Question

find the value of y.
12.
13.
14.
15.

1. graph the triangle with vertices ( d(2,0), e(2,4) ), and ( f(6,2) ). then graph a triangle congruent to ( \triangle def ).
2. given ( \triangle h j k cong \triangle t r s ), find the values of ( a ) and ( b )

Explanation:

Step1: Use the triangle angle - sum property

The sum of the interior angles of a triangle is \(180^{\circ}\).
For \(\triangle ABC\), \(\angle A = 65^{\circ}\), \(\angle C=90^{\circ}\). Let \(\angle B = y\). Then \(y + 65+90=180\).
For \(\triangle FDE\), \(\angle F = 90^{\circ}\). Let \(\angle D=y\), \(\angle E\) is also \(y\) (since the two non - right angles in a right - isosceles triangle are equal). Then \(y + y+90 = 180\), \(2y=90\), \(y = 45\).

For \(\triangle OKM\), \(\angle K\) is unknown, \(\angle O = 65^{\circ}\). Using the angle - sum property \(y+65+\angle K=180\). But if we assume it's a different triangle (maybe the user intended a simple angle - sum), if we consider a non - right triangle with two known angles, say \(\angle K\) is found from another relation (but since the problem is about finding \(y\) in a triangle, assume a basic triangle).

For \(\triangle PRS\) (the quadrilateral part):
First, find the angle adjacent to \(70^{\circ}\) in the triangle with \(55^{\circ}\) and \(65^{\circ}\). The sum of angles in a triangle is \(180^{\circ}\). In the triangle with \(55^{\circ}\) and \(65^{\circ}\), the third angle is \(180-(55 + 65)=60^{\circ}\). Then in the other triangle (with \(70^{\circ}\) and \(5y\)), using the angle - sum property of a triangle: \(70+5y + 60=180\).
\[ \begin{align*} 5y+130&=180\\ 5y&=180 - 130\\ 5y&=50\\ y& = 10 \end{align*} \]

For the quadrilateral in problem 15:
The sum of angles in a quadrilateral is \(360^{\circ}\). Let's consider the two triangles that make up the quadrilateral. The sum of angles around the intersection of the diagonals:
First, find the angle at \(W\) in the left - hand triangle. Using the angle - sum property of a triangle (in the triangle with \(20^{\circ}\) and the other angles). But if we use the quadrilateral angle - sum formula \(360^{\circ}\). Let's assume we use the triangle relations. If we consider the two triangles:
In one triangle, we know some angles. Using the angle - sum property for the quadrilateral (sum of all interior angles \(=360^{\circ}\)). If we assume the two triangles:
Let's first find the angle at \(W\) (using the triangle with \(20^{\circ}\)). Then for the quadrilateral: \((2y + 2)+120+\text{other angles}=360\). But if we consider the triangle relations (assuming the two triangles are congruent in a way):
If we use the triangle angle - sum for the left - hand triangle: assume the angle at \(W\) is found from \(180-( \text{other angles})\). But a better approach is:
If we consider the two triangles that form the quadrilateral. Let's assume the sum of angles in the two triangles (since a quadrilateral can be divided into two triangles, sum of angles \(=2\times180 = 360^{\circ}\)).
If we assume one triangle has angles \(20^{\circ}\), \(x\), \(z\) and the other has \((2y + 2)\), \(120^{\circ}\), \(w\). But if we use the property of the figure (maybe the two triangles are such that we can equate angles).
Alternatively, if we consider the triangle with \(2y+2\) and use the angle - sum:
Let's assume a different approach. If we consider the fact that in a triangle (part of the quadrilateral), if we use the angle - sum:
Suppose we have a triangle with angles \((2y + 2)\), \(x\), \(z\) and another with \(120^{\circ}\), \(m\), \(n\). But if we use the property that the sum of angles in the quadrilateral is \(360^{\circ}\) and assume some angle equalities (from the figure's symmetry, though not clear). A better step - by - step (assuming the user made a typo and it's a triangle - related \(y\) in 15, but since it's a quadrilateral, we'll skip and focus on 14 (as 12 and 13 are not clear in figure - based angle - sum without proper labels).

Step2: Solve the equation for \(y\) in problem 14

We had the equation \(5y+70+(180-(55 + 65))=180\) (using the angle - sum of a triangle).
\[ \begin{align*} 5y+70 + 60&=180\\ 5y&=180-(70 + 60)\\ 5y&=50\\ y&=10 \end{align*} \]

Answer:

If we assume the problem is 14 (since 12 and 13 have unclear figure - based angle - sum without proper labels), \(y = 10\)