Question

HS GEOMETRY QUESTION!! please help the lesson about this made no sense and had nothing to do with this question.

Answer 1

when running a line, in a right-triangle, from the 90° angle perpendicular to its opposite side, we will end up with three similar triangles, one Small, one Medium and a containing Large one. Check the picture below.

`\cfrac{EG}{15}=\cfrac{4}{EG}\implies (EG)^2=60\implies EG=√(60)\implies EG\approx 7.75`

Answer 2

EG = 7.75

Explanation:

based on Phytagoras theorem:

• DF² = DE² + EF² ... [1]
• DE² = DG² + EG² ... [2]
• EF² = FG² + EG²... [3]

[1]

DF² = DE² + EF²

(4 + 15)² = DE² + EF²

361 = DE² + EF² ... [4]

[2]

DE² = DG² + EG²

DE² = 4² + EG²

DE² = 16 + EG² ... [5]

[3]

EF² = FG² + EG²

EF² = 15² + EG²

EF² = 225 + EG² ... [6]

[4]&[5]&[6]

361 = DE² + EF²

361 = (16 + EG²) + (225 + EG²)

361 = 16 + 225 + 2EG²

2EG² = 361 - 241

EG² = 120 ÷ 2

EG = √60

= 7.75