Question

the diagram shows isosceles trapezoid lmnp. it also shows how line segment no was drawn to form parallelogram lmno. what is the measure of angle onp? 50° 65° 80° 130°

Explanation:

Step1: Properties of parallelogram

In parallelogram \(LMNO\), \(LM\parallel NO\) and \(LM = NO\). Also, \(\angle L=\angle MON = 50^{\circ}\) (opposite angles of a parallelogram are equal, and consecutive angles are supplementary. But here we use the property that \(LM\parallel NO\) so \(\angle L\) and \(\angle MON\) are equal as they are corresponding angles).

Step2: Properties of isosceles trapezoid

Since \(LMNP\) is an isosceles trapezoid, \(LM\parallel NP\), and \(LN = MP\). Also, \(NO\parallel LM\) (from parallelogram \(LMNO\)), so \(NO\parallel NP\) is wrong, actually \(LM\parallel NP\) and \(LM\parallel NO\) implies \(NO\parallel NP\) is incorrect, we know that in isosceles trapezoid \(LMNP\), \(LO + OP=LP\) and \(LM = NO\), \(LM = ON\), \(LN = MP\). Another property: in isosceles trapezoid \(LMNP\), the non - parallel sides \(LN\) and \(MP\) are equal. Also, since \(LMNO\) is a parallelogram \(LM = NO\) and \(LM = ON\).
Since \(LMNP\) is an isosceles trapezoid, \(\triangle NOP\) is isosceles. \(\angle MON\) is the exterior angle for \(\triangle NOP\). Let \(\angle ONP=\angle OPN = x\). Using the angle - sum property of a triangle in \(\triangle NOP\): \(\angle MON+\angle ONP+\angle OPN = 180^{\circ}\). Substituting \(\angle MON = 50^{\circ}\) and \(\angle ONP=\angle OPN\) (because \(NO = LM\) and \(LMNP\) is isosceles trapezoid, \(NO = NP\) (since \(LM = NP\) in isosceles trapezoid and \(LM = NO\) from parallelogram)).

$$50^{\circ}+x + x=180^{\circ}$$

$$2x=180^{\circ}- 50^{\circ}$$

$$2x = 130^{\circ}$$

$$x = 65^{\circ}$$

Answer:

\(65^{\circ}\)