For quadrilateral ABCD with AC ⊥ BD , find the area using the information in the pictures below. Derive a formula for the area of a quadrilateral with perpendicular diagonals using the diagonals only. ANSWER: The numeric value of area is __ sq units**FIRST CORRECT ANSWER GETS BRAINLIEST ANSWER*

Area of triangle ABD = [tex] \frac{b \times h}{2} [/tex]Area of triangle ABD = [tex] \frac{(10+20) \times 15}{2} [/tex]Area of triangle ABD = [tex] \frac{30 \times 15}{2} [/tex]Area of triangle ABD = (15) (15) = 225 sq. unitsArea of triangle BDC = [tex] \frac{b \times h}{2} [/tex]Area of triangle BDC = [tex] \frac{(10+20) \times 25}{2} [/tex]Area of triangle BDC = [tex] \frac{30 \times 25}{2} [/tex]Area of triangle BDC = (15) (15) = 375 sq. unitsArea of Quadrilateral ABCD = Area of triangle ABD + Area of triangle BDC Area of Quadrilateral ABCD = 225 + 375 = 600 sq. units-------------------------------------------------------------------------------------------------------------BD = d1 and AC = d2Area of triangle ABD = [tex] \frac{b \times h}{2} [/tex]Area of triangle ABD = [tex] \frac{d1 \times AO}{2} [/tex]Area of triangle BDC = [tex] \frac{b \times h}{2} [/tex]Area of triangle BDC = [tex] \frac{d1 \times OC}{2} [/tex]Area of Quadrilateral ABCD = Area of triangle ABD + Area of triangle BDC Area of Quadrilateral ABCD = [tex] \frac{d1 \times AO}{2} [/tex] + [tex] \frac{d1 \times OC}{2} [/tex]Area of Quadrilateral ABCD = [tex] \frac{1}{2}\times d1(AO + OC) [/tex]But AO + OC = AC = d2Area of Quadrilateral ABCD = [tex] \frac{1}{2}\times d1 \times d2 [/tex]