Answer:
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
