Let O be the point of intersection of diagonals.
Consider triangles FOE and EOD:
FO = OD, ∠FOE = ∠EOD and OE is a common side ⇒
triangles FOE and EOD are congruent.
In congruent triangles all corresponding sides and are congruent ⇒
EF = DE = 6
Answer:
x = 7
y = 0
Step-by-step explanation:
Close, but not quite.
Answer:
√2=1.414
then :√8 +2√32 +3√128+4√50
√8=√2³ =2√2
2√32=√2^5 = 4*2√2 = 8√2
3√128 = 3√2^6*2=8*3√2 =24√2
4√50 =4√5²*2= 20√2
add results : 2√2+8√2 +24√2+20√2=54√2
<h2>
54√2=54×1.414=76.356 ( it is not in the options)</h2>
x=7-4√3
√x+ 1/√x
√(7-4√3) +1/√(7-4√3) =
(8-4√3)/√(7-4√3)
(8-6.93)/√(7-6.93) = 4 ( after rounded to the nearest whole number)
4 is your answer
Answer:
Step-by-step explanation:
The length of AB = 1 unit.
area of square = 1² unit² = 1 square unit.
