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melomori [17]
3 years ago
11

In the rectangle below, AE = 3x + 6, CE =6x-9, and mZEDA= 55°. Find BE and m ZEAB.

Mathematics
1 answer:
tekilochka [14]3 years ago
7 0

Answer:

BE = 21

m∠EAB = 45°

Step-by-step explanation:

The given dimensions in the rectangle are;

The length of AE = 3·x + 6

The length of CE = 6·x - 9

m∠EDA = 55°

From the properties of a rectangle, we have;

The diagonals of a rectangle bisect each other,

The lengths of the diagonals are equal

Therefore;

AC = BD

Diagonal BD bisects diagonal AC, therefore, AE = CE

Plugging in the values of AC and AE, we have;

3·x + 6 = 6·x - 9

∴ 6·x - 3·x = 6 + 9 = 15

3·x = 15

∴ x = 15/3 = 5

Therefore, we have;

AE = CE = 3·x + 6 = 3×5 + 6 = 21

AC = AE + CE = 21 + 21 = 42 = BD

BD = BE + DE and for rectangle ABCD, BE = DE

∴ BD = BE + BE = 2·BE = 42

∴ BE = 42/2 = 21

<em>BE = 21</em>

m∠EDA = m∠EAD = 55° by base angle of isosceles triangle ΔAED

m∠DAB = 90° = m∠EAD + m∠EAB

∴ m∠EAB = 90° - m∠EAD = 90° - 55° = 45°

<em>m∠EAB = 45°.</em>

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