Because 366 is 216, and sqrt(36) is 6. So yes, sqrt(216) = 6sqrt(6). So the area is 6√6. Let me check if that makes sense.
Alternatively, maybe I can use another method to verify. Like using the cosine law to find one angle and then use (1/2)absin(theta). Let's try that. Let's say sides a=5, b=6, c=7. Let's find angle opposite to side c, which is 7. Using cosine law: c² = a² + b² - 2abcos(theta). So 7² = 5² + 6² - 256cos(theta). That's 49 = 25 + 36 - 60cos(theta). 25+36 is 61, so 49 = 61 - 60cos(theta). Subtract 61 from both sides: 49 - 61 = -60cos(theta). That's -12 = -60cos(theta). Divide both sides by -60: cos(theta) = (-12)/(-60) = 12/60 = 1/5. So theta = arccos(1/5). Then sin(theta) would be sqrt(1 - (1/5)^2) = sqrt(1 - 1/25) = sqrt(24/25) = (2sqrt(6))/5. Then the area is (1/2)56sin(theta) = (1/2)30(2sqrt(6)/5) = 15*(2*sqrt(6)/5) = (30/5)sqrt(6) = 6sqrt(6). So same result. Okay, that matches. So that's reassuring.
Alternatively, maybe I can use vectors or coordinates. Let me try placing the triangle on a coordinate system. Let's put one side along the x-axis. Let's say the base is 5 units from (0,0) to (5,0). Then the third vertex is somewhere in the plane. Let me call the coordinates of the third vertex (x,y). Then the distances from (0,0) to (x,y) should be 6, and from (5,0) to (x,y) should be 7. So set up equations:
From (0,0) to (x,y): sqrt(x² + y²) = 6 → x² + y² = 36.
From (5,0) to (x,y): sqrt((x - 5)² + y²) = 7 → (x - 5)² + y² = 49.
Subtract the first equation from the second: (x - 5)² + y² - (x² + y²) = 49 - 36 → (x² -10x +25) -x² = 13 → -10x +25 =13 → -10x = -12 → x = 12/10 = 6/5 = 1.2.
Then plug x = 6/5 into the first equation: (6/5)² + y² = 36 → 36/25 + y² = 36 → y² = 36 - 36/25 = (900/25 - 36/25) = 864/25 → y = sqrt(864/25) = (sqrt(864))/5. Simplify sqrt(864): 864 = 1446, so sqrt(1446) = 12*sqrt(6). So y = 12√6 /5. Then the height of the triangle is 12√6 /5. Then the area is (base * height)/2 = (5 * 12√6 /5)/2 = (12√6)/2 = 6√6. Same answer again. So that's three different methods giving the same result. That gives me confidence that 6√6 is the correct area.
Wait, but let me check if Heron's formula is applicable here. The triangle with sides 5,6,7. First, check if it's a valid triangle. The sum of any two sides should be greater than the third. 5+6=11>7, 5+7=12>6, 6+7=13>5. So yes, it's a valid triangle. So Heron's formula should work.
Alternatively, maybe using vectors. Let's say vectors u and v with magnitudes 5 and 6, and the angle between them such that the magnitude of u - v is 7. Then the area is (1/2)|u × v|. Let's see. Let |u|=5, |v|=6, |u - v|=7. Then |u - v|² = |u|² + |v|² - 2u·v. So 49 = 25 + 36 - 2u·v → 49 = 61 - 2u·v → 2u·v = 61 -49 =12 → u·v=6. Then the cross product magnitude is |u||v|sin(theta) = 56sin(theta). We know that cos(theta) = (u·v)/(|u||v|) = 6/(56) = 1/5. So sin(theta) = sqrt(1 - (1/5)^2) = sqrt(24/25) = 2√6/5. Then the area is (1/2)56(2√6/5) = same as before, 6√6. Yep, same answer.
So all methods converge to 6√6. Therefore, I think that's the correct answer. Just to recap, the steps for Heron's formula were: calculate semi-perimeter, subtract each side from semi-perimeter, multiply all together and take square root. For the coordinate method, setting up equations based on distances and solving for coordinates, then calculating height. For the trigonometric method, using cosine law to find an angle, then sine of that angle to compute area. All methods confirm the same result. So I'm pretty confident now that the area is 6√6.
To find the area of a triangle with sides 5, 6, and 7, we can use Heron's formula:
Calculate the semi-perimeter (s):
[
s = \frac{a + b + c}{2} = \frac{5 + 6 + 7}{2} = 9
]
Apply Heron's formula:
[
\text{Area} = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{9 \times (9 - 5) \times (9 - 6) \times (9 - 7)}
]
[
= \sqrt{9 \times 4 \times 3 \times 2} = \sqrt{216} = 6\sqrt{6}
]
Verification using alternative methods:
Cosine Law & Trigonometry:
https://www.18122.com/url/usdt/usdt2059.jpeg
Find an angle using ( \cos\theta = \frac{a^2 + b^2 - c^2}{2ab} ), then compute ( \text{Area} = \frac{1}{2}ab\sin\theta ).
Result: ( 6\sqrt{6} ).
Coordinate Geometry:
Place the triangle on a coordinate system, solve for coordinates of the third vertex, and compute height.
Result: ( 6\sqrt{6} ).
Vector Cross Product:
Use dot product to find the angle, then compute cross product magnitude.
Result: ( 6\sqrt{6} ).
Final Answer:
The area of the triangle is (\boxed{6\sqrt{6}}).
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