5) The hypotenuse of the "12 by 16" triangle, using pythagoras' theorem, is √122+162=√400=20
Then, the hypotenuse of the "12 by x" triangle, using the big triangle, is √(x+16)2−202
But we also know that that side is √122+x2
So we can make an equation since the two are equivalent:
√(x+16)2−202=√122+x2
(x+16)2−202=122+x2
(x+16)(x+16)−x2=144+400
x2−x2+32x+256=544
32x=288
x=9
6) This question is similar to the previous one so I won't explain in depth:
√602−362=48
√(x−36)2+482=√x2−602
(x−36)(x−36)−x2=−602−482
x2−x2−72x+1296=−5904
−72x=−4608
x=64
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