Question

the diagram shows quadrilateral mnpq. what is the length of line segment mq? 8 units 10 units 11 units 15 units

Explanation:

Step1: Analyze the right - angled triangle

Since \(NP\parallel MQ\) (given by the equal marks on the sides indicating parallelism) and \(PQ\perp MQ\), \(NP\perp PQ\). Let's consider the right - angled triangle formed.

Step2: Apply the Pythagorean theorem

We know that in a right - angled triangle, if we assume the length of \(MQ=x\), and we have \(MN = 10\), \(NP = 3\), \(PQ=6\). By the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Here, if we consider the right - angled triangle with hypotenuse \(MN\) and the other two sides related to \(MQ\) and \(PQ\).
Let's first find the length of the base of the right - angled triangle (assuming we complete the right - angled triangle). The base of the right - angled triangle (let's call it \(a\)) is \(x - 3\) (by the parallel sides \(NP\) and \(MQ\)) and the height \(b = 6\), hypotenuse \(c=10\).
Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), we substitute \(a=x - 3\), \(b = 6\), \(c = 10\) into the formula: \((x - 3)^{2}+6^{2}=10^{2}\).
Expand \((x - 3)^{2}\): \(x^{2}-6x + 9+36=100\).
Simplify the equation: \(x^{2}-6x+45 = 100\), then \(x^{2}-6x-55 = 0\).
Factor the quadratic equation: \((x - 11)(x + 5)=0\).
We get \(x=11\) or \(x=-5\). Since length cannot be negative, \(x = 11\).

Answer:

11 units