In isosceles trapezoid ABCD, AB=22, CD=34 and AC=10. Find BD
A)28
B)5
C)10
D)35
Based on the labeling of the sides,
I am assuming AC and BD are Diagonals.
The diagonals of an Isosceles Trapezoid
are equal. Therefore, AC = BD = 10.
ans. = C.
To find the length of BD in the isosceles trapezoid ABCD, we can use the property of the trapezoid that its bases (AB and CD) are parallel.
Since AB and CD are parallel, we can draw a line segment from A to C, which will be perpendicular to both AB and CD. Let's call the point where this perpendicular line intersects BD as E.
Now, we have two right-angled triangles: AEB and CED.
In triangle AEB, we know that AE = AC = 10 (given). And since AB = 22, we can find BE using Pythagoras' theorem:
BE^2 + AE^2 = AB^2
BE^2 + 10^2 = 22^2
BE^2 + 100 = 484
BE^2 = 384
BE ≈ √384
BE ≈ 19.6
Similarly, in triangle CED, we know that CE = AC = 10 (given). And since CD = 34, we can find DE using Pythagoras' theorem:
DE^2 + CE^2 = CD^2
DE^2 + 10^2 = 34^2
DE^2 + 100 = 1156
DE^2 = 1056
DE ≈ √1056
DE ≈ 32.5
Since BD = BE + DE, we can now find the length of BD:
BD = 19.6 + 32.5
BD ≈ 52.1
So, the correct answer is not listed among the options provided.