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ABCD is a rectangle of dimensions $80 \text{ cm x 60 cm}$. Another rectangle PQRS is drawn inside ABCD leaving space of equal width $x \text{ cm}$ along the edges of ABCD. If area PQRS is half of the area ABCD, then find the value of $x$.
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Solution:
(a) $\text{PQ = (80 - 2x) cm}$, $\text{QR = (60 - 2x) cm}$ (2 Marks)
Area $\text{PQRS} = \frac{1}{2}$ Area $\text{ABCD}$
$(80 - 2x)(60 - 2x) = \frac{1}{2} \times 80 \times 60$ (1 Mark)
$2(80 - 2x)(60 - 2x) = 80 \times 60$
$4x^2 - 280x + 2400 = 0$ or $x^2 - 70x + 600 = 0$ (1 Mark)
$(x-10)(x-60) = 0 \Rightarrow x = 10, 60$ (rejected) (1 Mark)
$\therefore x = 10 \text{ cm}$
(a) $\text{PQ = (80 - 2x) cm}$, $\text{QR = (60 - 2x) cm}$ (2 Marks)
Area $\text{PQRS} = \frac{1}{2}$ Area $\text{ABCD}$
$(80 - 2x)(60 - 2x) = \frac{1}{2} \times 80 \times 60$ (1 Mark)
$2(80 - 2x)(60 - 2x) = 80 \times 60$
$4x^2 - 280x + 2400 = 0$ or $x^2 - 70x + 600 = 0$ (1 Mark)
$(x-10)(x-60) = 0 \Rightarrow x = 10, 60$ (rejected) (1 Mark)
$\therefore x = 10 \text{ cm}$