Question

in the given parallelogram, find the value of a and b.
options:
○ a = 7; b = 4
○ a = 4; b = 7
○ a = 11; b = 3
○ a = 3; b = 11

Explanation:

Step1: Recall parallelogram diagonal property

In a parallelogram, the diagonals bisect each other. So, the segments of one diagonal are equal, and the segments of the other diagonal are equal.
For the diagonal with segments \(17\) and \(3a - 6\), we have \(3a-6 = 17\)? Wait, no, wait. Wait, looking at the diagram, the diagonals intersect, so the two parts of one diagonal should be equal. Wait, the lengths given: one diagonal has segments \(15\) and \(5b - 3\), and the other has \(17\) and \(3a - 6\)? Wait, no, maybe I misread. Wait, in a parallelogram, diagonals bisect each other, so the two halves of each diagonal are equal. So, if one diagonal is split into \(17\) and \(3a - 6\), then \(3a - 6=17\)? No, wait, maybe the other diagonal: the segments are \(15\) and \(5b - 3\)? Wait, no, let's check the options. Wait, maybe the correct approach is: in a parallelogram, diagonals bisect each other, so the length of one part of a diagonal equals the other part. So, for the diagonal with segments \(17\) and \(3a - 6\), wait, no, maybe the segments are \(15\) and \(5b - 3\), and \(17\) and \(3a - 6\). Wait, let's solve for \(a\) first. Let's assume that \(3a - 6 = 17\)? No, that would give \(3a=23\), which is not an integer. Wait, maybe I got the segments wrong. Wait, looking at the options, let's test the options. Let's take option \(a = 7\), \(b = 4\): \(3a - 6=3*7 -6=21 -6 = 15\), and \(5b - 3=5*4 -3=20 -3 = 17\). Wait, no, that's not matching. Wait, option \(a = 4\), \(b = 7\): \(3a -6=3*4 -6=12 -6 = 6\), no. Wait, wait, maybe I mixed up the segments. Wait, maybe the diagonal segments are \(15\) and \(5b - 3\), and \(17\) and \(3a - 6\). Wait, no, let's re - examine. Wait, the problem is a parallelogram, so diagonals bisect each other. So, the two parts of each diagonal are equal. So, if one diagonal is split into lengths \(17\) and \(3a - 6\), then \(3a - 6=17\)? No, that gives \(a=(17 + 6)/3=23/3\), not integer. Wait, maybe the other diagonal: segments \(15\) and \(5b - 3\), so \(5b-3 = 15\)? Then \(5b=18\), \(b = 18/5\), not integer. Wait, maybe I have the segments reversed. Wait, looking at the diagram, the intersection of the diagonals: so, the two parts of one diagonal are \(17\) and \(3a - 6\), and the two parts of the other diagonal are \(15\) and \(5b - 3\). Wait, no, in a parallelogram, diagonals bisect each other, so the length of one part of a diagonal is equal to the length of the other part of the same diagonal. So, for the diagonal with segments \(17\) and \(3a - 6\), we have \(3a - 6=17\)? No, that's not. Wait, maybe the segments are \(15\) and \(5b - 3\), so \(5b - 3 = 15\), then \(5b=18\), no. Wait, maybe I made a mistake. Wait, let's check the options. Let's take option \(a = 7\), \(b = 4\): \(3a - 6=3*7 -6=15\), \(5b - 3=5*4 -3=17\). Ah! So, the two parts of the diagonals: one diagonal has parts \(15\) and \(3a - 6\) (when \(a = 7\), \(3a -6 = 15\)), and the other diagonal has parts \(17\) and \(5b - 3\) (when \(b = 4\), \(5b - 3=17\)). So, that works because in a parallelogram, diagonals bisect each other, so each part of a diagonal is equal to the other part of the same diagonal. Wait, so \(3a - 6 = 15\) (since the other part is \(15\)) and \(5b - 3 = 17\) (since the other part is \(17\)). Let's solve \(3a - 6 = 15\): \(3a=15 + 6=21\), so \(a = 7\). Solve \(5b - 3 = 17\): \(5b=17 + 3=20\), so \(b = 4\). So that's option \(a = 7\), \(b = 4\).

Step2: Verify the solution

For \(a = 7\): \(3a-6=3\times7 - 6=21 - 6 = 15\), which matches the other segment of the same diagonal (length \(15\)). For \(b = 4\): \(5b - 3=5\times4 - 3=20 - 3 = 17\), which matches the other segment of the same diagonal (length \(17\)). So this satisfies the property of parallelogram diagonals bisecting each other.

Answer:

\(a = 7\); \(b = 4\) (the option with \(a = 7\); \(b = 4\))