Question

please solve que 1 and 2. plzzz it's urgent

Answer 1

1. Given,
∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.
In diagram,
BD is joined.
Now,
In ΔBCD,
By applying Pythagoras theorem,
BD^2 = BC^ 2 + CD^2
⇒ BD^2 = 12^2 + 5^2
⇒ BD^2 = 169
⇒ BD = 13 m
Area of ΔBCD = 1/2 × 12 × 5 = 30 m2
Now,
Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m
Using heron's formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
= √15(15 - 13) ( 15 - 9) (15 - 8) m 2
= √15 × 2 × 6 × 7 m2
= 6√35 m 2 = 35.5 m2 (approx)
Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m 2 + 35.5m 2 = 65.5m 2

2. Given,
AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm
In ΔABC,
By applying Pythagoras theorem,
AC 2 = AB2 + BC 2
⇒ 5 2 = 3 2 + 4 2
⇒ 25 = 25
Thus, ΔABC is a right angled at B.
Area of ΔBCD = 1/2 × 3 × 4 = 6 cm 2
Now,
Semi perimeter of ΔACD(s) = (5 + 5 + 4)/2 cm = 14/2 cm = 7 m
Using heron's formula,
Area of ΔABD = √s (s-a) (s-b) (s-c)
= √7(7 - 5) (7 - 5 ) (7 - 4 ) cm 2
= √7 × 2 × 2 × 3 cm 2
= 2√21 cm 2 = 9.17 cm2 (approx)
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔABD = 6 cm 2 + 9.17 cm 2 = 15.17 cm 2