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The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.

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Solution

Let the smaller side of the triangle be x cm

Let BC = x cm

According to the question,

One side of a triangle is 4 cm longer than the smaller side

Let this side be AC = x + 4

Also, the third side is 6 cm less than twice the smaller side

Let this side be AB = (2x - 6) cm

Given the perimeter of $\Delta$ ABC = 50 cm

x + x + 4 + 2x - 6 = 50

$4x - 2 = 50$

$4x = 50+2$

$4x = 52$

$x = \frac{52}{4}$

$x = 13 \; cm$

$So\; the\; side \; AC = (x + 4) = (13 + 4) = 17\; cm$

$Side\; AB = \left ( 2x-6 \right )\; cm= \left ( 26-6 \right )\; cm= 20\; cm$

Now in $\Delta$ ABC, a = 13 cm, b = 17 cm, and c = 20 cm

Using Heron’s formula

$S= \frac{a+b+c}{2}= \frac{13+17+20}{2}= \frac{50}{2}= 25\; cm$

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

$=\sqrt{25\left ( 25-13 \right )\left ( 25-17 \right )\left ( 25-20 \right )}$

$=\sqrt{25\times 12\times 8\times 5}$

$=\sqrt{5\times 5\times 2\times 2\times 3\times 2\times 2\times 2\times 5}$

$=\sqrt{5\times 5\times 5\times 2\times 2\times 2\times 2\times 2\times 3}$

$= 5\times 2\times 2\sqrt{30}$

$Area \; of \; \Delta ABC = 20\sqrt{30}\; cm^{2}$

$Hence\; the \; area\; of\; triangle\; is \; 20\sqrt{30}\; cm^{2}$

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