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Since AB = CD the trapezoid is isosceles, which means that ∡A = ∡D
Therefore also ∡2 = ∡3 (they are half of the congruent angles)
For the properties of parallel lines (BD and AD) crossed by a transversal (BD) we have ∡3 = ∡CBD.
Now consider triangles AOD and BCD:
∡OAD (2) = ∡ADO (3) = ∡CBD (3) = ∡CDB (4)
T<span>he sum of the angles of a triangle must be 180°, t</span>herefore:
∡AOD = 180 - ∡2 - ∡3
∡BCD = 180 - ∡3 - ∡4
∡AOD = ∡BCD because their measure is the difference of congruent angles.
V1 + V2
V1 = 1*2*1 = 2
V2 = 5*2*2 = 20
V1 + V2
2+20
22
For rhombus A:
base = 7 in
area = 35 in²
Area of rhombus = b * h
⇒ 35 = 7 * h
⇒ h = 35/7
⇒ h = 5 in
Rhombus B:
height = 3*5 = 15 in
base = 7*3 = 21 in
Area = 15 * 21 = 315 in²
Area
of rhombus B is 9 times area of rhombus A. When the dimensions are
increased to 3 times the initial dimension, area became 9 times.
