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Question

The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonals is 2 m?


Solution

We have, side of a square =4 m and one diagonal of a square =2 m

Area of the rhombus= Area of the square of side 4 m
12×AC×BD=4m2
12×AC×2=16
AC=16 m

We know that the diagonals of a rhombus are perpendicular bisectors of each other.
AO=12AC
=8 m and 
BO=12BD=1 m
From right angled triangle ΔAOB
We have Pythagoras theorem, AO2+BO2=AB2
AB2=82+12
=64+1
Therefore, side of a rhombus AB=65 m
Let DX be the altitude.
 Area of the rhombus=AB×DX 
16=65×DX
Therefore, Altitude =DX=1665 m


Mathematics
RD Sharma (2014)
Standard VII

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