Question

If CDE ~ GDF, find ED

Answer 1

Answer:

10

Step-by-step explanation:

$\triangle CDE \sim \triangle GDF.. (given) \\\\

\therefore \frac{CD}{GD} =\frac{DE}{DF}.. (csst) \\\\

\therefore \frac{15}{x+3} =\frac{3x+1}{4}\\\\ \therefore \: 15 \times 4 = (x + 3)(3x + 1) \\ \\ \therefore \: 60 = 3 {x}^{2} + x + 9x + 3 \\ \\ \therefore 3 {x}^{2} + 10x + 3 - 60 = 0 \\ \therefore 3 {x}^{2} + 10x - 57 = 0 \\ \therefore 3 {x}^{2} + 19x - 9x - 57 = 0 \\ \therefore \: x(3x + 19) - 3(3x + 19) = 0 \\\therefore \: (3x + 19)(x - 3) = 0 \\ \therefore \: 3x + 19 = 0 \: \: or \: \: x - 3 = 0 \\ \therefore \: x = - \frac{19}{3} \: \: or \: \: x = 3 \\ \because \: x \: can \: not \: be \: - ve \\ \therefore \: x = 3 \\ ED = 3x + 1 = 3 \times 3 + 1 \\ \huge \red{ \boxed{ ED= 10}}$

Answer 2

If triangle CDE ~ GDF , then length of ED is 10 unit.

Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion

In Triangles CDE and GDF,  sides CD and DG , sides ED and DF are corresponding sides.

So, these corresponding sides will be in proportion.

               $\frac{CD}{DG}=\frac{ED}{DF}\\\\\frac{15}{x+3}=\frac{3x+1}{4}\\\\(3x+1)*(x+3)=60\\\\3x^{2} +10x-57=0\\\\3x^{2} +19x-9x-57=0\\\\x(3x+19) -3(3x+19)=0\\\\(x-3)(3x+19)=0\\\\x=3,x=-19/3$

Since, side can not be negative. Therefore, x= 3

         $ED=3x+1=3(3)+1=10 unit.$

Thus,  length of ED is 10 unit.

Learn more:

https://brainly.com/question/19051610