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A 10-foot ladder is leaning against a tree. The bottom of the ladder is 4 feet away from the bottom of the tree. Approximately how high up the tree does the top of the ladder reach? A. 2.4 feet B. 6 feet C. 9.2 feet D. 10.8 feet

Sagot :

Trimidn
a^2 + b^2 = c^2

4^2 + b^2 = 10^2

16 + b^2 = 100

b^2 = 100 - 16

b^2 = 84

b = SqrRt 84 = 9.2

A 10-foot ladder is leaning against a tree. The bottom of the ladder is 4 feet away from the bottom of the tree. the top of The ladder reachupto 9.2 feet high approximately

Given: Ladder=10foot

          Bottom of ladder = 4 feet away from the bottom of tree.          

Now, we will assume length of the sides of ladder as a,b,c

According to the property of Right angled triangle ,

[tex]\rm a^2 + b^2 = c^2[/tex]

On putting the values of a,b,c in the formula we get,

[tex]\rm 4^2 + b^2 = 10^2\\16 + b^2 = 100\\b^2 = 100 - 16\\b^2 = 84b = \sqrt{84} = 9.2[/tex]

Therefore, the top of the ladder reach upto 9.2 feet high.

Learn more about Properties of Right angle triangle here: https://brainly.com/question/2510444

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