-3/4*2/5 what is that answer to this
1 answer:
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Answer:
Hence proved that m∠AOD = m∠BCD.
Step-by-step explanation:
Given:
A Trapezoid with AC and BD as diagonals with O as intersection point to it.
Four angles formed at A and D are equal.
To Prove:
m∠AOD = m∠BCD
Proof:
Consider a Quadrilateral ABCD as trapezoid with diagonals as AC and BD
(Refer the Attachment).
And AB=CD which means the Quadrilateral ABCD isosceles in nature
i.e. angle A=angle D
Now
Consider triangle AOD and triangle BCD
Apply sum of all angles of triangle is 180.
And also
m∠2+∠AOD+m∠3=180
∠AOD=180-m∠2-m∠3..............equation(1)
And for Triangle BCD
m∠BCD=180-m∠3-m∠4,.....................Equation(2)
∠A=∠D................ as Quadrilateral is Isosceles.
i.e.∠2=∠3
That means by transitive property,
∠1=∠4=∠3=∠2
Using this values in above equation(1) we get ,
∠AOD=180-∠3-∠4.......................(3)
Now comparing equation 2 and 3 we get
RHS is equal for both equation that means LHS is also equal .
Therefore ∠AOD=∠BCD
Hence proved.
a) Parallel for both line segments have the same slope = -5/4
b) 55º
Step-by-step explanation:
a) To verify the parallelism of these line segments, It's necessary to find the slope of each one. Parallel lines share the same slope value:
b) If ∠DAB = 125°, what is the measure of ∠CBA? Justify your reasoning.
Those beams do not form a parallelogram because:
Using the formula of distance between those points
So this is a trapezoid. And two adjacent angles within a trapezoid are supplementary, so If ∠DAB = 125º then 180 - ∠CBA=55º
