Pole trapezu o podstawach a i b ( a > 0, b > 0), jest równe 2ab + 2b². Wysokość tego trapezu wynosi ? MA WYJŚĆ 4B
P=1/2(a+b)h
2ab+2b²=1/2(a+b)h
4ab+4b²=(a+b)h
4b(a+b)=(a+b)h
h=4b
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P=1/2(a+b)h
2ab+2b²=1/2(a+b)h
4ab+4b²=(a+b)h
4b(a+b)=(a+b)h
h=4b