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The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. Find the area of the triangular field.

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$[2100\sqrt{15}\; m^{2} ]$

Given perimeter of a triangular field = 420 m and ratio of sides = 6 : 7 : 8

Let the sides of triangular field be = 6x, 7x and 8x

$\therefore Perimeter\; of\; triangular\; field = 6x + 7x +\; 8x$

$420 = 6x + 7x + 8x$

$420 = 21x$

$x=\frac{420}{21}$

$x = 20$

$Then\; sides\; are\; 6 \times 20 = 120m, 7 \times 20 = 140m\; and \; 8 \times 20 = 160m$

Using Heron’s formula for finding the area of $\Delta$ ABC

a = 120m, b = 140m, c = 160m

$S=\frac{a+b+c}{2}=\frac{120+140+160}{2}=\frac{420}{2}=210m$

$Area \; of\; \; triangle\; \Delta ABC =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{210\left ( 210-120 \right )\left ( 210-140 \right )\left ( 210-160\right )}$

$= \sqrt{210\times 90\times 70\times 50}$

$= \sqrt{7\times 30\times 3\times 30\times 7\times 10\times 5\times 10}$

$= \sqrt{7\times 7\times 10\times 10\times 30\times 30\times 3\times 5}$

= $2100\sqrt{15}\; m^{2}$

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infoexpert21

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