Given $\dfrac{\log a}{p}=\dfrac{\log b}{q}=\dfrac{\log c}{r}=\log x,$ all logarithms to the same base and $x\ne1.$ If $\dfrac{b^2}{ac}=x^y,$ then $y$ is:
A triangle is circumscribed about a circle of radius $r$ inches. If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is:
Let $K,$ in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then:
$K$ must be an integer
$K$ must be a rational fraction
$K$ must be an irrational number
$K$ must be an integer or a rational fraction
taken alone neither $(A)$ nor $(B)$ nor
$(C)$ nor $(D)$ is true
A triangle $ABC$ is to be constructed given side $a$ (opposite angle $A$), angle $B,$ and $h_c,$ the altitude from $C.$ If $N$ is the number of noncongruent solutions, then $N$
The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triangle, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is:
Let the product $(12)(15)(16),$ each factor written in base $b,$ equal $3146$ in base $b.$ Let $s=12+15+16,$ each term expressed in base $b.$ Then $s,$ in base $b,$ is:
The area of a rectangle remains unchanged when it is made $2\dfrac12$ inches longer and $\dfrac23$ inch narrower, or when it is made $2\dfrac12$ inches shorter and $\dfrac43$ inch wider. Its area, in square inches, is:
A circle is inscribed in a square of side $m,$ then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If $S_n$ is the sum of the areas of the first $n$ circles so inscribed, then, as $n$ grows beyond all bounds, $S_n$ approaches: