No it is not a prime number
Answer:
Hence proved △ABE∼△CBF.
Step-by-step explanation:
Given,
ABCD is a parallelogram.
BF ⊥ CD and
BE ⊥ AD
To Prove : △ABE∼△CBF
We have drawn the diagram for your reference.
Proof:
Since ABCD is a parallelogram,
So according to the property of parallelogram opposite angles are equal in measure.
⇒1
And given that BF ⊥ CD and BE ⊥ AD.
So we can say that;
⇒2
Now In △ABE and △CBF
∠A = ∠C (from 1)
∠E = ∠F (from 2)
So by A.A. similarity postulate;
△ABE∼△CBF
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Perpendicular from the center to the chord, bisects it!
DE = DF/2 = 12
To find R, use Pyth Th in triangle AOB. ( AB = 10)
10^2 + 14^2 = R^2
100+ 196 = 296 = R^2
In triangle DOE
x^2 + 12^2 = 296
x^2 = 296-144 = 152
x = √152
x = 12.3