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Observe the figure given above. It shows six identical rectangular enclosures made by using fencing wire mesh. These enclosures are used to protect baby animals in a zoo. Dimensions of each enclosure is $x$ feet $x$ $y$ feet. The total length of fencing required is $152$ feet and area of each enclosure is $80$ square feet.
Based on the above, answer the following questions :
(i) Write an expression for length of fencing required in terms of $x$ and $y$.
(ii) Write the area of each enclosure in terms of $x$.
(iii) (a) Write the above equation in quadratic equation form and thus find the dimensions of each enclosure using factorisation method.
OR
(b) Using above equation in quadratic form, solve the equation and find the dimensions of each enclosure using quadratic formula.
Based on the above, answer the following questions :
(i) Write an expression for length of fencing required in terms of $x$ and $y$.
(ii) Write the area of each enclosure in terms of $x$.
(iii) (a) Write the above equation in quadratic equation form and thus find the dimensions of each enclosure using factorisation method.
OR
(b) Using above equation in quadratic form, solve the equation and find the dimensions of each enclosure using quadratic formula.
Show SolutionHide Solution↓
Solution: (i) $8x + 9y = 152$ (1)
(ii) $x(\frac{152-8x}{9}) = 80$ (1)
(iii) (a) $x^2 - 19x + 90 = 0$ (
frac{1}{2})
$(x - 10) (x - 9) = 0$ (
frac{1}{2})
$x = 10$ or $9$ (
frac{1}{2})
$\Rightarrow y = 8$ or $\frac{80}{9}$ (
frac{1}{2})
OR
(b) $x^2 - 19x + 90 = 0$ (
frac{1}{2})
$x = \frac{19\pm\sqrt{361-360}}{2}$ (
frac{1}{2})
$x = 10$ or $9$ (
frac{1}{2})
and $y = 8$ or $\frac{80}{9}$ (
frac{1}{2})
(ii) $x(\frac{152-8x}{9}) = 80$ (1)
(iii) (a) $x^2 - 19x + 90 = 0$ (
frac{1}{2})
$(x - 10) (x - 9) = 0$ (
frac{1}{2})
$x = 10$ or $9$ (
frac{1}{2})
$\Rightarrow y = 8$ or $\frac{80}{9}$ (
frac{1}{2})
OR
(b) $x^2 - 19x + 90 = 0$ (
frac{1}{2})
$x = \frac{19\pm\sqrt{361-360}}{2}$ (
frac{1}{2})
$x = 10$ or $9$ (
frac{1}{2})
and $y = 8$ or $\frac{80}{9}$ (
frac{1}{2})