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Sample Question 16:
Step 1: Draw a right triangle with leg lengths \(AB=20\) and \(BC=21\). The hypotenuse \(AC\) has length \(29\). Draw a line from \(B\) perpendicular to \(AC\). Let the point of intersection be \(P\).
Step 2: Calculate the length of the altitude \(BP\) from \(B\) to \(AC\). This is done by writing the area of the triangle in two ways: \(\frac{1}{2}\times AB\times BC\) and \(\frac{1}{2} \times AC\times BP\). Equating both expressions and rearranging, we find \(BP=\frac{20\times 21}{29}\), which lies between \(14\) and \(15\).
Step 3: Consider a line segment \(BX\) from \(B\) to a point \(X\) on \(AC\). As \(X\) is moved along the hypotenuse from \(A\) to \(P\), the length of \(BX\) strictly decreases, hitting all the integer values from \(20, 19, ..., 15\).
Step 4: Similarly, if we move \(X\) along the hypotenuse from \(P\) to \(C\), the length of \(BX\) hits all integer values from \(15, 16, ..., 21\).
Step 5: Count all the distinct integer line segments obtained in steps 3 and 4. From \(20\) to \(15\), we get \(6\) distinct lengths, and from \(15\) to \(21\) we get \(7\) distinct lengths. Adding these up, we get a total of \(13\). So, the answer is \(13\) distinct line segments with integer length can be drawn from \(B\) to a point on hypotenuse \(AC\). The answer is thus \(\textbf{(D)} 13\).
Right triangle \(ABC\) has leg lengths \(AB=20\) and \(BC=21\). Including \(\overline{AB}\) and \(\overline{BC}\), how many line segments with integer length can be drawn from vertex \(B\) to a point on hypotenuse \(\overline{AC}\)?
\(\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }15 \qquad\)
Answer Keys
Question 16: DSolutions
Question 16Step 1: Draw a right triangle with leg lengths \(AB=20\) and \(BC=21\). The hypotenuse \(AC\) has length \(29\). Draw a line from \(B\) perpendicular to \(AC\). Let the point of intersection be \(P\).
Step 2: Calculate the length of the altitude \(BP\) from \(B\) to \(AC\). This is done by writing the area of the triangle in two ways: \(\frac{1}{2}\times AB\times BC\) and \(\frac{1}{2} \times AC\times BP\). Equating both expressions and rearranging, we find \(BP=\frac{20\times 21}{29}\), which lies between \(14\) and \(15\).
Step 3: Consider a line segment \(BX\) from \(B\) to a point \(X\) on \(AC\). As \(X\) is moved along the hypotenuse from \(A\) to \(P\), the length of \(BX\) strictly decreases, hitting all the integer values from \(20, 19, ..., 15\).
Step 4: Similarly, if we move \(X\) along the hypotenuse from \(P\) to \(C\), the length of \(BX\) hits all integer values from \(15, 16, ..., 21\).
Step 5: Count all the distinct integer line segments obtained in steps 3 and 4. From \(20\) to \(15\), we get \(6\) distinct lengths, and from \(15\) to \(21\) we get \(7\) distinct lengths. Adding these up, we get a total of \(13\). So, the answer is \(13\) distinct line segments with integer length can be drawn from \(B\) to a point on hypotenuse \(AC\). The answer is thus \(\textbf{(D)} 13\).