Question

1. construct a triangle with one side - length of 5 cm and an angle measure of 70°. 2) solve for x in the following: (mangle ijp=x + 46), (mangle pjk=x + 102) and (mangle ijk = 138^{circ}). find x. 3) pick two sets of points, create two line - segments, and find the distance of the line - segments. you must use the distance formula for each line - segment (d=sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}). 4) reflect the triangle first over the y - axis, and then over the x - axis. then, draw the resulting triangle with the coordinates of the triangle written to the right of the graph.

Explanation:

1. Construct a triangle with one - side length of 5 cm and an angle measure of 70°:

• Explanation:
• Step 1: Draw a line segment
• Use a ruler to draw a line segment of length 5 cm. Let's call this line segment \(AB = 5\mathrm{cm}\).
• Step 2: Construct the angle
• Place the center of a protractor at point \(A\). Align the baseline of the protractor with the line segment \(AB\). Mark a point \(C\) at the 70 - degree mark on the protractor. Then draw a ray \(AC\) from point \(A\) through the marked point \(C\).
• Step 3: Complete the triangle
• To complete the triangle, we can choose any point on the ray \(AC\) (other than \(A\)) and connect it to point \(B\) with a straight - line segment. For a non - degenerate triangle, we can use a compass to create a more precise construction if needed. But a simple way is to just draw a line from \(B\) to a point on \(AC\) to form \(\triangle ABC\).

1. Solve for \(x\) given \(m\angle IJP=x + 46\), \(m\angle PJK=x + 102\) and \(m\angle IJK = 138^{\circ}\):

• Explanation:
• Step 1: Use the angle - addition postulate
• According to the angle - addition postulate, \(m\angle IJP+m\angle PJK=m\angle IJK\). So, \((x + 46)+(x + 102)=138\).
• Step 2: Simplify the left - hand side of the equation
• Combine like terms: \(x+x+46 + 102=138\), which simplifies to \(2x+148 = 138\).
• Step 3: Isolate the variable \(x\)
• Subtract 148 from both sides of the equation: \(2x=138−148\), so \(2x=-10\).
• Then divide both sides by 2: \(x=\frac{-10}{2}=-5\).

1. Pick two sets of points, create two line segments, and find the distance of the line segments using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\):

• Let's assume two points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) on the coordinate - grid.
• Explanation:
• Step 1: Select the points
• For example, let \(P(1,1)\) and \(Q(4,5)\).
• Step 2: Identify the coordinates
• Here, \(x_1 = 1,y_1 = 1,x_2 = 4,y_2 = 5\).
• Step 3: Apply the distance formula
• \(d=\sqrt{(4 - 1)^2+(5 - 1)^2}=\sqrt{3^2+4^2}=\sqrt{9 + 16}=\sqrt{25}=5\).

1. Reflect the triangle first over the \(y\) - axis, and then over the \(x\) - axis:

• Let the vertices of the original triangle be \((x,y)\).
• Explanation:
• Step 1: Reflection over the \(y\) - axis
• The rule for reflecting a point \((x,y)\) over the \(y\) - axis is \((-x,y)\). So if the vertices of the original triangle are \((x_1,y_1)\), \((x_2,y_2)\) and \((x_3,y_3)\), after reflection over the \(y\) - axis, the vertices become \((-x_1,y_1)\), \((-x_2,y_2)\) and \((-x_3,y_3)\).
• Step 2: Reflection over the \(x\) - axis
• The rule for reflecting a point \((x,y)\) over the \(x\) - axis is \((x,-y)\). So, if the vertices after the first reflection are \((-x_1,y_1)\), \((-x_2,y_2)\) and \((-x_3,y_3)\), after reflection over the \(x\) - axis, the vertices become \((-x_1,-y_1)\), \((-x_2,-y_2)\) and \((-x_3,-y_3)\). Then plot these new vertices to draw the resulting triangle.

Answer:

1. Triangle constructed as described above.
2. \(x=-5\)
3. Distance calculated as described above (example distance is 5 for points \((1,1)\) and \((4,5)\)).
4. Triangle reflected as described above with new vertices \((-x_1,-y_1)\), \((-x_2,-y_2)\) and \((-x_3,-y_3)\) (where \((x_1,y_1)\), \((x_2,y_2)\) and \((x_3,y_3)\) are the original vertices).