the lenght of diognal is 40cm and 42cm. find the perimeter of rohmbus
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Solution-
Let AC AND BD BISECT EACH OTHER AT 0
NOW,DIAGONALS OF RHOMBUS BISECT EACH OTHER AT RIGHT ANGLES.
THUS,WE HAVE
AO=1/2*40=20CM
BO=1/2*42=21CM
SINCE AOB IS A RIGHT ANGLED TRIANGLE , BY PYTHAGORAS THEOREM
AB^2=AO^2+BO^2
AB^2=400+441
AB^2=841
Square root of 841 will be the answer
Root 841=29
29 is the length of each side of the rhombus
Now perimeter will be,
P=4*29
P=116cm
Let AC AND BD BISECT EACH OTHER AT 0
NOW,DIAGONALS OF RHOMBUS BISECT EACH OTHER AT RIGHT ANGLES.
THUS,WE HAVE
AO=1/2*40=20CM
BO=1/2*42=21CM
SINCE AOB IS A RIGHT ANGLED TRIANGLE , BY PYTHAGORAS THEOREM
AB^2=AO^2+BO^2
AB^2=400+441
AB^2=841
Square root of 841 will be the answer
Root 841=29
29 is the length of each side of the rhombus
Now perimeter will be,
P=4*29
P=116cm
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