check understanding
1. how can you construct a perpendicular bisector of a segment rs by folding a piece of paper?
2. in the figure, is $\overline{bd}$ perpendicular to $\overline{ac}$? explain why or why not.
find the value of the variable. explain your reasoning.
3.
4.

Question

Accepted Answer

### Problem 1
# Explanation:
## Step1: Understand the goal
We need to find a way to construct the perpendicular bisector of segment $ RS $ by folding paper.
## Step2: Recall folding properties
Folding the paper so that point $ R $ coincides with point $ S $ will align the two endpoints. The crease formed will be the perpendicular bisector because folding over this line maps $ R $ to $ S $, so it bisects $ RS $ and is perpendicular (since the fold creates a right angle with the segment).
# Answer:
To construct the perpendicular bisector of segment $ RS $ by folding paper, fold the paper so that point $ R $ lies exactly on top of point $ S $. The crease formed by this fold is the perpendicular bisector of $ RS $. This is because the fold line is equidistant from $ R $ and $ S $ (bisecting $ RS $) and forms a right angle with $ RS $ (perpendicular) due to the symmetry of the fold.

### Problem 2
# Explanation:
## Step1: Analyze the triangle
In the given triangle $ ABC $, $ AB = BC = 13 $ and $ AD = DC = 10 $. So, $ \triangle ABC $ is isosceles with $ AB = BC $, and $ D $ is the midpoint of $ AC $.
## Step2: Apply the Isosceles Triangle Theorem
The Isosceles Triangle Theorem states that in an isosceles triangle, the median from the apex (vertex $ B $ here) to the base ($ AC $) is also the altitude (perpendicular to the base) and the angle bisector. Since $ D $ is the midpoint ($ AD = DC $) and $ AB = BC $, $ BD $ is the median, so it must also be the altitude. Thus, $ \overline{BD} \perp \overline{AC} $.
# Answer:
Yes, $ \overline{BD} $ is perpendicular to $ \overline{AC} $. The triangle $ ABC $ is isosceles with $ AB = BC = 13 $. Point $ D $ is the midpoint of $ AC $ (since $ AD = DC = 10 $). By the Isosceles Triangle Theorem, the median from the apex $ B $ to the base $ AC $ (which is $ BD $) is also the altitude, meaning it is perpendicular to $ AC $.

### Problem 3
# Explanation:
## Step1: Identify the triangle type
In $ \triangle ABC $, $ BD \perp AC $, $ AD = DC = 4 $, and $ AB = 6 $. Triangles $ ABD $ and $ CBD $ are right triangles (since $ \angle ADB = \angle CDB = 90^\circ $) and congruent (by SAS, as $ AD = DC $, $ \angle ADB = \angle CDB $, $ BD = BD $). So, $ AB = BC $, and we can use the Pythagorean theorem in $ \triangle ABD $ to find $ BD $, but we need to find $ b $ (which is $ BC $, equal to $ AB $). Wait, $ AB = 6 $, and since $ \triangle ABC $ is isosceles with $ AB = BC $, $ b = AB = 6 $? Wait, no, wait: Wait, $ AB = 6 $, $ AD = 4 $, $ BD $ is perpendicular. Wait, actually, in $ \triangle ABD $, $ AB = 6 $, $ AD = 4 $, so by Pythagoras, $ BD^2 + AD^2 = AB^2 $, but we need $ b $, which is $ BC $. But since $ \triangle ABD \cong \triangle CBD $ (right triangles, $ AD = DC $, $ BD $ common), $ BC = AB = 6 $. Wait, but let's check: Wait, $ AB = 6 $, $ AD = 4 $, so $ BD = \sqrt{AB^2 - AD^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} $. Then in $ \triangle CBD $, $ BC = \sqrt{BD^2 + DC^2} = \sqrt{20 + 16} = \sqrt{36} = 6 $. So $ b = 6 $.
## Step2: Apply congruence or Pythagoras
Since $ BD \perp AC $ and $ AD = DC $, $ \triangle ABD \cong \triangle CBD $ (SAS: $ AD = DC $, $ \angle ADB = \angle CDB = 90^\circ $, $ BD = BD $). Therefore, $ BC = AB = 6 $, so $ b = 6 $.
# Answer:
The value of $ b $ is $ 6 $. In right triangle $ ABD $, $ AB = 6 $, $ AD = 4 $, and $ BD \perp AC $. Triangles $ ABD $ and $ CBD $ are congruent (SAS), so $ BC = AB = 6 $, hence $ b = 6 $.

### Problem 4
# Explanation:
## Step1: Identify the triangle type
In $ \triangle ABC $, $ AB = BC = 12 $, $ BD \perp AC $, and $ DC = 8 $. Triangles $ ABD $ and $ CBD $ are right triangles ( $ \angle ADB = \angle CDB = 90^\circ $) and congruent (by SAS, as $ AB = BC $, $ \angle ADB = \angle CDB $, $ BD = BD $). So, $ AD = DC $? Wait, no, wait: Wait, $ AB = BC = 12 $, so $ \triangle ABC $ is isosceles with $ AB = BC $, so the median from $ B $ to $ AC $ (which is $ BD $) should bisect $ AC $. Wait, but $ DC = 8 $, so $ AD = DC = 8 $? Wait, no, wait the diagram shows $ DC = 8 $ and $ AD = r $. Wait, no, actually, in an isosceles triangle with $ AB = BC $, the altitude $ BD $ bisects $ AC $, so $ AD = DC $. Wait, but $ AB = BC = 12 $, $ DC = 8 $, so $ AD = 8 $? Wait, no, that can't be. Wait, no, let's use the Pythagorean theorem in $ \triangle CBD $. $ BC = 12 $, $ DC = 8 $, so $ BD^2 + DC^2 = BC^2 $, so $ BD^2 = 12^2 - 8^2 = 144 - 64 = 80 $. Then in $ \triangle ABD $, $ AB = 12 $, $ BD^2 = 80 $, so $ AD^2 + BD^2 = AB^2 $, so $ AD^2 = 144 - 80 = 64 $, so $ AD = 8 $. Wait, so $ r = 8 $.
## Step2: Apply the Pythagorean theorem
In right triangle $ CBD $, $ BC = 12 $, $ DC = 8 $. By the Pythagorean theorem:
$ BD^2 + DC^2 = BC^2 $
$ BD^2 + 8^2 = 12^2 $
$ BD^2 = 144 - 64 = 80 $

In right triangle $ ABD $, $ AB = 12 $, $ BD^2 = 80 $. By the Pythagorean theorem:
$ AD^2 + BD^2 = AB^2 $
$ r^2 + 80 = 144 $
$ r^2 = 144 - 80 = 64 $
$ r = \sqrt{64} = 8 $
# Answer:
The value of $ r $ is $ 8 $. In the isosceles triangle $ ABC $ with $ AB = BC = 12 $, $ BD \perp AC $. Using the Pythagorean theorem in $ \triangle CBD $ ( $ BC = 12 $, $ DC = 8 $) gives $ BD^2 = 80 $. Then in $ \triangle ABD $, solving $ r^2 + 80 = 12^2 $ gives $ r^2 = 64 $, so $ r = 8 $.

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Understand the goal

Step2: Recall folding properties

Step1: Analyze the triangle

Step2: Apply the Isosceles Triangle Theorem

Step1: Identify the triangle type

Step2: Apply congruence or Pythagoras

Answer:

Problem 2