# Diagonals of a rhombus

One of the diagonals of a rhombus is twice as the other. If the sum of the lengths of the diagonals is 24, find the area of the rhombus.

Result

S =  64

#### Solution:

$u_{1}+u_{2}=24 \ \\ u_{1}=2 \ u_{2} \ \\ 2u_{2}+u_{2}=24 \ \\ u_{2}=24/3=8 \ \\ u_{1}=2 \cdot \ u_{2}=2 \cdot \ 8=16 \ \\ S=u_{1} \cdot \ u_{2}/2=16 \cdot \ 8/2=64$

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