Triangle Properties and Angle Theorems

46 soru

Soru 21Soru

A triangle has two sides of lengths 55 centimeters and 1111 centimeters. Which of the following could be the perimeter of the triangle, in centimeters?

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Cevap: 27

Cevap

27
The correct answer is 2727 centimeters. According to the Triangle Inequality Theorem, the length of the third side, xx, of a triangle must be strictly greater than the difference of the other two sides (115=611 - 5 = 6) and strictly less than their sum (11+5=1611 + 5 = 16). This gives the inequality range 6<x<166 < x < 16. The perimeter is the sum of all three sides, which is P=5+11+x=16+xP = 5 + 11 + x = 16 + x. Applying the bounds of xx, we find that the perimeter must satisfy 16+6<P<16+1616 + 6 < P < 16 + 16, which simplifies to 22<P<3222 < P < 32. Among the options, 2727 is the only value that is strictly within this range.

Adım Adım Çözüm

1
Apply the Triangle Inequality Theorem to find the limits for the third side.
Let the third side be xx. The length of xx must satisfy: 115<x<11+5    6<x<1611 - 5 < x < 11 + 5 \implies 6 < x < 16.
The length of any side of a triangle must be strictly between the positive difference and the sum of the lengths of the other two sides.
2
Set up the equation for the perimeter of the triangle.
Perimeter P=5+11+x=16+xP = 5 + 11 + x = 16 + x.
The perimeter of a triangle is defined as the sum of the lengths of all three of its sides.
3
Find the range of possible values for the perimeter PP by applying the inequality bounds of xx.
Add 1616 to all parts of the inequality 6<x<166 < x < 16: 16+6<16+x<16+16    22<P<3216 + 6 < 16 + x < 16 + 16 \implies 22 < P < 32.
We must shift the inequality bounds of the third side by the sum of the two known sides to find the range of the perimeter.
4
Identify the option that falls strictly inside the range 22<P<3222 < P < 32.
The value 2727 is the only option that satisfies 22<27<3222 < 27 < 32.
Only a value strictly between 2222 and 3232 can represent a mathematically valid perimeter for this triangle.

Anahtar Kavram

Triangle Inequality Theorem and Perimeter
Tahmini Süre:1m 0s
Soru 22Soru

A triangle has two sides of length 88 and 1515. If the length of the third side, ss, is a prime number, how many possible values are there for ss?

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Cevap: 44

Cevap

There are 4 possible values for the third side length s.
The correct answer is 44. According to the Triangle Inequality Theorem, the length of the third side ss of a triangle with sides of 88 and 1515 must satisfy 158<s<15+815 - 8 < s < 15 + 8. This simplifies to the open interval 7<s<237 < s < 23. The prime numbers strictly between 77 and 2323 are 1111, 1313, 1717, and 1919. Counting these gives exactly 44 possible prime values for ss.

Adım Adım Çözüm

1
Apply the Triangle Inequality Theorem to determine the bounds for the third side ss.
158<s<15+815 - 8 < s < 15 + 8, which simplifies to 7<s<237 < s < 23.
The Triangle Inequality Theorem states that the length of any side of a triangle must be strictly greater than the difference of the other two sides and strictly less than their sum.
2
Identify all prime numbers that lie strictly within the range (7,23)(7, 23).
The prime numbers in this range are 1111, 1313, 1717, and 1919.
A prime number is an integer greater than 1 that has no positive divisors other than 1 and itself.
3
Count the number of identified prime numbers.
There are 44 prime numbers (1111, 1313, 1717, 1919).
This counts the total number of possible valid lengths for ss.

Anahtar Kavram

Triangle Inequality Theorem and basic number properties
Soru 23Soru

In a triangle, two of the sides have lengths 1313 and 2020. The third side has a length of ss, where ss is an integer. If the side of length 2020 is the longest side of the triangle, and the triangle is obtuse, what is the number of possible values for ss?

