the diagnals of a rhomus measure 14cm and 48cm.find its perimeter
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Answered by
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a² =[(d1)/2]² +[(d2/2)/2]²,
a²=(14/2)² + (48/2)2,
a²=7²+24²,
a²=49+576,
a²=625,
then
a=25 cm,
therefore
perimeter of rhombus=4×a,
=4×25=100cm
a²=(14/2)² + (48/2)2,
a²=7²+24²,
a²=49+576,
a²=625,
then
a=25 cm,
therefore
perimeter of rhombus=4×a,
=4×25=100cm
Answered by
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d1=14 cm
d2=48 cm
from the figure
∆AOB is a right triangle
according to Pythagoras theorem
AB^2=AO^2+OB^2
AB=side of rhombus
AO=48/2=24cm
OB=14/2=7cm
AB^2=24^2+7^2
AB^2=576+49=625
AB=√625=25cm
perimeter of rhombus=4 × side
=4×25=100cm
d2=48 cm
from the figure
∆AOB is a right triangle
according to Pythagoras theorem
AB^2=AO^2+OB^2
AB=side of rhombus
AO=48/2=24cm
OB=14/2=7cm
AB^2=24^2+7^2
AB^2=576+49=625
AB=√625=25cm
perimeter of rhombus=4 × side
=4×25=100cm
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