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Zorluk: ZorRight Triangles and the Pythagorean Theorem

In the xyxy-plane, a circle with its center at the origin O(0,0)O(0,0) is tangent to a line at the point T(3.6,4.8)T(3.6, 4.8). The line intersects the xx-axis at point PP and the yy-axis at point QQ. What is the perimeter of triangle OPQOPQ?

  1. A
    14.4
  2. B
    24.0
  3. 30.0Cevap
  4. D
    35.0

Cevap

The perimeter of triangle OPQOPQ is 30.030.0.
The correct answer is 30.0. By finding the distance from the origin to the point of tangency T(3.6,4.8)T(3.6, 4.8), we find the radius is 66. Since the tangent line is perpendicular to this radius, we can determine its equation to be y=−0.75x+7.5y = -0.75x + 7.5. The intercepts are P(10,0)P(10,0) and Q(0,7.5)Q(0,7.5), representing the legs of right triangle OPQOPQ. Applying the Pythagorean theorem, the hypotenuse is PQ=102+7.52=12.5PQ = \sqrt{10^2 + 7.5^2} = 12.5. The sum of the sides is 10+7.5+12.5=3010 + 7.5 + 12.5 = 30.

Adım Adım Çözüm

1
Find the radius of the circle, which is the segment OTOT from the origin O(0,0)O(0,0) to the point of tangency T(3.6,4.8)T(3.6, 4.8).
The radius OT=3.62+4.82=12.96+23.04=36=6OT = \sqrt{3.6^2 + 4.8^2} = \sqrt{12.96 + 23.04} = \sqrt{36} = 6.
Since the line is tangent to the circle at TT, the radius OTOT is perpendicular to the tangent line at TT.
2
Find the equation of the tangent line. The slope of the radius OTOT is 4.83.6=43\frac{4.8}{3.6} = \frac{4}{3}.
The slope of the tangent line is the negative reciprocal, −34-\frac{3}{4}. Using the point-slope form with T(3.6,4.8)T(3.6, 4.8), the equation is y−4.8=−34(x−3.6)y - 4.8 = -\frac{3}{4}(x - 3.6), which simplifies to y=−0.75x+7.5y = -0.75x + 7.5.
The tangent line is perpendicular to the radius at the point of tangency.
3
Calculate the lengths of the legs of the right triangle OPQOPQ by finding the xx- and yy-intercepts of the tangent line.
Setting y=0y = 0 gives the xx-intercept P(10,0)P(10, 0), so OP=10OP = 10. Setting x=0x = 0 gives the yy-intercept Q(0,7.5)Q(0, 7.5), so OQ=7.5OQ = 7.5.
The vertices PP and QQ lie on the axes, forming a right angle at the origin OO, so OPOP and OQOQ are the legs of right triangle OPQOPQ.
4
Use the Pythagorean theorem to find the length of the hypotenuse PQPQ, and then calculate the perimeter.
PQ=102+7.52=100+56.25=156.25=12.5PQ = \sqrt{10^2 + 7.5^2} = \sqrt{100 + 56.25} = \sqrt{156.25} = 12.5. The perimeter of triangle OPQOPQ is OP+OQ+PQ=10+7.5+12.5=30OP + OQ + PQ = 10 + 7.5 + 12.5 = 30.
The perimeter is the sum of all three side lengths of the right triangle.

Anahtar Kavram

Right Triangles, the Pythagorean Theorem, and Tangent Lines in the Coordinate Plane
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Konular

Right Triangles and the Pythagorean Theorem

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