How to solve this problem and find the area
1 answer:
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Answer:
6
Step-by-step explanation:
We have to use Pythagoras' identity twice to find y
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
In right triangle MUT to find TU
TU² + 3² = 6², that is
TU² + 9 = 36 ( subtract 9 from both sides )
TU = 27 ( take the square root of both sides )
TU = =
= 3
In right triangle NUT to find y
y² = 9² + (3 )² = 81 + 27 = 108 ( take the square root of both sides )
y = =
= 6