Reason about lengths of collinear segments and midpoints given algebraically (no coordinate plane): add/subtract sub-segment lengths, use midpoint relationships, express a length in terms of others.
Free, timed SHSAT practice sets on segment lengths & midpoints— every question includes a full worked explanation, and there's no account required.
Loading questions…
Two ideas solve these problems.
With algebra: if \(M\) is the midpoint of \(AB\), \(AM = 2x + 5\) and \(MB = 4x - 3\), then \(2x + 5 = 4x - 3\), so \(x = 4\). Then \(AM = 13\) and \(AB = 26\).
Draw the points on a line in order and label every length you know before you start.
Common traps: stopping at \(x\) when the question asks for a length, giving half the segment when it asks for the whole, and with more than two parts, subtracting only one of them from the total.
\(AB + BC = AC\), so \(BC = AC - AB = 22 - 6 = 16\) cm. (Giving the whole segment \(AC\) without subtracting gives 22. Adding \(22 + 6\) instead of subtracting gives 28. Assuming \(B\) is the midpoint, \(22 \div 2\), gives 11.)
Since \(M\) is the midpoint, \(AM = MB\): \(3x + 40 = 5x + 6\), so \(2x = 34\) and \(x = 17\). Then \(AM = MB = 91\) m and \(AB = 91 + 91 = 182\) m. (Stopping at \(x = 17\) gives 17. Stopping at \(AM = 91\) gives only half of \(AB\). Forgetting to divide by 2 (so \(x = 34\)) gives 284.)
\(AB = 2(14) + 10 = 38\) and \(BC = 3(14) + 2 = 44\). Since \(AD = AB + BC + CD\), \(CD = 126 - 38 - 44 = 44\) in. (Subtracting only \(AB\) and forgetting \(BC\) gives 88. Stopping at \(AC = 38 + 44\) gives 82. Adding before multiplying, \(AB = 2(14 + 10) = 48\) and \(BC = 3(14 + 2) = 48\), gives 30.)