Question

find the area of the quadilateral whose sides measures 9cm,40cm,28cm and 15cm. The angle between the first two sides of the quadilateral is a right angle

Answer 1

The area of quadrilateral is equal to sum of areas of two triangles in which it's diagonals divide it.

So, area =0.5×40×9+√s(s-a)(s-b)(s-c)

a=√{40²+9²}=41 
b=28 and c=15 
And s=(41+15+28)/2

Answer 2

AnswEr:

Clearly, ∆DAB is a right - angles triangle. Therefore,

$\tt \: {db}^{2} = {da}^{2} + {ab}^{2} \\ \\ \rightarrow \tt {db}^{2} = {9}^{2} + {40}^{2} \\ \\ \rightarrow \tt \: db = \sqrt{81 + 1600} \: m = 41 \: m$

$\rule{170}3$

In ∆DAB, we have

$\tt 2s = da + ab + bd = (9 + 40 + 41) = 90 \: m \\ \\ \tt \rightarrow \: s = 45m \\ \\ \tt \rightarrow \: a_1 = area \: of \triangle dab \\ = \tt \sqrt{45(45 - 9)(45 - 40)(45 - 41)} \: {m}^{2} \\ \\ = \tt \sqrt{45 \times 36 \times 5 \times 4} \: {m}^{2} \\ \\ = \tt \sqrt{5 \times 9 \times 36 \times 5 \times 4} \: {m}^{2} \\ \\ \tt = \sqrt{ {5}^{2} \times {3}^{2} \times {6}^{2} \times {2}^{2} } \: {m}^{2} \\ \\ \tt = (5 \times 3 \times 6 \times 2) \: {m}^{2} = 180 \: {m}^{2}$

$\rule{170}3$

In ∆DCB, we have

$\tt = 2s = DC + CB + BD \\ = \tt \: 20s = 28 + 15 + 41 = 84 \\ \tt \: s = 42$

$\tt area\: of \: \triangle \: DCB \\ \\ \tt \: = \sqrt{42(42 - 28)(42 - 15)(42 - 41)} \: {m}^{2} \\ \\ \tt \: = \sqrt{42 \times 14 \times 27 \times 1} \: {m}^{2} \\ \\ \tt = \sqrt{7 \times 2 \times 3 \times 7 \times 2 \times 3 \times 3 \times 3} \: {m}^{2} \\ \\ \tt = \sqrt{ {7}^{2} \tines {2}^{2} \times {3}^{4} } \: {m}^{2} \\ \\ \tt = (7 \times 2 \times {3}^{2} ) = 126 \: {m}^{2}$

$\rule{170}3$

Hence, Area of the field

= (180 + 126) m² = 306 m².

#BAL

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