In a `Delta` ABC, BE = 9 CF = 6 . Intersect at `90^(@)` Find AC

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the triangle and given values
We have a triangle ABC with points B and C. The segments BE and CF intersect at a right angle (90 degrees). We are given that BE = 9 cm and CF = 6 cm.

### Step 2: Determine the ratio of the segments
Since BE and CF intersect at a right angle, we can use the property of the areas of triangles. The segments divide the triangle into smaller triangles, and the areas of these triangles will be in a ratio of 2:1.

### Step 3: Calculate the area of triangle BFC
To find the area of triangle BFC, we can use the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base FC = 6 cm and the height BO (which we will find) will be in the ratio of 2:1.

Let’s denote the height BO as \( h \). Since BE is 9 cm, we can say that BO is also 6 cm (because of the 2:1 ratio).

Now, we can calculate the area of triangle BFC:
\[ \text{Area of } BFC = \frac{1}{2} \times FC \times BO = \frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2 \]

### Step 4: Calculate the area of triangle ABC
Since the area of triangle BFC is 18 cm² and the ratio of the areas of triangle ABC to triangle BFC is 2:1, we can find the area of triangle ABC:
\[ \text{Area of } ABC = 2 \times \text{Area of } BFC = 2 \times 18 = 36 \text{ cm}^2 \]

### Step 5: Use the area to find AC
Now, we can use the area of triangle ABC to find the length of AC. The formula for the area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \times AC \times BE \]
We know the area is 36 cm² and BE is 9 cm. Plugging in the values:
\[ 36 = \frac{1}{2} \times AC \times 9 \]

### Step 6: Solve for AC
To find AC, we can rearrange the equation:
\[ 36 = \frac{9}{2} \times AC \]
Multiplying both sides by 2:
\[ 72 = 9 \times AC \]
Now, divide both sides by 9:
\[ AC = \frac{72}{9} = 8 \text{ cm} \]

### Conclusion
The length of AC is 8 cm.

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