First, let's make sure you know what Pythagoras said.
It's very important, mainly because it's so useful:
In any RIGHT triangle, the square of the longest side
is the sum of the squares of the other two sides.
Now let's look at #30:
This is two right triangles back-to-back.
'x' is the longest side of the one on the left, but we can't
solve for 'x' yet because we only know one of the other sides.
But look at the right triangle on the right side. We know the
longest side, and the vertical one, so we can find the base of it.
(longest side)² = (one short side)² + (the other short side)²
(15)² = (9)² + (the base)²
Subtract (9)² from each side:
(The base)² = (15)² - (9)²
(The base)² = 225 - 81 = 144
The base = √144 = 12 .
Do you see where we are now ?
The base of the right half is 12,
so the base of the left half is (25 - 12) = 13 .
Pythagoras again, on the left half:
x² = (13)² + (9)²
x² = 169 + 81 = 250
x = √250 = (√25)·(√10) = 5 √10 .
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#31:
Again, two right triangles back to back.
You know two sides in each one.
So you can find the base of each one, then add the bases to get 'x'.
Left half: (Base)² = (10)² - (6)²
(Base)² = 100 - 36 = 64
Base = √64 <== you know what that is
Right half: (Base)² = (7)² - (6)²
(Base)² = 49 - 36 = 13
Base = √13
x = sum of the two bases = √64 + √13 .
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#32:
Two right triangles again.
Not back to back.
Right half: (Missing leg)² = (7)² - (2)²
(Missing leg)² = 49 - 4 = 45
Missing leg = √45 = (√9)·(√5) = 3 √5 .
Left half: The longest side is 3√5 .
x² + (5)² = (3√5)²
I'm sure you got the idea by now.