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Cevap: 8

Cevap

There are 8 possible integer values for ss.
To find the number of possible integer values for ss, we combine the Triangle Inequality Theorem (13+s>20    s>713 + s > 20 \implies s > 7) and the condition for an obtuse triangle with 2020 as the longest side (202>132+s2    s2<231    s1520^2 > 13^2 + s^2 \implies s^2 < 231 \implies s \leq 15). This limits ss to integers in the range [8,15][8, 15], which contains exactly 88 values.

Adım Adım Çözüm

1
Apply the Triangle Inequality Theorem to find the lower bound for ss.
s>7s > 7, so the minimum integer value is 88.
The sum of the two shorter sides of a triangle must be strictly greater than the longest side.
2
Set up the obtuse triangle inequality with 2020 as the longest side.
202>132+s220^2 > 13^2 + s^2
In any obtuse triangle with longest side cc, the inequality c2>a2+b2c^2 > a^2 + b^2 must hold.
3
Solve the inequality 202>132+s220^2 > 13^2 + s^2 for ss.
s2<231    s15s^2 < 231 \implies s \leq 15
Simplifying the inequality gives 400>169+s2    s2<231400 > 169 + s^2 \implies s^2 < 231. The largest integer whose square is less than 231231 is 1515.
4
Determine the number of integers in the range [8,15][8, 15].
8 possible values
The integers satisfying both conditions are {8,9,10,11,12,13,14,15}\{8, 9, 10, 11, 12, 13, 14, 15\}, which count to 88.

Anahtar Kavram

Triangle Inequality Theorem and obtuse triangle classification using side lengths

Alternatif Yöntem

List the perfect squares and verify which ones satisfy both s2<231s^2 < 231 and the Triangle Inequality Theorem s>7s > 7.
Tahmini Süre:2m 0s
Soru 24Soru

A non-degenerate triangle has side lengths of 55, 1212, and xx. A second non-degenerate triangle has side lengths of xx, 1010, and yy. If xx and yy must be integers, and the perimeter of the second triangle is the minimum possible integer value, what is the sum of all possible values of yy?

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Cevap: 6

Cevap

6
The correct answer is 6. By applying the Triangle Inequality Theorem, the shared side length xx of the first triangle must satisfy 125<x<12+512 - 5 < x < 12 + 5, which simplifies to 7<x<177 < x < 17. Since xx must be an integer, its possible values are {8,9,10,11,12,13,14,15,16}\{8, 9, 10, 11, 12, 13, 14, 15, 16\}. For the second triangle with side lengths xx, 1010, and yy, the Triangle Inequality Theorem requires x10<y<x+10|x - 10| < y < x + 10. To minimize the perimeter P=x+10+yP = x + 10 + y, we minimize x+yx + y. Checking the possible values of xx, we find that when x=8x = 8, the minimum integer value for yy is 33 (giving P=21P = 21); when x=9x = 9, the minimum integer value for yy is 22 (giving P=21P = 21); and when x=10x = 10, the minimum integer value for yy is 11 (giving P=21P = 21). For any x11x \ge 11, the minimum value of yy is x9x - 9, resulting in a perimeter of at least 2323. Therefore, the minimum perimeter of the second triangle is 2121, which is achieved when yy is 33, 22, or 11. The sum of these values of yy is 1+2+3=61 + 2 + 3 = 6.

Adım Adım Çözüm

1
Apply the Triangle Inequality Theorem to the first triangle to find the range of the shared side length xx.
7<x<177 < x < 17
The sum of any two side lengths of a triangle must be strictly greater than the third side length. Thus, 125<x<12+512 - 5 < x < 12 + 5.
2
List the possible integer values of xx based on the range.
x{8,9,10,11,12,13,14,15,16}x \in \{8, 9, 10, 11, 12, 13, 14, 15, 16\}
The problem states that xx must be an integer.
3
Apply the Triangle Inequality Theorem to the second triangle with sides xx, 1010, and yy to express the range of yy in terms of xx.
x10<y<x+10|x - 10| < y < x + 10
The third side length yy must be strictly between the difference and the sum of the other two sides (xx and 1010).
4
Determine the minimum integer value of the perimeter of the second triangle, P=x+10+yP = x + 10 + y, by testing the possible values of xx and finding the minimum integer yy for each.
The minimum perimeter is 2121, achieved when (x,y)=(8,3)(x, y) = (8, 3), (9,2)(9, 2), or (10,1)(10, 1).
Minimizing the perimeter P=x+y+10P = x + y + 10 is equivalent to minimizing the sum x+yx + y for integer values of y>x10y > |x - 10|.
5
Sum the possible integer values of yy that yield the minimum perimeter.
3+2+1=63 + 2 + 1 = 6
We need to find the sum of all possible values of yy that result in the minimum perimeter of 2121.

