document.write( "Question 1094389: In rectangle ABCD, diagonal BD = 26 cm and cotangent of angle ABD = 1.5. Find the area and the perimeter of the rectangle ABCD. \n" ); document.write( "
Algebra.Com's Answer #708965 by hgv (5)\"\" \"About 
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Given in the question;\r
\n" ); document.write( "\n" ); document.write( "Cotangent (Cot)ABD = 1.5
\n" ); document.write( "can be written as Cot ABD = AB/AD => 3/2\r
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\n" ); document.write( " [Cot = Perpendicular/Base]\r
\n" ); document.write( "\n" ); document.write( "=> AB = 3x, AD = 2x\r
\n" ); document.write( "\n" ); document.write( "=> By Pythagoras theorem in right angled triangle ABD\r
\n" ); document.write( "\n" ); document.write( "AB2 + AD2 = BD2
\n" ); document.write( "(3x)^2 + (2x)^2 = (26)^2
\n" ); document.write( "9x^2 + 4x^2 = 676
\n" ); document.write( "x^2 = 676/13
\n" ); document.write( "x^2 = 52
\n" ); document.write( "x = 2 sqrt13 cm
\n" ); document.write( "Thus, AB = 3x => 3 × 2 sqrt13
\n" ); document.write( "AB = 6 sqrt13
\n" ); document.write( "AD = 2x => 2 × 2 sqrt13
\n" ); document.write( "AD = 4 sqrt13\r
\n" ); document.write( "\n" ); document.write( "Now Area of rectangle ABCD = AB × AD (length × breadth)
\n" ); document.write( "i.e. 6 sqrt13 × 4 sqrt13
\n" ); document.write( "=> 24 × 13
\n" ); document.write( "=> 312 cm^2\r
\n" ); document.write( "\n" ); document.write( "And Perimeter of the rectangle = 2 (AB + AD) [2(length + breadth)]
\n" ); document.write( "=> 2 (6 sqrt13 + 4 sqrt13)
\n" ); document.write( "=> 2 × 10 sqrt13
\n" ); document.write( "=> 20 sqrt13 cm
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