Respuesta :
Answer:
42
Step-by-step explanation:
We can make use of Heron's formula for the area of the triangle. Let x represent half the length of BC:
MB = MC = x
Then the perimeter of ΔAMB is ...
p1 = 7 + 10 + x
and the value used in Heron's formula, the semi-perimeter is ...
s1 = (17+x)/2
The square of the area of ΔAMB is then ...
A1² = s1·(s1 -x)·(s1 -10)·(s1 -7) = (-1/16)(x^4 -298x^2 +2601)
Similarly, the square of the area of ΔABC is ...
A2² = s2·(s2 -x)·(s2 -10)·(s2 -15) = (-1/16)(x^4 -650x^2 +15625)
We want these two triangle areas to be the same, so we can solve for the value of x that makes it so.
A1² - A2² = 0 = (-1/16)((x^4 -298x^2 +2601) -(x^4 -650x^2 +15625))
352x^2 -13024 = 0 . . . . multiply by -16 and collect terms
x^2 -37 = 0 . . . . . . . . . . . divide by 352
x^2 = 37 . . . . . this is as far as we need to take it in order to find the area.
Substituting the value for x^2 into the expression for A1², we get ...
A1² = (-1/16)((x^2 -298)x^2 +2601) = (-1/16)((37 -298)37 +2601) = 441
So, the area of ΔAMB is √441 = 21 square units. The area of ΔAMC is the same, so the area of ΔABC is ...
area(ΔABC) = 2×area(ΔAMB) = 2×21 units²
area(ΔABC) = 42 units²
_____
Addendum to the answer
After thinking about this a little more, I realized the area is that of half a parallelogram with AB and AC as two sides and AM as half the diagonal. In other words, the area is the same as that of a triangle with sides 7, 15, and 2×10=20. A single straightforward application of Heron's formula gives the area as ...
A = √(21(21-7)(21-15)(21-20)) = √(21(14)(6)) = 42
The base length 2·x can be obtained from the given side length through
the law of cosines and the area can then be calculated.
Response:
- The area of triangle ΔABC is 42
Which method can be used to find the area of the triangle?
From cosine law, we have;
7² = x² + 10² - 2 × x × 10 × cos(∠BMA)
15² = x² + 10² - 2 × x × 10 × cos(∠CMA)
∠BMA and ∠CMA are supplementary angles, therefore;
cos(∠CMA) = -cos(∠BMA)
Which gives;
- 7² = x² + 10² - 20·x·cos(∠BMA)
- 15² = x² + 10² + 20·x·cos(∠BMA)
Adding the two equations gives;
7² + 15² = x² + 10² - 20·x·cos(∠BMA) + x² + 10² + 20·x·cos(∠BMA)
274 = 2·x² + 200
2·x² = 274 - 200
[tex]x^2 = \dfrac{274 - 200}{2} = 37[/tex]
x = [tex]\sqrt{37}[/tex]
From cosine law, we have;
- [tex]\left(2 \cdot \sqrt{37} \right)^2[/tex] = 7² + 15² - 2× 7 × 15 × cos(∠BAC)
148 = 274 - 210 × cos(∠BAC)
[tex]cos(\angle BAC) = \dfrac{148 - 274}{-210} = \mathbf{ 0.6}[/tex]
∠BAC = arccos(0.6)
[tex]The \ area \ of \ a \ triangle \ is \ \mathbf{\dfrac{1}{2} \times a \times b \times sin(\angle C)}[/tex]
Where a and b are adjacent sides and C is the included angle between sides a and b.
Area of the triangle ΔABC is, A, is therefore;
A = 0.5 × 7 × 15 × sin(arccos(0.6)) = 42
- The area of the triangle ΔABC is 42
Learn more about the law of cosines here:
https://brainly.com/question/11625914
