in figure DE parallel to BC …
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Ram Kushwah 5 years, 3 months ago
Given That AD=2 cm , DB=3 cm DE= 4 cm BC=x
{tex}\begin{array}{l}As\;DE\parallel\;BC\\So\\\frac{AD}{AB}=\frac{DE}{BC}\\\frac2{2+3}=\frac4x\\2x=4\times5=20\\x=10\\\end{array}{/tex}
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