If ABCD is a rhombus AD=4x+2 DC=7x-13 and BD=34 find AE (E is in the middle)

Line AE equals 14 units (approximately)

Step-by-step explanation: If you have a rhombus given, it is important to begin by highlighting the properties of a rhombus. A rhombus first of all is four-sided shape with all sides congruent, then the two opposite sides are parallel and congruent. Also, the diagonals bisect each other into two equal halves each.

Therefore, we shall examine a rhombus labelled ABCD, with side AD measuring 4x + 2, and side DC measuring 7x - 13. Then BD is a diagonal from point B to point D, and it measures 34 units. Also a diagonal is drawn halfway from point A and stops at point E (E is in the middle), where it bisects diagonal BD into two halves. Remembering that all sides of a rhombus are equal, then note that AD equals DC, that is;

4x + 2 = 7x - 13

2 + 13 = 7x - 4x

15 = 3x

Divide both sides of the equation by 3

5 = x

If AD = 4x + 2, then when x  = 5

AD = 4(5) + 2

AD = 20 + 2

AD = 22

Therefore triangle ADE can now be isolated with side AD measuring 22 units and side ED measuring 17 units (midpoint between points B and D). Line AE can be calculated using the Pythagoras theorem which states as follows;

AC² = AB² + BC²

Where AC is the hypotenuse, and AB and BC are the two other sides

Substituting for values as given in the question, we now have;

AD² = ED² + AE²

22² = 17² + AE²

484 = 289 + AE²

484 - 289 = AE²

195 = AE²

Add the square root sign to both sides of the equation

√195 = √AE²

13.96 = AE

Therefore line AE ≈ 14 units (to the nearest whole number)

can you rephrase the question ? it is unclear : -)

step-by-step explanation:

hi! i think the answer is 1/2 : )