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svp [43]
3 years ago
11

ABCD is a parallelogram with diagonals AC and BC intersecting O Prove that triangle AOB is congruent to triangle DOC

Mathematics
1 answer:
OleMash [197]3 years ago
7 0

Given: ABCD is a parallelogram with diagonals AC and BD intersecting at O.

To prove: ∆AOB ≅ ∆DOC

Proof: Since ABCD is a parallelogram.

Therefore, AB || CD and AB = CD, AD || BC and AD = BC

Now, AB || CD and transversal AC intersects them a A and C respectively.

Therefore, Angle BAC = Angle DCA [Since alternate interior angles are equal]

or, Angle BAO = Angle DCO

Now, in ∆AOB and ∆DOC,

Angle BAO = Angle DCO [proved above]

AB = CD [opposite sides of a parallelogram are equal]

Angle AOB = Angle COD [vertically opposite angles are equal]

So, by ASA congruency,

∆AOB ≅ ∆DOC.

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Please help me to solve this . Thank you so much .
worty [1.4K]

Answer:

a) N

b) L

c) area I

d) area II

e) area VI

Step-by-step explanation:

a) the points that are 2cm from R are Q, N, M, S. Then, points that are 4cm from P are K, N, R. So, the only one point that works for both is N.

b) the points that are >2cm from R are P, K, L, T. We do not count those are exactly 2cm from R. Then, points that are 4cm from T are R, M, L. Ans is L.

c) <4cm from P, are area I and II. Then area that are >2cm from R are I, VI, and V. So, the only area that works for both is I.

d) <2cm from R, are areas II, III, and IV. Then, <4cm from P, are areas I and II. So, the only one works for both is area II.

e) >4cm from T, are areas I, II, III, VI. Then, >4cm from P, are III, IV, V, VI. Finally, >2cm from R, are areas I, VI, V. The only one that works for all three conditions is area VI.

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4 years ago
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The circle below is centered at the point (1, 2) and has a radius of length 3.
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4 years ago
A math teacher instructed students to graph the following equations on a coordinate plane. y=2.5x+2 and y=2x+4. If graphed corre
algol [13]

Both the graphical lines will intersect at (4,12)

Step-by-step explanation:

Step 1; We must find the point in the graph where the lines of y=2.5x+2 and y=2x+4 intersect. To find out the intersecting point we must substitute values of x in both equations and see in which co-ordinate they are the same.

Step 2; We plot various values of x on the line governed by the equation y=2.5x+2

For a value of x=1 , y=2.5(1) + 2 = 4.5

For a value of x=2 , y=2.5(2) + 2 = 7

For a value of x= 3, y=2.5(3) + 2 = 9.5

For a value of x= 4, y=2.5(4) + 2 = 12

For a value of x= 5, y=2.5(5) + 2 = 14.5

Step 3; Now we also plot the values of x on the other line governed by y=2x+4

For a value of x= 1, y=2(1) + 4 = 6

For a value of x= 2, y=2(2) + 4 = 8

For a value of x= 3, y=2(3) + 4 = 10

For a value of x= 4, y=2(4) + 4 = 12

For a value of x= 5, y=2(5) + 4 = 14

Step 4; For both lines we must see if any of the values repeat at a particular point. In this case at the point (4,12) repeats for both so that becomes the point of intersection.

8 0
3 years ago
(PLEASE HELP ME IM STRUGGLING) (picture is shown) Here is an illustration of a four way stop. those roads are two lines and they
Montano1993 [528]

Answer:

they intersect at a right angle

Step-by-step explanation:

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