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WARRIOR [948]
3 years ago
13

Find the area of the triangle that divides the parallelogram in half.​

Mathematics
1 answer:
nignag [31]3 years ago
4 0

equation: 1/2bh

workings...

13x9= 117

117/2=58.5

answer: 58.5cm^2

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neonofarm [45]

Answer:

Δ BEC ≅ Δ AED

Step-by-step explanation:

Consider triangles BCA and ADB. Each of them share a common side, AB. Respectively each we should be able to tell that AD is congruent to BC, and DB is congruent to CA, so by SSS the triangles should be congruent.

_________

So another possibility is triangles BEC, and AED. As you can see, by the Vertical Angles Theorem m∠BEC = m∠ADE, resulting in the congruency of an angle, rather a side. As mentioned before AD is congruent to BC, and perhaps another side is congruent to another in the same triangle. It should be then, by SSA that the triangles are congruent - but that is not an option. SSA does is one of the exceptions, a rule that is not permitted to make the triangles congruent. Therefore, it is highly unlikely that triangles BEC and AED are congruent, but that is what our solution, comparative to the rest.

Δ BEC ≅ Δ AED .... this is our solution

7 0
3 years ago
Please help
nalin [4]

Answer:

12870ways

Step-by-step explanation:

Combination has to do with selection

Total members in a tennis club = 15

number of men = 8

number of women = 7

Number of team consisting of women will be expressed as 15C7

15C7 = 15!/(15-7)!7!

15C7 = 15!/8!7!

15C7 = 15*14*13*12*11*10*9*8!/8!7!

15C7 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2

15C7  = 15*14*13*12*11/56

15C7 = 6,435 ways

Number of team consisting of men will be expressed as 15C8

15C8 = 15!/8!7!

15C8 = 15*14*13*12*11*10*9*8!/8!7!

15C8 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2

15C8 = 6,435 ways

Adding both

Total ways = 6,435 ways + 6,435 ways

Total ways = 12870ways

Hence the required number of ways is 12870ways

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2 years ago
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fijate tu modem a ver si se soluciona si no al tecnico

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