Question

ABCD is a kite with AB = AD and CB = CD. The diagonals AC and BD meet at O. If BD = 12 cm, AO = 8 cm and OC = 4.5 cm, find the area of the kite

Answer 1

Answer:

75 cm^2

Step-by-step explanation:

Area of a kite =( Length of diagonal AC×Length of diagonal BD) ÷ 2

= (8+4.5×12)÷2

= 150÷2

= 75

Answer 2

The area of the kite=$75cm^2$

Step-by-step explanation:

ABCD is a kite

AB=AD

CB=CD

OA=8 cm

OC=4.5 cm

BD=12 cm

Area of triangle=$\frac{1}{2}\times base\times height$

Using the formula

Area of triangle ABD=$\frac{1}{2}\times BD\times OA=\frac{1}{2}\times 8\times 12=48cm^2$

Area of triangle BCD=$\frac{1}{2}\times BD\times OC=\frac{1}{2}\times 12\times 4.5=27cm^2$

Area of the kite=Area of triangle ABD+area of triangle BCD=48+27=75 square cm

Hence, the area of the kite=$75cm^2$

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