+2446 Another option is to utilize something called the Exterior Angle Theorem. This theorem states that the measure of the exterior angle of a triangle is equal to sum of the two nonadjacent interior (also known as remote) angles. The diagram below does all the speaking for me.
Now that this theorem is established, we can save one step. In the given diagram, \(m\angle CAB+m\angle ABC=m\angle BCD\)
| \(m\angle CAB+m\angle ABC=m\angle BCD\) | Exterior Angle Theorem |
| \(m\angle CAB+85=120\) | Substitution Property of Equality |
| \(m\angle CAB=35^{\circ}\) | Subtraction Property of Equality |
Why is this the case? Well, I'm happy to show you why!
In this triangle, one is given that this is \(\triangle ABC\) with an exterior angle \(\angle BAD\). Our goal here is to prove that \(m\angle B+m\angle C=m\angle BAD\). I will utilize a two-column proof.
Let's assume that \(m\angle BAC=x^{\circ}\).
| \(m\angle BAC+m\angle B+m\angle C=180^{\circ}\) | Triangle Sum Theorem |
| \(x+m\angle B+m \angle C=180^{\circ}\) | Substitution Property of Equality |
| \(m\angle B+m\angle C=(180-x)^{\circ}\) | Subtraction Property of Equality |
| \(\angle BAC\) and \(\angle BAD\) form a linear pair | Definition of a linear pair |
| \(\angle BAC\) and \(\angle BAD\) are supplementary | \(\)Linear Pair Theorem |
| \(m\angle BAC+m\angle BAD=180^{\circ}\) | Definition of supplementary angles |
| \(x^{\circ}+m\angle BAD=180^{\circ}\) | Substitution Property of Equality |
| \(m\angle BAD=(180-x)^{\circ}\) | Subtraction Property of Equality |
| \(m\angle B+m\angle C=m\angle BAD\) | Transitive Property of Equality |
+2446 A midsegment is a segment joining the midpoint of two sides of a triangle. Each triangle has 3 midsegments. The Triangle Midsegment Theorem states that a midsegment of a triangle is parallel to the nonintersecting side, and its length is half the length of the nonintersecting side.
In this diagram here, \(\overline{BD}\), the midsegment, bisects both of its intersected sides. Also, \(\overline{BD}\parallel\overline{AE}\) and \(\frac{1}{2}\text{AE}=\text{BD}\).
Knowing this information, we can then determine where the coordinates of the endpoints of the midsegment will be.
Let's find the midpoint of both of the segments.
| \(\text{MDPT}_{\overline{PR}}\left(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2}\right)\) | Find the midpoint of the segment PR using this formula. |
| \(\text{MDPT}_{\overline{PR}}\left(\frac{-3-4}{2},\frac{3-2}{2}\right)\) | Simplify from here to find the midpoint. |
| \(\text{MDPT}_{\overline{PR}}\left(\frac{-7}{2},\frac{1}{2}\right)\) | Since the answers must be in decimal format, I will convert them. |
| \(\text{MDPT}_{\overline{PR}}\left(-3.5,0.5\right)\) |
Now, find the other midpoint.
| \(\text{MDPT}_{\overline{QR}}\left(\frac{2-4}{2},\frac{1-2}{2}\right)\) | Simplify from here. |
| \(\text{MDPT}_{\overline{PR}}\left(\frac{-2}{2},\frac{-1}{2}\right)\) | |
| \(\text{MDPT}_{\overline{PR}}\left(-1,\frac{-1}{2}\right)\) | |
| \(\text{MDPT}_{\overline{PR}}\left(-1,-0.5\right)\) |
+2446 It does not matter if the parallelogram is a rhombus; it makes no difference.
A property of a parallelogram is that opposite sides are parallel. This means that \(\overline{JK}\parallel\overline{ML}\). By the alternate interior angles theorem, \(\angle MLJ\cong\angle LJK\). This indicates that both angles are also of equal measure. Therefore, \(m\angle MLK=m\angle LJK=25^{\circ}\)
By the triangle sum theorem, the sum of all the angles in a triangle is equal to 180 degrees. Using this rule, we can solve for the measure of the remaining angle.
| \(m\angle KLJ+m\angle LJK+m\angle JKL=180\) | Substitute the known values in for the angles. |
| \(m\angle KLJ+25+130=180\) | Now, solve for the only unknown. |
| \(m\angle KLJ+155=180\) | |
| \(m\angle KLJ=25^{\circ}\) | |
Yes, you have to do basic addition and subtraction to get the answer to this problem. And as helperid1839321 mentioned, because the diagonals of the parallelogram bisect a pair of opposite angles, this figure is a rhombus.
+2446 The argument within the square root must be evaluated before taking the square root of something. Therefore, simplify the fraction first and then take the square root.
| \(\sqrt{\frac{80+1}{68-4}}\) | Simplify the numerator and denominator before proceeding. |
| \(\sqrt{\frac{81}{64}}\) | A square root rule worth knowing is that the square root of a fraction is equivalent to the square root of the numerator divided by the square root of the denominator. In other words, \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\). It is like "distributing" the square root, if you want to think of it that way. |
| \(\sqrt{\frac{81}{64}}=\frac{\sqrt{81}}{\sqrt{64}}\) | Now, simplify the numerator and the denominator. |
| \(\frac{\sqrt{81}}{\sqrt{64}}=\frac{9}{8}=1.125\) | This is your simplified answer. |
