Let alpha and beta be integers satisfying 0

Question

(D) The coordinates of the points are: $P(\alpha, \beta)$,$Q(\beta, \alpha)$,$R(-\beta, \alpha)$,$S(-\beta, -\alpha)$,and $T(\beta, -\alpha)$. The pen

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) The coordinates of the points are: $P(\alpha, \beta)$,$Q(\beta, \alpha)$,$R(-\beta, \alpha)$,$S(-\beta, -\alpha)$,and $T(\beta, -\alpha)$. The pentagon $PQRST$ can be divided into a rectangle with vertices $(-\beta, \alpha), (\beta, \alpha), (\beta, -\alpha), (-\beta, -\alpha)$ and a triangle with vertices $(\beta, \alpha), (\alpha, \beta), (\beta, -\alpha)$. The area of the rectangle is $(2\beta) 	imes (2\alpha) = 4\alpha\beta$. The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2\alpha) 	imes (\alpha - \beta) = \alpha(\alpha - \beta)$. Total area $= 4\alpha\beta + \alpha^2 - \alpha\beta = \alpha^2 + 3\alpha\beta = \alpha(\alpha + 3\beta) = 187$. Since $187 = 11 	imes 17$,we have $\alpha = 11$ and $\alpha + 3\beta = 17$. Substituting $\alpha = 11$,we get $11 + 3\beta = 17$,so $3\beta = 6$,which gives $\beta = 2$. We need to find $\alpha + \beta^2 = 11 + 2^2 = 11 + 4 = 15$.

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