question no 22 . pls do it fast
Answers
In triangle ABD
BD = 15 cm, AD = 17 cm and ∠ABD = 90°
To find AB we can use Pythagoras theorem which is
Hypotenuse² = Base² + Perpendicular²
Here Hypotenuse = AD = 17 cm and Perpendicular BD = 15 cm, we need to find base AB
∴ AD² = BD² + AB²
17² = 15² + AB²
AB² = 289 - 225
AB² = 64
AB = 8 cm
∴ Area of ΔABD
= 1/2 (Base)(perpendicular)
= 1/2 (8 cm) (15 cm)
= 60 cm²
Also, In ΔBCD
CD= 9 cm, BC = 12 cm and BD = 15 cm
If you'll observe
BD² = BC² + CD²
Means these are sides of right angled triangle in which ∠C = 90 (opposite angle of longest side BD)
∴ Base of this triangle can be considered BC and Perpendicular can be CD
∴ Area of ΔBCD = 1/2 (12 cm) (9 cm)
= 54 cm²
Area of Quadrilateral ABCD = Area of ΔBCD + Area of ΔABD
Area of Quadrilateral ABCD = 60 cm² + 54 cm² = 114 cm²
Perimeter = AB + BC + CD + DA
= 8 cm + 12 cm + 9 cm + 17 cm
Perimeter of ABCD = 46 cm