In triangle ABC, ∠ABC = 90◦ . Point M is on side AB so that AM = MB, and CM = 4 and AC = 7. Find BC.

To solve the problem, let's analyze the given triangle \(ABC\):

- \( \triangle ABC \) is a right triangle with \( \angle ABC = 90^\circ \).
- \( AM = MB \), meaning that \( M \) is the midpoint of \( AB \).
- \( CM = 4 \) and \( AC = 7 \).

We need to find the length of \( BC \).

### Step 1: Label the triangle
Let:
- \( AB = c \)
- \( BC = a \) (which we need to find)
- \( AC = b = 7 \)
- \( AM = MB = \frac{c}{2} \)

### Step 2: Apply the Pythagorean Theorem in \( \triangle ABC \)
Since \( \triangle ABC \) is a right triangle, the Pythagorean theorem states:

\[
a^2 + c^2 = b^2
\]

Substituting the known value of \( b = 7 \):

\[
a^2 + c^2 = 49
\]

### Step 3: Consider the triangle \( \triangle AMC \)
Since \( M \) is the midpoint of \( AB \), \( CM \) can be considered as the median to the hypotenuse in \( \triangle ABC \). In a right triangle, the length of the median to the hypotenuse is half the length of the hypotenuse.

However, we can also use the distance formula:

In \( \triangle AMC \):

- \( AM = MB = \frac{c}{2} \)
- \( CM = 4 \)
- \( AC = 7 \)

Apply the distance formula:

\[
CM^2 = AM^2 + (AC - AM)^2
\]

Substitute the given values:

\[
16 = \left(\frac{c}{2}\right)^2 + \left(7 - \frac{c}{2}\right)^2
\]

### Step 4: Solve the quadratic equation
First, expand and simplify the equation:

\[
16 = \frac{c^2}{4} + \left(7 - \frac{c}{2}\right)^2
\]

Now expand the square:

\[
16 = \frac{c^2}{4} + \left(49 - 7c + \frac{c^2}{4}\right)
\]

Combine like terms:

\[
16 = \frac{c^2}{2} - 7c + 49
\]

Multiply through by 2 to eliminate the fraction:

\[
32 = c^2 - 14c + 98
\]

Subtract 32 from both sides:

\[
c^2 - 14c + 66 = 0
\]

### Step 5: Use the quadratic formula to solve for \( c \)
The quadratic formula is:

\[
c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the coefficients:

\[
c = \frac{14 \pm \sqrt{196 - 264}}{2}
\]

\[
c = \frac{14 \pm \sqrt{-68}}{2}
\]

Since the discriminant is negative, the equation has no real solution under these specific conditions. There seems to be a miscalculation in the problem setup. Given the structure, this indicates either additional constraints or another geometric method might be necessary, such as applying trigonometric identities or more geometric insights into the relationship between the lengths. Recheck problem conditions or steps might be essential.

Would you like me to recheck any particular part or provide further details?

---

Here are some additional questions for practice:

1. What would happen if \( CM \) were larger than \( AC \)?
2. How would the problem change if \( AM \neq MB \)?
3. What if \( \triangle ABC \) were not a right triangle?
4. How could you solve the problem using trigonometric ratios?
5. What is the significance of the median to the hypotenuse in a right triangle?
6. How would you solve for \( BC \) if \( CM \) were given as 3 instead of 4?
7. What additional information would simplify finding \( BC \)?
8. How can geometric constructions help in solving such problems?

**Tip:** Always double-check the conditions and try different methods, such as the Pythagorean theorem, trigonometry, or the use of medians, to explore all possible solutions.

Explore how to find the length of BC in triangle ABC where ∠ABC = 90°, AM = MB, CM = 4, and AC = 7. This problem involves applying the Pythagorean theorem, distance formula, and quadratic equations. Suitable for high school students learning geometry.

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