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GalinKa [24]
4 years ago
8

A ver y Strong wind is Blowing over a big lake. A flock of geese begins their 5(10)3 meter journey home over the lake. For every

5 (10)2 meters the geese fly over the lake, they are blown back 5 (10)1 meters by the wind. It takes three minutes for the geese to fly 5 (10)2 meters forward and be blown back 5 (10)1 meters. If the geese begin flying across the lake at eleven-thirty in the morning, what time sólo they reach Home? Show all of your mathematical thinking.
Mathematics
1 answer:
maksim [4K]4 years ago
7 0
<span>The geese are required to make a 5000 meter journey. For every 500 meters they travel forward they are blown back 50 meters. Therefore rather than traveling forward 500 meters, they travel forward only 450 meters. Presumably this happens during the whole course of each section of the flight, so when they travel fractions of the distance they are blown back in the same proportion. Since each 450 meter gain requires 3 minutes, and it will take 11.11111 450 meter sections of flight, it will take 33.333333 minutes. Since they leave at 11:30, they will arrive at 12:03 and roughly twenty seconds.</span>
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3 years ago
In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some poin
Mademuasel [1]

Answer:  7\frac{5}{7} unit and 18 unit

Step-by-step explanation:

Let ABCD is a trapezoid where AB and CD are the bases. ( In which AB is greatest base which shown in below figure)

AD and BC are the legs of the trapezoid ABCD.

Now, we have ( According to the question ),

AB = 18 unit, BC = 7 unit, AD = 3 unit and DC = 11 unit.

Here the leg AD extends from point D.

Similarly leg BC extends from point C.

Let they meet at point P ( shown in below diagram)

Since In triangles PAB and PDC,

∠PDC ≅ ∠PAB ( because DC ║ AB )

And, ∠ PAB ≅ ∠ PBA

∠DPC ≅ ∠ APB ( reflexive)

Therefore, By AAA similarity postulate,

\triangle PDC \sim \triangle PAB

Thus, By the definition of similarity,

\frac{PD}{PA} = \frac{DC}{AB}

\frac{PD}{PD+3} = \frac{11}{18} ( because PA = PD+DA)

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⇒ PD = 33/7

Again by the definition of similarity,

\frac{PC}{PB} = \frac{DC}{AB}

\frac{PC}{PC+7} = \frac{11}{18} ( because PB = PC + CB)

⇒ 18 PC = 11 PD +77

⇒7PC = 77

⇒ PC = 11

Thus, PA =  PD+DA = 33/7 + 3 = 7\frac{5}{7}

And, PB = PC + CB = 11 + 7 = 18


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