Question

In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, AE = 3x - 5, and CE = x + 11.

Answer 1

Answer:

38

Step-by-step explanation:

For the case of a parallelogram, when the diagonals ( in our case AC & BD) are intersected, they are basically bisected ( divided into two equal halves ). Therefore when the diagonals intersect at point E, we can say that the diagonal AC is divided in two equal halves which in our case are AE and CE. since AE and CE are equal , we can say that,

$AE= CE \\or\\3x-5 = x+11 ---- (1)\\now solving (1) for x, \\3x-x=11+5\\2x=16\\x=8-----> (2)\\\\as \\ AC = AE+CE\\AC = 3x-5+x+11\\AC = 4x+6\\\\\\from (2)\\\\\\AC= 4(8)+6\\AC = 32+6\\AC = 38\\\\since AC = BD ( Diagonals are of same Length )\\therefore \\BD =38$