Question

use the diagram shown. if bc = 14, ab = 5, and the perimeter of △abc is 33, then what is the value of bd?
the value of bd is □
(simplify your answer.)

Explanation:

Step1: Find the length of AC

The perimeter of \(\triangle ABC\) is \(P = AB + BC+AC\). Given \(P = 33\), \(AB = 5\), and \(BC = 14\).
Substitute into the formula: \(33=5 + 14+AC\).
Solve for \(AC\): \(AC=33-(5 + 14)=33 - 19=14\).

Step2: Use the Pythagorean theorem in \(\triangle BDC\)

Since \(AC = BC = 14\), \(\triangle ABC\) is isosceles. And \(CD\perp AB\), so \(AD = BD\) (by the property of isosceles triangles: the altitude from the vertex to the base bisects the base). Let \(BD=x\), then \(AD=x\) and \(AB=AD + BD=2x\). But \(AB = 5\), so \(x=\frac{5}{2}\) is wrong. Wait, no, we made a mistake above.

Let's start over.
The perimeter of \(\triangle ABC\) is \(P=AB + BC+AC\). Given \(AB = 5\), \(BC = 14\), \(P = 33\). So \(AC=33-(5 + 14)=14\).
Since \(CD\perp AB\), in right - triangle \(BDC\), using the Pythagorean theorem. But also, since \(AC = BC = 14\), \(\triangle ABC\) is isosceles and \(CD\) is the altitude, so \(AD=BD\) (by the property of isosceles triangle: the altitude from the vertex to the base bisects the base). Let \(BD=x\), then \(AD=x\) and \(AB = 2x\) is wrong. Wait, no, \(AB = 5\).
Let's use the Pythagorean theorem in \(\triangle ADC\) and \(\triangle BDC\). Let \(BD=x\), then \(AD = 5 - x\).
In \(\triangle ADC\), \(CD^{2}=AC^{2}-AD^{2}\), in \(\triangle BDC\), \(CD^{2}=BC^{2}-BD^{2}\). Since \(AC = BC = 14\), we have \(14^{2}-(5 - x)^{2}=14^{2}-x^{2}\).
Expand: \(196-(25 - 10x+x^{2})=196 - x^{2}\).
\(196-25 + 10x-x^{2}=196 - x^{2}\).
Cancel out \(196\) and \(-x^{2}\) on both sides: \(10x=25\), \(x=\frac{5}{2}\).

Another way:
Since \(AC = BC = 14\), \(\triangle ABC\) is isosceles and \(CD\perp AB\). Let \(BD=x\), then \(AD = 5 - x\).
By the Pythagorean theorem:
In \(\triangle BDC\), \(CD^{2}=BC^{2}-BD^{2}=14^{2}-x^{2}\).
In \(\triangle ADC\), \(CD^{2}=AC^{2}-(AB - BD)^{2}=14^{2}-(5 - x)^{2}\).
Set them equal: \(14^{2}-x^{2}=14^{2}-(25 - 10x+x^{2})\).
\(-x^{2}=-25 + 10x-x^{2}\).
\(10x = 25\), \(x=\frac{5}{2}\).

Answer:

\(\frac{5}{2}\)