What is the Altitude from A on BC in an Isosceles Triangle ( triangle ABC ) with ABAC25cm and BC14cm?
The given triangle ABC is an isosceles triangle with ABAC25cm and BC14cm. In an isosceles triangle, the altitude drawn from the vertex angle (which can be any of the angles formed by the equal sides) to the base will also bisect the base and form two right triangles. In this case, the altitude from A to BC will bisect BC at point D, dividing it into two equal segments of 7cm each (BDDC14/27cm).
Using the Pythagorean Theorem
Since AD forms a right angle with BC at D, AD is the height of the triangle, and we can use the Pythagorean theorem to find its length. Consider (triangle ADB): [ AB^2 AD^2 BD^2 ] Substituting the known values from the triangle: [ 25^2 AD^2 7^2 ] Solving for AD: [ AD^2 25^2 - 7^2 24^2 ] [ AD 24 text{ cm} ] Thus, the altitude from A to BC is 24 cm.
Calculating the Area of the Triangle Using Heron's Formula
We can also verify the altitude by using Heron's formula to calculate the area of the triangle and then finding the altitude from the area. First, calculate the semi-perimeter s of the triangle: [ s frac{AB AC BC}{2} frac{25 25 14}{2} 32 text{ cm} ] Using Heron's formula, the area A of the triangle is: [ A sqrt{s(s-a)(s-b)(s-c)} ] Substituting the values: [ A sqrt{32(32-25)(32-25)(32-14)} sqrt{32 cdot 7 cdot 7 cdot 18} sqrt{28224} ] [ A 168 text{ cm}^2 ] The area of the triangle can also be calculated as: [ A frac{1}{2} times text{base} times text{height} frac{1}{2} times 14 times text{AD} ] Setting these equal to each other: [ 168 frac{1}{2} times 14 times text{AD} ] Solving for AD: [ AD frac{168 times 2}{14} 24 text{ cm} ] This confirms our earlier calculation that the altitude from A to BC is 24 cm.
Summary
In this example, we confirmed the height of the altitude from A to BC in an isosceles triangle (triangle ABC) with side lengths ABAC25cm and BC14cm using both the Pythagorean theorem and Heron's formula. The altitude was found to be 24 cm, which verifies the consistency of the result.