The square of hypotenuse of a right angles triangle is 80cm and one side is half the other. Find the length of the two sides
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Let in △ABCx△ABCx be the length of ABAB and x2x2be the length of BCBC. Applying pythagoras theorem,
(AB)2+(BC)2=(AC)2(AB)2+(BC)2=(AC)2
It is given that (AC)2=80(AC)2=80
x2+(x2)2=80x2+(x2)2=80
x2+x24=80x2+x24=80
5x24=805x24=80
Multiplying both sides by 4545,
x2=80∗45=5∗16∗45=64x2=80∗45=5∗16∗45=64
x=64−−√=8x=64=8
hope this will help u :-))
Hence AB=8cmAB=8cm and BC=82=4cm
(AB)2+(BC)2=(AC)2(AB)2+(BC)2=(AC)2
It is given that (AC)2=80(AC)2=80
x2+(x2)2=80x2+(x2)2=80
x2+x24=80x2+x24=80
5x24=805x24=80
Multiplying both sides by 4545,
x2=80∗45=5∗16∗45=64x2=80∗45=5∗16∗45=64
x=64−−√=8x=64=8
hope this will help u :-))
Hence AB=8cmAB=8cm and BC=82=4cm
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