The area of quadrilateral AB CD in which
AB =3cm, BC=4cm
DA =25cm and AC=5cm is
Answers
Answer:
since triangle ABC satisfies the pythagorean theorem
area of triangle ABC = 1/2× base(AB) ×height (BC)
= 1/2×3×4
= 6 cm
now draw a perpendicular from C to AD marking as P such that DP=PA=25/2 =12.5cm
therefore PC = √(AC²-PA²) By pythagorean theorem
= √(5²-12.5²)
= √(-525/4)
area of triangle PAC = 1/2× base(AP) ×height (PC)
=1/2 × 12.5 ×√(-525/4)
=25/4 ×√(-525/4)
area of triangle PDC = 1/2× base(PD) ×height (PC)
= 1/2 × 12.5 ×√(-525/4)
=25/4 ×√(-525/4)
THEREFORE
area of the quadrilateral ABCD =area of triangle ABC + area of triangle PAC + area of triangle PDC
= 6 +25/4 ×√(-525/4) +25/4 ×√(-525/4)
= 6 + 25/2×√(-525/4)
=6 + 12.5×5√21 /2 i
=6 + 6.25×5√21 icm (answer)
you may get some clue by this method.........