Question

Chocolate is in the form of a quadrilateral with sides 6cm and 10cm,5cm and 5cm(as shown in the figure)is

cut into two parts on one of its diagonal by a lady. Part-I is given to her maid and Part-II is equally divided

among a driver and Gardner.



ii. Area of ∆ABD

a)24cm2

b) 12cm2

c) 42cm2

d) 21cm2
​

Answer 1

Answer:

ans is c

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hope it will be helpful to you

Answer 2

Answer:

There is a figure that goes with this question. The figure is attached.

ΔABD is a right angled triangle

∴ AB² + BD² = AD²

or BD² = AD² - AB²

or BD² = 10² - 6²

or BD² = 100 - 36

or BD² = 64

or BD = 8

Area of ΔABD = 1/2 × 6 × 8 = 24 cm²

Area of ΔBCD can be calculated using Heron's formula

Area = \sqrt{s(s-a)(s-b)(s-c)}

s(s−a)(s−b)(s−c).

semi-parameter s = (5+5+8)/2 = 9

Area = \sqrt{9(9-8)(9-5)(9-5)}

9(9−8)(9−5)(9−5)

= \sqrt{9(4)(4)}

9(4)(4)

= 12 cm²

ΔBCD forms Part-I and ΔABD forms Part-II

Part-2 is equally divided between driver and gardener so they both get the chocolate of area 12 cm²

Thus the distribution is fair

The lady has distributed the chocolate among the three indiscriminately. This shows that she cares for all three and there is no biased.