Super-Hard SAT Geometry Question 2 | Free SAT Prep

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Question 2·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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A circle has center $C$ and radius $25$ centimeters. Chord $AB$ has length $40$ centimeters. The tangent lines to the circle at $A$ and $B$ intersect at point $T$ . What is the perimeter, in centimeters, of quadrilateral $CATB$ ?

Your answer

When a problem includes a chord and two tangents, draw or identify the midpoint of the chord. The center and the intersection of the tangents both lie on the chord's perpendicular bisector, often creating two similar right triangles. Use that similarity to find a missing tangent length, then add only the outer sides named in the requested perimeter.

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Locate a useful midpoint

Let $M$ be the midpoint of chord $AB$ . The segment from the center to a chord's midpoint is perpendicular to the chord.

Connect the center to the tangent intersection

Use the facts that $CA=CB$ and $TA=TB$ to determine why $C$ , $M$ , and $T$ lie on the same line.

Look for similar triangles

Compare right triangles $CMA$ and $CAT$ . They share the angle at $C$ , so write a proportion involving $CM$ , $CA$ , and $CT$ .

Desmos Guide

Use geometry first

The similarity approach is faster by hand. For a coordinate-based verification, place $C$ at $(0,0)$ and use $CM=15$ and $AM=20$ to place $A$ at $(15,20)$ and $B$ at $(15,-20)$ .

Graph the circle and tangent lines

Enter $x^2+y^2=625$ . Then enter the tangent lines $y-20=-\frac{3}{4}(x-15)$ and $y+20=\frac{3}{4}(x-15)$ . Their slopes are perpendicular to the slopes of radii $CA$ and $CB$ , respectively.

Find the tangent length and perimeter

Click the intersection of the tangent lines to obtain point $T$ . Enter $d=\sqrt{(x_T-15)^2+(y_T-20)^2}$ , using the coordinates displayed for $T$ , to verify $AT$ . Then evaluate $2(25+d)$ for the perimeter.

Step-by-step Explanation

Find the distance from the center to the chord

Let $M$ be the midpoint of $AB$ . Since $CA=CB$ and tangent segments from $T$ are equal, both $C$ and $T$ are equidistant from $A$ and $B$ . Therefore, $CT$ is the perpendicular bisector of $AB$ , so $CM\perp AB$ and $AM=20$ .

In right triangle $CMA$ ,

CM=\sqrt{25^2-20^2}=15.

Use similar right triangles

Triangles $CMA$ and $CAT$ are similar: each is a right triangle, and they share the angle at $C$ because $M$ lies on $CT$ . Therefore,

\frac{CM}{CA}=\frac{CA}{CT}.

Substitute $CM=15$ and $CA=25$ :

\frac{15}{25}=\frac{25}{CT},

so $CT=\frac{125}{3}$ .

Find a tangent segment

A radius is perpendicular to a tangent at the point of tangency, so triangle $CAT$ is right at $A$ . Thus,

AT=\sqrt{\left(\frac{125}{3}\right)^2-25^2}=\frac{100}{3}.

Also, $AT=BT$ because tangent segments from the same external point are equal.

Calculate the perimeter

The sides of quadrilateral $CATB$ are $CA$ , $AT$ , $TB$ , and $BC$ . Therefore,

25+\frac{100}{3}+\frac{100}{3}+25=\frac{350}{3}.

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Step-by-step Explanation

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