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Yogita Ingle 7 years, 4 months ago
Here the length of the sides of the quadrilateral is given as
AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m
Diagonal AC is joined.
Now, in triangle ADC
By applying Pythagoras theorem
<nobr>AC2=AD2+CD2</nobr><amp-mathml data-formula="\[AC^{2} = AD^{2}+CD^{2}\]" inline="" layout="container"></amp-mathml>
<nobr>AC2=242+72</nobr><amp-mathml data-formula="\[AC^{2} = 14^{2}+7^{2}\]" inline="" layout="container"></amp-mathml>
AC = 25 m
Now area of triangle ABC
Perimeter = 2s = AB + BC + CA
2s = 26 m + 27 m + 25 m
s = 39 m
By using Herons Formula
The area of a triangle = √ <nobr>s×(s−a)×(s−b)×(s−c) </nobr><amp-mathml data-formula="\[\sqrt{s\times (s-a)\times(s-b) \times(s-c) }\]" inline="" layout="container"></amp-mathml>
= <nobr>√ 39×(39−26)×(39−27)×(39−25)</nobr><amp-mathml data-formula="\[\sqrt{39\times (39-26)\times (39-27) \times (39-25)}\]" inline="" layout="container"></amp-mathml>
= <nobr>291.84m2</nobr><amp-mathml data-formula="\[291.84m^{2}\]" inline="" layout="container"></amp-mathml>
Thus, the area of a triangle is <nobr>291.84m2</nobr><amp-mathml data-formula="\[291.84m^{2}\]" inline="" layout="container"></amp-mathml>
Now for area of triangle ADC
Perimeter = 2s = AD + CD + AC
= 25 m + 24 m + 7 m
s = 28 m
By using Herons Formula
The area of a triangle = <nobr>√ s×(s−a)×(s−b)×(s−c)</nobr><amp-mathml data-formula="\[\sqrt{s\times (s-a)\times(s-b) \times(s-c) }\]" inline="" layout="container"></amp-mathml>
= <nobr>√ 28×(28−24)×(28−7)×(28−25)</nobr><amp-mathml data-formula="\[\sqrt{28\times (28-24)\times (28-7) \times (28-25)}\]" inline="" layout="container"></amp-mathml>
= <nobr>84m2</nobr><amp-mathml data-formula="\[84m^{2}\]" inline="" layout="container"></amp-mathml>
Thus, the area of a triangle is <nobr>84m2</nobr><amp-mathml data-formula="\[84m^{2}\]" inline="" layout="container"></amp-mathml>
Therefore, Area of rectangular field ABCD
= Area of triangle ABC + Area of triangle ADC
= 291.84 + 84
= 375.8 m2
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