find the value of the variable. the diagram is not to scale.
45. use the information in the figure. find m∠d
drawing not to scale
46. find the value of x. the diagram is not to scale.
47. find the value of x. the diagram is not to scale.
48. find the value of x.
49. list the sides in order from shortest to longest. the diagram is not to scale.

Question

Accepted Answer

### Problem 45: Find $ m\angle D $
# Explanation:
## Step 1: Identify Triangle Type
The triangle $ DEF $ has two equal sides (marked with ticks), so it's isosceles with $ \angle E = \angle D $ or $ \angle E = \angle F $? Wait, no, the angle at $ F $ is $ 106^\circ $, which is obtuse. In a triangle, the sum of angles is $ 180^\circ $, and in an isosceles triangle, the base angles are equal. Wait, actually, the sides $ DE $ and $ EF $? No, the marks: $ DF $ and $ EF $? Wait, the diagram shows $ DF $ with one tick, $ EF $ with one tick? Wait, no, the user's diagram: $ D $, $ F $ with a tick on $ DF $ and $ EF $? Wait, no, the angle at $ F $ is $ 106^\circ $. Wait, maybe it's an isosceles triangle with $ \angle D = \angle E $, and $ \angle F = 106^\circ $. Wait, sum of angles in a triangle is $ 180^\circ $. So $ \angle D + \angle E + \angle F = 180^\circ $. If $ \angle D = \angle E $ (since sides $ DF $ and $ EF $ are equal? Wait, no, the ticks: maybe $ DE $ and $ DF $? Wait, maybe I misread. Wait, the problem is to find $ m\angle D $. Let's assume it's an isosceles triangle with $ \angle D = \angle E $, and $ \angle F = 106^\circ $. Then:

## Step 2: Calculate $ \angle D $
Let $ \angle D = \angle E = x $. Then $ x + x + 106^\circ = 180^\circ $. So $ 2x = 180 - 106 = 74^\circ $, so $ x = 37^\circ $. Wait, but maybe the triangle is isosceles with $ \angle F $ as the vertex angle. Wait, no, $ 106^\circ $ is obtuse, so it must be the vertex angle, and the base angles are $ \angle D $ and $ \angle E $. So yes, $ 2\angle D = 180 - 106 $, so $ \angle D = (180 - 106)/2 = 37^\circ $.

# Answer:
$ 37^\circ $

### Problem 46: Find $ x $
# Explanation:
## Step 1: Identify Triangle Properties
Triangle $ RST $ is isosceles (sides $ RS $ and $ ST $ are equal, marked with ticks), so $ \angle R = \angle S $? Wait, no, the base is $ RT $, and the exterior angle at $ T $ is $ (5x)^\circ $. The interior angle at $ T $ is $ 180 - 5x $ (since exterior and interior angles are supplementary). In triangle $ RST $, $ \angle R = (5x - 60)^\circ $, $ \angle T = (180 - 5x)^\circ $, and since it's isosceles with $ RS = ST $, $ \angle R = \angle T $? Wait, no, $ RS = ST $, so the base angles are $ \angle R $ and $ \angle T $? Wait, no, $ RS = ST $, so the angles opposite them: $ \angle T $ is opposite $ RS $, and $ \angle R $ is opposite $ ST $. So $ \angle R = \angle T $. Wait, $ \angle R = 5x - 60 $, $ \angle T = 180 - 5x $ (since exterior angle is $ 5x $, so interior is $ 180 - 5x $). So set them equal:

## Step 2: Solve for $ x $
$ 5x - 60 = 180 - 5x $

Add $ 5x $ to both sides: $ 10x - 60 = 180 $

Add 60 to both sides: $ 10x = 240 $

Divide by 10: $ x = 24 $

Wait, but let's check: $ \angle R = 5(24) - 60 = 120 - 60 = 60^\circ $, $ \angle T = 180 - 5(24) = 180 - 120 = 60^\circ $, so triangle $ RST $ is equilateral? Wait, maybe. Alternatively, the exterior angle theorem: the exterior angle is equal to the sum of the two non-adjacent interior angles. The exterior angle at $ T $ is $ 5x $, which should equal $ \angle R + \angle S $. But since $ RS = ST $, $ \angle R = \angle S $? Wait, no, $ RS = ST $, so $ \angle R = \angle T $. Wait, maybe I made a mistake. Wait, the exterior angle at $ T $ (angle $ T U $) is $ 5x $, so the interior angle at $ T $ is $ 180 - 5x $. In triangle $ RST $, angles sum to $ 180 $: $ (5x - 60) + \angle S + (180 - 5x) = 180 $. Simplify: $ 120 + \angle S = 180 $, so $ \angle S = 60^\circ $. Then, since $ RS = ST $, $ \angle R = \angle T $, so $ 5x - 60 = 180 - 5x $, which gives $ 10x = 240 $, $ x = 24 $. That works.

# Answer:
$ 24 $

### Problem 47: Find $ x $
# Explanation:
## Step 1: Identify Similar Triangles
The diagram shows a triangle with a line segment parallel to the base, creating two similar triangles (by the Basic Proportionality Theorem, or AA similarity). The smaller triangle has sides 20, 30, 41, and the larger triangle has base 20 + 20 = 40, and the other sides proportional. Wait, the left side of the larger triangle is $ x $, the smaller triangle has left side 30, and the hypotenuse 41. Wait, the segments on the base are 20 and 20, so the base of the larger triangle is 40, and the base of the smaller is 20. So the ratio of similarity is 2 (since 40 / 20 = 2). Therefore, the sides of the larger triangle are twice the sides of the smaller triangle. The smaller triangle has a side of 30, so the larger triangle's corresponding side is $ x = 30 \times 2 = 60 $. Wait, that matches the handwritten 60. Let's verify:

## Step 2: Apply Similar Triangles Ratio
Since the line is parallel to the base, the triangles are similar with a scale factor of 2 (because the base of the larger triangle is 20 + 20 = 40, and the base of the smaller is 20, so 40 / 20 = 2). Therefore, the side $ x $ in the larger triangle is twice the corresponding side (30) in the smaller triangle: $ x = 30 \times 2 = 60 $.

# Answer:
$ 60 $

### Problem 48: Find $ x $
# Explanation:
## Step 1: Identify Midsegment Theorem
The diagram shows a triangle with a midsegment (a line segment connecting the midpoints of two sides). The Midsegment Theorem states that the midsegment is parallel to the third side and half its length. Here, the midsegment has length $ 3x - 7 $, and the third side (the base) has length 16. Wait, no, the sides: the triangle has sides with ticks, indicating midpoints. So the midsegment should be half the length of the third side. Wait, the third side (the one with length 16) is parallel to the midsegment? Wait, no, the midsegment connects the midpoints of the two sides with double ticks, so it's parallel to the side with length 16, and its length is half of 16? Wait, no, the midsegment length is $ 3x - 7 $, and the third side is 16. Wait, maybe the midsegment is parallel to the side of length 16, so $ 3x - 7 = \frac{1}{2} \times 16 $? Wait, no, the Midsegment Theorem: the midsegment is half the length of the third side. Wait, if the two sides with double ticks are midpoints, then the midsegment is parallel to the side with length 16, so midsegment length = $ \frac{1}{2} \times 16 = 8 $? No, that can't be. Wait, maybe the side with length 16 is the third side, and the midsegment is $ 3x - 7 $, so $ 3x - 7 = \frac{1}{2} \times 16 $? Wait, no, 16 is one of the equal sides? Wait, the triangle has two sides with single ticks (equal) and two sides with double ticks (equal), so it's an isosceles triangle with a midsegment. Wait, the midsegment connects the midpoints of the two equal sides (with double ticks) and the two equal sides (with single ticks). So the midsegment length should be equal to half the length of the base? Wait, no, let's re-express:

The Midsegment Theorem: In a triangle, the segment connecting the midpoints of two sides is parallel to the third side and half as long. So if the midsegment is $ 3x - 7 $, and the third side (the one not connected by the midsegment) is 16, then $ 3x - 7 = \frac{1}{2} \times 16 $? Wait, no, 16 is one of the sides, and the midsegment is parallel to it. Wait, maybe the side with length 16 is the third side, so midsegment length is $ \frac{1}{2} \times 16 = 8 $, but that would make $ 3x - 7 = 8 $, so $ 3x = 15 $, $ x = 5 $. But that seems too small. Wait, maybe I got the sides reversed. Wait, the triangle has a side of 16, and the midsegment is $ 3x - 7 $, and the midsegment is parallel to the side of length 16, so midsegment length is half of 16? No, the Midsegment Theorem says the midsegment is half the length of the third side. Wait, maybe the third side is 16, so midsegment is $ \frac{1}{2} \times 16 = 8 $, so $ 3x - 7 = 8 $, so $ 3x = 15 $, $ x = 5 $. But let's check: $ 3(5) - 7 = 15 - 7 = 8 $, which is half of 16. Yes, that works.

## Step 2: Solve for $ x $
Using the Midsegment Theorem: midsegment length = $ \frac{1}{2} \times $ third side length. Here, third side length is 16, so midsegment length = $ \frac{1}{2} \times 16 = 8 $. Set $ 3x - 7 = 8 $:

$ 3x = 8 + 7 = 15 $

$ x = \frac{15}{3} = 5 $

# Answer:
$ 5 $

### Problem 49: List sides from shortest to longest
# Explanation:
## Step 1: Find the Angles
In triangle $ JKL $, the angles are $ \angle J = 125^\circ $, $ \angle K = 23^\circ $, $ \angle L = 32^\circ $. Wait, sum of angles: $ 125 + 23 + 32 = 180 $, which checks out.

## Step 2: Relate Angles to Sides
In a triangle, the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle. So:

- Smallest angle: $ \angle K = 23^\circ $, so the side opposite $ \angle K $ (which is $ JL $) is the shortest.
- Next angle: $ \angle L = 32^\circ $, so the side opposite $ \angle L $ (which is $ JK $) is next.
- Largest angle: $ \angle J = 125^\circ $, so the side opposite $ \angle J $ (which is $ KL $) is the longest.

Wait, let's confirm:

- Side opposite $ \angle K $ (23°): $ JL $
- Side opposite $ \angle L $ (32°): $ JK $
- Side opposite $ \angle J $ (125°): $ KL $

So from shortest to longest: $ JL $, $ JK $, $ KL $.

# Answer:
$ JL $, $ JK $, $ KL $ (or using the standard notation, if the triangle is labeled $ J $, $ K $, $ L $, then sides: $ JL $ (opposite $ K $), $ JK $ (opposite $ L $), $ KL $ (opposite $ J $))

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 45: Find \( m\angle D \)

Step 1: Identify Triangle Type

Step 2: Calculate \( \angle D \)

Step 1: Identify Triangle Properties

Step 2: Solve for \( x \)

Step 1: Identify Similar Triangles

Step 2: Apply Similar Triangles Ratio

Answer:

Problem 46: Find \( x \)