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# In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, BE=2x2−3x , and DE=x2+10 .

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In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, BE=2x2−3x , and DE=x2+10 .

What is BD ?

AngelRay  Nov 4, 2017

#2
+85859
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The diagonals of a parallelogram bisect each other....therefore...

2x^2  - 3x  = x^2  + 10    simplify as

x^2 - 3x  - 10  = 0     factor

(x - 5) ( x + 2)  = 0

Setting each factor to 0 and solving for x, we have that x = 5  or x  = -2

Therefore we have two possible values for BD

If  x  = 5  ...BD  =  2 [ (5)^2 + 10]  =  2 [ 35 ]  = 70 units

If x = - 2, .....BD  = 2 [ (-2)^2 + 10  ] =  2 [ 14 ]  =  28 units

CPhill  Nov 5, 2017
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#1
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BD = $$\sqrt{3x^{2}-3x+10}$$ . This is because, with the Pythagorean theorem, you add the 2 legs of a right triangle and take the square root to get the hypotenuse of it. This only applies if ABCD is a rhombus.

helperid1839321  Nov 4, 2017
#2
+85859
+1

The diagonals of a parallelogram bisect each other....therefore...

2x^2  - 3x  = x^2  + 10    simplify as

x^2 - 3x  - 10  = 0     factor

(x - 5) ( x + 2)  = 0

Setting each factor to 0 and solving for x, we have that x = 5  or x  = -2

Therefore we have two possible values for BD

If  x  = 5  ...BD  =  2 [ (5)^2 + 10]  =  2 [ 35 ]  = 70 units

If x = - 2, .....BD  = 2 [ (-2)^2 + 10  ] =  2 [ 14 ]  =  28 units

CPhill  Nov 5, 2017

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