Question

Allie and Ezra are sitting on flagpoles, throwing beanbags at each other. Allie's flagpole is $35$ feet tall. Ezra's flagpole is $27$ feet tall. The flagpoles stand $15$ feet apart. How far, in feet, does Ezra have to throw a beanbag to reach Allie

Answer 1

Ezra has to throw a beanbag 17 feet far to reach Allie

According to the question,

The distance between Allie and Ezra's flagpoles is 15 feet

length of Allie's flagpole = 35 feet

length of Ezra's flagpole = 27 feet

Difference between Allie and Ezra's flagpole = (35 - 27) feet

= 8 feet

Now, there forms a right angled triangle, where hypotenuse is the distance that Ezra's bean bad has to cover.

Using Pythagoras theorem,

$H^{2} = B^{2} + A^{2}$

where H = hypotenuse,

B= Base of the triangle

A = Altitude of the triangle

we are given,

B = 15 feet,

A = 8 feet

On substituting the values,

$H^{2} = 15^{2} + 8^{2}$

$H^{2} = 225 + 64\\H^{2} = 289\\H = \sqrt{289} \\H = 17$

So hypotenuse = 17 feet

Ezra has to throw the beanbag 17 feet far in order to reach it to Allie.

To learn more about Pythagoras theorm from the given link

brainly.com/question/343682

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