Obseruation Take a = 5cm, b 3cm. AB=atle- units sq units sq units Area of ABCD = (a+b) Area of each rectangle = ab- 4 units Area of square Pars co-b)²- = (2 - Area ABCD - 4Carea of rectangulay piłce.) + Area of - UC + c > 4C 1.e. (a+b)2 > Yab Square PARS . ) 041 (af b) 2 sab . AM GM.
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Given: a = sum, b = 3cm
To find: The value of given equations
Explanation:
AB = a + b = 8 cm units
Area of ABCD,
= (a + b)² = 64cm sq units.
Area of each rectangle,
= ab = 15 cm sq units
Area of Square PARS,
= (a - b) ² = 4 cm sq units.
Area ABCD,
= 4 (area of rectangular piece) + area of square PQRS
= 64 = 4(15) + 4
=64 > 4 (15)
i.e (64) > 4 (15)
or,
[tex]=(\frac{a+b}{2} )^{2} >ab\\ = (\frac{8}{2})^{2}>15\\ = 16>15[/tex]
or,
[tex]= \frac{a+b}{2} >\sqrt{ab} \\ = \frac{8}{2} >\sqrt{15} \\ = 4 >\sqrt{15} [/tex]
Hence AM> GM
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