ACT · Math · Preparing for Higher Math
Triángulos, círculos, geometría analítica, trigonometría y volúmenes.
Triángulos, círculos, coordenadas, trigonometría y volúmenes.
$x2+y2-6x+8y=0$ → $(x-3)2+(y+4)2=25$ → $r=5$.
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
What is the area, in square centimeters, of a circle with a diameter of 10 centimeters?
Respuesta correcta: B. $25\pi$
The radius is half the diameter, so $r=5$ cm. The area is $\pi r2=\pi\cdot52=25\pi$ square centimeters.
$10\pi$ is the circumference ($\pi d$). $50\pi$ uses $2\pi r2$ or doubles the area. $100\pi$ uses the diameter in place of the radius $(\pi\cdot102)$.
In the figure above, line $\ell$ is parallel to line $m$. What is the value of $x$ ?
Respuesta correcta: A. $60$
Alternate interior angles: $\angle PQR=50\circ$. The $110\circ$ angle at $R$ is supplementary to $\angle PRQ$, so $\angle PRQ=70\circ$. In triangle $PQR$: $x=180-50-70=60$.
$70$ is $\angle PRQ$, not $\angle QPR$. $130$ is the supplement of the $50\circ$ angle, the angle between $\overline{PQ}$ and line $\ell$ on the other side. $110$ is the exterior angle at $R$, which equals $50+x$ (exterior angle theorem), not $x$.
Points $A$ and $B$ lie on a circle with center $O$ and radius 9. The length of chord $\overline{AB}$ is $9\sqrt{3}$. What is the length of the minor arc $AB$ ?
Respuesta correcta: C. $6\pi$
Drop the perpendicular from $O$ to the chord: it bisects $\overline{AB}$, giving half-chord $\frac{9\sqrt{3}}{2}$ and hypotenuse 9, so $\sin(\tfrac12\angle AOB)=\frac{\sqrt3}{2}$ and $\tfrac12\angle AOB=60\circ$. The central angle is $120\circ$, so the arc is $\frac{120}{360}\cdot2\pi(9)=6\pi$.
$3\pi$ uses $60\circ$ as the central angle (stops at the half-angle). $9\sqrt{3}$ is the chord, not the arc. $12\pi$ is the major arc ($240\circ$).