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find the area of a pentagon whose vertices are (4,3);(-5,6);(0,-7);(3,-6)and(-7,-2)

Asked by tanishadhankhar2 | 01 Nov, 2023, 08:09: PM

Expert Answer

Figure shows the pentagon drawn with coordinates of given vertices.

Pentagon can be divided into three triangles ΔABC , ΔACD and ΔADE as shown in figure.

For each triangle, area can be calculated from the coordinates of vertices .

If coordinates of vertices P, Q and R are ( x_1,y₁ ), (x₂, y₂ ) and (x₃, y₃ ) , then area of triangle is

Area =

begin mathsize 14px style 1 half space left square bracket space x subscript 1 left parenthesis y subscript 2 minus y subscript 3 right parenthesis space plus x subscript 2 left parenthesis y subscript 3 minus y subscript 1 right parenthesis space plus x subscript 3 left parenthesis y subscript 1 minus y subscript 2 right parenthesis space right square bracket end style

Area of ΔABC = (1/2) [ 3 ( 3+7) + 4 (-7+6) ] = 13 sq.units

Area of ΔACD = (1/2) [ 4(-2+7)-7(-7-3) ] = 45 sq.units

Area of ΔADE = (1/2) [ -7(3-6)+4(6+2)-5(-2-3) ] = 39 sq.units

Area of pentagon = ( 13+45+39) sq. units = 97 sq. units

Answered by Thiyagarajan K | 01 Nov, 2023, 10:05: PM

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