Match the two columns for the figure shown below.
Column - IColumn - II(A) Area of □DABC(p) 41(B) Area of ΔADC(q) 126(C) Area ofΔABC(r) 306(D) AC(s) 180
Using Pythagorus theorem in ΔABC
AC=√AB2+BC2
=√92+402
=√81+1600=41 m
(A) Area of \(\square DABC = area~(\Delta ABC) + area~(\Delta ADC)\
area (ΔABC)=12(a)(40)=180 m2forΔADC
S=28+15+412=42 m
area (ΔADC)=√42(42−28)(42−15)(42−41)
=√42(14)(27)
=√2 × 3 × 7 × 2 × 7 × 3 × 3 ×3
= 2 x 3 x 3 x 7 = 126 m2
∴area□DABC=180+126=306 m2
(B) Area of ΔABC=126 m2
(C) Area of ΔABC=180 m2
(D) AC=41m
(A)→r;(B)→q;(C)→s;(D)→p