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Find the area of a parallelogram given in figure. Also find the length of the altitude from vertex A on the side DC.

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$Area = 180 cm^{2}$

$Length\; of\; the\; altitude \; from\; vertex\; A\; on\; the\; side\; DC = 15\; cm$

$We\; know\; AB = CD\; and\; AD = BC \; \; \; \; \; \; [\because ABCD \ is \ a\ parallelogram]$

$Area \; of\; parallelogram = 2 \times area \; of\; \Delta DBC$

$So \; here\; \Delta DBC\; sides\ having\; DB = 25 cm, BC = 17cm\;and\;CD = 12 cm$

Using Heron’s formula

$S=\frac{25+17+12}{2}=\frac{54}{2}=27cm$

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

$=\sqrt{27\left ( 27-25 \right )\left ( 27-17 \right )\left ( 27-12 \right )}$

$= \sqrt{27\times 2\times 10\times 15}$

$= \sqrt{9\times 3\times 2\times 5\times 2\times 5\times 3}$

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

$= 2\times 3\times 3\times 5=90\; cm^{2}$

$\\ Area\; of\; parallelogram \; ABCD = 2 \times Ar(\Delta DBC) = 2 \times 90 cm^{2} = 180 cm^{2}.$

$We \; know \; that\; area\; of\; parallelogram = base \times height$

$= base\; CD \times length\; of\; the \; altitude\; from \; vertex\; A \; on\; the\; side\; DC$

$180\; cm^{2} = 12\; cm \; (length\; of\; the \; altitude\; from\; vertex\; A\; on\; the\; side\; DC)$

$\\Length\; of\; the \; altitude\; from \; vertex\; A\; on\; the\; side\; DC=180/12 = 15 cm.$

Hence, length of the altitude from vertex A on the side DC = 15 cm

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