In the figure ( 2 ) given below, is right-angled triangle at A. AGFB is a square on the side and is a square on the hypotenuse BC. If A N \perp ED, prove that: (i) (ii) area of square ABFG = area of rectangle BENM.
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Solution
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Given: ABC is a right-angled triangle at A. Squares AGFB and BCDE are drawn on the side AB and hypotenuse BC of △ABC From
△ABE and rectangle BENM ar(ΔABE)=1/2
ar (rectangle BENM) ……(6)
ar (rectangle BENM) Hence proved.
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