Question

ABC is an isosceles triangle with AB=AC=12cm and BC=8cm. Find the altitude on BC and hence calculate its area

Answer 1

By using herons formula first, then by area of triangle formula.....

Answer 2

Answer:

The altitude is 8√2 cm and area is 32√2cm²

Step-by-step explanation:

In isosceles AABC, AB = AC = 12 cm and BC = 8 cm.

Draw AD 1 BC

⇒ZADB = <ADC = 90°

The altitude on the ual side touches the midpoint of the side in an isosceles triangle.

Therefore, BD = DC = BC÷2 = 8÷2 = 4

1. In AADC, ZADC = 90°, AC = 12 cm and DC = 4 cm. By Pythagoras Theorem: AC² = AD² + DC²
2. Put the values of AC and DC in the above the equation: 12² = AD + 4²
3. Solve for AD: AD² = 12² – 4² = 144 – 16 = 128
4. ⇒ AD = √128 = 8√/2 cm

Find area of the isosceles AABC:

A = 1/2 × base × altitude

A = 1/2 × BC × AD

A = 32√2cm²