Question

ABCD is a field in the form if a quadilateral whose sides are DAis 9mDC is 28.cm AB isv40 cm,CB is 15cm . angle DAB is 90 degree. Find thevarea of the field

Answer 1

Hey there!!!

Here is the answer to your question...


We are given that,

In quadrilateral ABCD, AB = 40cm, BC = 15cm, CD = 28cm, DA = 9cm, and and angle DAB = 90°
Let us join DB
Then, in triangle DAB,
DB^2 = DA^2 + AB^2

(by Pythagoras theorem)

DB^2 = (9)^2 + (40)^2
DB^2 = 81 + 1600
DB^2 = 1681
DB = _/1681
DB = 41cm

Now,
area of triangle DAB = 1/2×b×h
= 1/2×AB×DA
= 1/2×40×9
= 180cm^2
We can find the area of triangle BCD by the herons formula...
s = A + B + C/2
= 15+28+41/2
= 84/2
= 42 cm
The area of the triangle BCD is given by
Area = \sqrt{s(s-a)(s-b)(s-c)}
= \sqrt{42(42-15)(42-28)(42-41)}
= \sqrt{42(27)(14)(1)}
= \sqrt{15876}
= 126 cm^2

Now, area of quadrilateral ABCD
= area of triangle DAB + area of triangle BCD
= 180 + 126 cm^2
= 306cm^2


Hope it helps!!!
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