Serphidn
Answered

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What is the area of a rectanglewhose sides are twice as long as the sides of a smaller rectangle whose area us 17 cm²

Sagot :

Chenie8idn
A = length x width

let x be the length and
    y be the width

Area of the smaller rectangle = 17 cm²

so, A = l x w
     17 = xy
      [tex] y = \frac{17}{x} [/tex]

Area of the bigger rectangle = 2x(2y)   --> for it is twice the sides of the smaller rectangle

A = 2x(2y)
   = 2x ([tex] \frac{2(17)}{x} [/tex])
   = 2x ([tex] \frac{34}{x} [/tex])
   = [tex] \frac{68x}{x} [/tex]
   = 68 cm²
Dominiidn
[tex]Formula; \\ A=lw \\ \\ Let\ l\ be\ the\ length \\ Let\ w\ be\ the\ width \\ \\ Area\ of\ Small\ Rectangle=17cm^{2} \\ \\ Area=lw \\ 17=lw \\ \boxed{w= \frac{17}{l}} (to\ give\ the\ value\ of\ w) \\ \\ \\ Area\ of\ Big\ Rectangle=2l(2w) \\ \\ Area=2l(2w) \\ Area=2l( \frac{2\times17}{l}) \\ Area=2l( \frac{34}{l}) \\ Area=\frac{65l}{l} \\ \boxed{\boxed{Area=68\ cm^{2}}} \\ \\ Hope\ it\ Helps:) \\ Domini [/tex]