PLEASE IM IN NEED OF AN ANSWER
theres 20 points
and give proper answers please

45.1 cm^2

Step-by-step explanation:

We can break up the area of quadrilateral ABCD into the sum of the areas of right triangle ABD and triangle BCD.

Triangle ABD is a right triangle, so the legs can be used as base and heightto find the area.

area of triangle ABD = bh/2

area of triangle ABD = (AD)(AB)/2

sin <ADB = opp/hyp

sin 55 = AB/10

AB = 10 sin 55

cos <ADB = adj/hyp

cos 55 = AD/10

AD = 10 cos 55

area of triangle ABD = (10 cos 55)(10 sin 55)/2

area of triangle ABD = 23.5 cm^2

Triangle BCD is a scalene triangle. We know the measures of 2 angles and the length of the included side.

Find the measure of angle C.

m<BDC + m<CBD + m<C = 180

44 + 38 + m<C = 180

m<C = 98

Use the law of sines to BC.

BD/(sin C) = BC/(sin <BDC)

10/(sin 98) = BC/(sin 44)

BC = (10 * sin 44)/(sin 98)

BC = 7.01485

Now use BC and m<CBD to find the length of the altitude from C to side BD. Call the altitude h.

sin 38 = opp/hyp

sin 38 = h/BC

sin 38 = h/7.01485

h = 7.01485 * sin 38

h = 4.3188

Now use BD as the base and h as the height to find the area of triangle BCD.

area of triangle BCD = base * height/2

area of triangle BCD = BD * h/2

area of triangle BCD = 10 * 4.3188/2

area of triangle BCD = 21.6 cm^2

Now we add the areas of the triangles.

area of quadrilateral ABCD = 23.5 cm^2 + 21.6 cm^2

area of quadrilateral ABCD = 45.1 cm^2

solve systems of equations by graphing. a system of linear equations contains two or more equations e.g. y=0.5x+2 and y=x-2. the solution of such a system is the ordered pair that is a solution to both equations. to solve a system of linear equations graphically we graph both equations in the same coordinate system