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Question
in parallelogram ( abcd ), diagonals ( overline{ac} ) and ( overline{bd} ) intersect at point ( e ), ( be = x^{2}-21 ), and ( de = 4x ). what is ( bd )? enter your answer in the box. ( bd=square ) units
Step1: Use the property of parallelogram diagonals
In a parallelogram, the diagonals bisect each other. So \(BE = DE\).
$$x^{2}-21 = 4x$$
Step2: Rearrange the equation to standard quadratic form
$$x^{2}-4x - 21=0$$
Step3: Factor the quadratic equation
$$(x - 7)(x+ 3)=0$$
Step4: Solve for \(x\)
Set each factor equal to zero:
- \(x - 7=0\Rightarrow x = 7\)
- \(x+3=0\Rightarrow x=-3\)
Since length cannot be negative, \(x = 7\).
Step5: Find \(BE\) and \(DE\)
- \(BE=x^{2}-21=(7)^{2}-21=49 - 21=28\)
- \(DE = 4x=4\times7 = 28\)
Step6: Calculate \(BD\)
Since \(BD=BE + DE\), \(BD=28+28\)
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