Find the area of an isosceles `DeltaABC` in which BC = 8 cm and AB = AC = 5 cm. If `CEbotAB`, find CE.

Answer

To find the area of isosceles triangle ABC where BC = 8 cm and AB = AC = 5 cm, and to find the length of CE (the height from C to AB), we can follow these steps:

### Step 1: Identify the lengths of the sides
We have:
- BC = 8 cm (base)
- AB = AC = 5 cm (equal sides)

### Step 2: Draw the height CE from point C to line AB
Since triangle ABC is isosceles, the height CE will bisect the base BC into two equal parts. Therefore:
- BE = AE = BC / 2 = 8 cm / 2 = 4 cm

### Step 3: Apply the Pythagorean theorem to triangle CBE
In triangle CBE, we can apply the Pythagorean theorem since we have a right triangle:
- Hypotenuse (CB) = 5 cm
- Base (BE) = 4 cm
- Height (CE) = ?

Using the Pythagorean theorem:
\[ CB^2 = BE^2 + CE^2 \]
Substituting the known values:
\[ 5^2 = 4^2 + CE^2 \]
\[ 25 = 16 + CE^2 \]

### Step 4: Solve for CE
Rearranging the equation:
\[ CE^2 = 25 - 16 \]
\[ CE^2 = 9 \]
Taking the square root of both sides:
\[ CE = \sqrt{9} = 3 \, \text{cm} \]

### Step 5: Calculate the area of triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is BC (8 cm) and the height is CE (3 cm):
\[ A = \frac{1}{2} \times 8 \times 3 \]
\[ A = \frac{1}{2} \times 24 \]
\[ A = 12 \, \text{cm}^2 \]

### Final Answers:
- Length of CE = 3 cm
- Area of triangle ABC = 12 cm²