Question

MACD - AABE, find the value of x,C3x + 8B20A X-1E27DA. 12.51B14,28c16D 11

Answer 1

Given that the triangle ACD and ABE are similar.

`\Delta ACD\text{ \textasciitilde }\Delta ABE`

Then their corresponding sides must be proportional;

`(BE)/(CD)=(AE)/(AD)`

Given;

`\begin{gathered} BE=20 \\ CD=3x+8 \\ AE=x-1 \\ AD=x-1+27=x+26 \end{gathered}`

substituting the given values;

`\begin{gathered} (BE)/(CD)=(AE)/(AD) \\ (20)/(3x+8)=(x-1)/(x+26) \end{gathered}`

cross multiply and solve;

`\begin{gathered} 20(x+26)=(3x+8)(x-1) \\ 20x+520=3x^2-3x+8x-8 \\ 20x+520=3x^2+5x-8 \\ 3x^2+5x-8-(20x+520)=0 \\ 3x^2-15x-528=0 \end{gathered}`

solving the quadratic equation, we have;

`\begin{gathered} 3(x-16)(x+11)=0 \\ so; \\ x=16 \\ or \\ x=-11 \\ \text{ since x cannot be negative then;} \\ x=16 \end{gathered}`

Therefore, the value of x is;