Question

1. find the value of a. show your work for full credit!!! type a response

Explanation:

Step1: Apply the Pythagorean theorem in right - triangle BCD

In right - triangle BCD with legs CD = 20 and BD (unknown) and hypotenuse BC = 27. By the Pythagorean theorem \(BD=\sqrt{BC^{2}-CD^{2}}\). So \(BD = \sqrt{27^{2}-20^{2}}=\sqrt{(27 + 20)(27 - 20)}=\sqrt{47\times7}=\sqrt{329}\).

Step2: Apply the Pythagorean theorem in right - triangle ABD

In right - triangle ABD, AD = 20, BD=\(\sqrt{329}\), and AB = 4a + 11. By the Pythagorean theorem \(AB^{2}=AD^{2}+BD^{2}\). So \((4a + 11)^{2}=20^{2}+329\).
Expand the left - hand side: \(16a^{2}+88a + 121=400+329\).
Simplify the right - hand side: \(16a^{2}+88a+121 = 729\).
Rearrange to get a quadratic equation: \(16a^{2}+88a+121 - 729=0\), i.e., \(16a^{2}+88a - 608 = 0\).
Divide through by 8: \(2a^{2}+11a - 76 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(Ax^{2}+Bx + C = 0\) (here \(A = 2\), \(B = 11\), \(C=-76\)), the quadratic formula is \(x=\frac{-B\pm\sqrt{B^{2}-4AC}}{2A}\).
First, calculate the discriminant \(\Delta=B^{2}-4AC=(11)^{2}-4\times2\times(-76)=121 + 608 = 729\).
Then \(a=\frac{-11\pm\sqrt{729}}{4}=\frac{-11\pm27}{4}\).
We have two solutions for a: \(a_1=\frac{-11 + 27}{4}=\frac{16}{4}=4\) and \(a_2=\frac{-11 - 27}{4}=\frac{-38}{4}=-\frac{19}{2}\).
Since length cannot be negative in the context of the geometric figure, we discard \(a =-\frac{19}{2}\).

Answer:

\(a = 4\)