# I have a quadrilateral ABCD. angles ABC and D are 70, 88, 98 and 104 degrees. Side AB is 200 + x, BC is 160 + x, CD is 150 + x, DA is 140 + x. I need to find x.

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### 1 Answer

You need to find what x is evaluating the area of quadrilateral such that:

`A_(ABCD) = A_(ABC) + A_(ADC)`

Notice that diagonal AC splits quadrilateral in two triangles ABC and ADC, hence you may find the area of quadrilateral adding the areas of both triangles.

Since the problem provides the lengths of sides of quadrilateral and the angles, you may find the area of triangles such that:

`A_(ABC) = (AB*BC*sin(ABC))/2=gtA_(ABC) = ((200+x)(160 + x)*sin88^o)/2`

`A_(ABC) = (32000 + 360x + x^2)*0.999/2`

`A_(ADC) = (AD*DC*sin(ADC))/2=gtA_(ADC) = ((150+x)(140+x)*sin 104^o)/2`

`A_(ADC) = ((21000 + 290x + x^2)*0.069)/2`

You need to find the area of quadrilateral adding the areas of triangles the diagonal BD forms.

`A_(ABCD) = A_(ABD) + A_(BDC) A_(ABD) = (BA*AD*sinBAD)/2=gt`

`A_(ABD) = (28000 + 340x + x^2)*0.939/2`

`A_(BDC) = (BC*CD*sinBCD)/2=gtA_(BDC) = (24000 + 310x + x^2)*0.990/2`

You need to equate `(32000 + 360x + x^2)*0.999/2 + ((21000 + 290x + x^2)*0.069)/2 = (28000 + 340x + x^2)*0.939/2 + (24000 + 310x + x^2)*0.990/2` `28800 + 324x + 0.9x^2 + 1260 + 17.4x + 0.06x^2 = 25200 + 3060x + 0.9x^2 + 21600 + 279x + 0.9x^2`

`-16740 - 2997.6 x - 0.84x^2 = 0`

`0.84x^2 + 2997.6 x + 16740 = 0`

You need to apply quadratic formula such that:

`x_(1,2) = (-2997.6 +- sqrt(8985605 - 56246))/1.68`

`x_(1,2) = (-2997.6 +-2998)/1.68 =gt x_1~~ 0.595`

**You need to keep only positive value for length of x, hence, evaluating the value of x yields `x_1 ~~ 0.595` .**