A triangular advertisement board has sides 11m, 15m and 6m. If the advertisement yields an earning of $R s 1000 / m^{2}$ per month. In 3 and half years, the company will earn $A \sqrt{2}$ . Find the value of A ___.

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Solution

Using the Heron's formula, Area (A)= $\sqrt{s (s - a) (s - b) (s - c)}$ .
Semi-Perimeter (s) = $\frac{a + b + c}{2}$ $= \frac{11 + 15 + 6}{2}$
$= 16 m$
$∴ A = \sqrt{16 (16 - 11) (16 - 15) (16 - 6)}$ $= 20 \sqrt{2} m^{2}$
3 and a half years = 12 x 3 + 6 = 42 months.
$∴$ earning in 3 and a half years $= 20 \sqrt{2} \times 1000 \times 42$ $= R s . 840000 \sqrt{2}$
The value of A is 840000.

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A triangular advertisement board has sides 11m, 15m and 6m. If the advertisement yields an earning of Rs1000/m2per month. In 3 and half years, the company will earn A√2. Find the value of A ___.

Using the Heron's formula, Area (A)= √s(s−a)(s−b)(s−c). Semi-Perimeter (s) = a+b+c2 =11+15+62 =16 m ∴A=√16(16−11)(16−15)(16−6) =20√2 m2 3 and a half years = 12 x 3 + 6 = 42 months. ∴ earning in 3 and a half years =20√2×1000×42 =Rs.840000√2 The value of A is 840000.

A triangular advertisement board has sides 11m, 15m and 6m. If the advertisement yields an earning of $R s 1000 / m^{2}$ per month. In 3 and half years, the company will earn $A \sqrt{2}$ . Find the value of A ___.