ΔABC is isosceles with AB=AC=7cm and BC=9cm. The height AD from A to BC, is 6cm. Find the area of ΔABC. What will be the height from C to AB i.e., CE?

Answer

To solve the problem step by step, we will first find the area of triangle ABC and then determine the height from point C to line AB.

### Step 1: Calculate the area of triangle ABC

Given:
- AB = AC = 7 cm (isosceles triangle)
- BC = 9 cm (base)
- Height AD from A to BC = 6 cm

The formula for the area of a triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

In triangle ABC, the base is BC and the height is AD.
\[ \text{Area} = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 9 \times 6 \]
Calculating this gives:
\[ \text{Area} = \frac{1}{2} \times 54 = 27 \text{ cm}^2 \]

### Step 2: Use the area to find height CE from C to AB

We know the area of triangle ABC is also given by:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we will use AB as the base and CE as the height. We can set up the equation:
\[ 27 = \frac{1}{2} \times AB \times CE \]
Substituting AB = 7 cm:
\[ 27 = \frac{1}{2} \times 7 \times CE \]

### Step 3: Solve for CE

To isolate CE, we can multiply both sides by 2:
\[ 54 = 7 \times CE \]
Now, divide both sides by 7:
\[ CE = \frac{54}{7} \approx 7.71 \text{ cm} \]

### Final Answers
- The area of triangle ABC is \(27 \text{ cm}^2\).
- The height from C to AB (CE) is approximately \(7.71 \text{ cm}\).