Anahtar Kavram

Triangle Inequality Theorem and Perimeter Optimization
Tahmini Süre:3m 0s
Soru 25Soru

The measures of the interior angles of a triangle are in the ratio 2:3:72:3:7. If the measure of the largest angle is decreased by 1515^\circ and the measure of the smallest angle is increased by 1515^\circ, what is the ratio of the interior angles of the new triangle, ordered from smallest to largest?

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Cevap: 1:1:21:1:2

Cevap

The ratio of the interior angles of the new triangle is 1:1:21:1:2.
The correct answer is the ratio 1:1:21:1:2. The sum of the interior angles in a triangle is always 180180^\circ. Given the ratio 2:3:72:3:7, the sum of the parts is 1212, which means each part represents 180/12=15180^\circ / 12 = 15^\circ. The original angles are therefore 3030^\circ, 4545^\circ, and 105105^\circ. Increasing the smallest angle by 1515^\circ gives 4545^\circ, and decreasing the largest by 1515^\circ gives 9090^\circ. The new angle measures are 4545^\circ, 4545^\circ, and 9090^\circ, which simplifies to 1:1:21:1:2.

Adım Adım Çözüm

1
Set up the equation for the sum of the interior angles of a triangle.
2x+3x+7x=1802x + 3x + 7x = 180^\circ
The sum of the interior angles of any triangle is always 180180^\circ.
2
Solve for the value of xx.
12x=180    x=1512x = 180^\circ \implies x = 15^\circ
Combining like terms simplifies the equation to find the value of one ratio unit.
3
Calculate the original measures of the three angles.
Smallest: 3030^\circ, Middle: 4545^\circ, Largest: 105105^\circ
Multiply each part of the ratio by x=15x = 15^\circ.
4
Apply the modifications to the smallest and largest angles.
New smallest: 30+15=4530^\circ + 15^\circ = 45^\circ; New largest: 10515=90105^\circ - 15^\circ = 90^\circ; Middle: 4545^\circ (unchanged).
Perform the operations described in the problem statement.
5
Order the new angle measures from smallest to largest and simplify the ratio.
45:45:90    1:1:245^\circ : 45^\circ : 90^\circ \implies 1 : 1 : 2
Divide each term in the ratio by the greatest common divisor, which is 4545.

Anahtar Kavram

Angle sum theorem of a triangle and ratio partition applications

Alternatif Yöntem

Instead of calculating the actual angle values, note that the sum of the ratio parts is 2+3+7=122+3+7 = 12, and the sum of the interior angles of a triangle is 180180^\circ. This means 11 ratio unit is equal to 180/12=15180^\circ / 12 = 15^\circ. Since the smallest angle is increased by 1515^\circ (exactly 11 ratio unit) and the largest is decreased by 1515^\circ (exactly 11 ratio unit), we can apply these modifications directly to the ratio terms. The new ratio terms are 2+1=32+1 = 3, 33 (unchanged), and 71=67-1 = 6. This gives a ratio of 3:3:63:3:6, which simplifies to 1:1:21:1:2.
Tahmini Süre:1m 30s
Soru 26Soru

In ABC\triangle ABC, the side lengths are AB=12AB = 12, BC=15BC = 15, and AC=18AC = 18. A point PP lies strictly inside ABC\triangle ABC. If the lengths of the segments BPBP and CPCP are both integers, what is the maximum possible value of the sum of these two lengths?

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Cevap: 28

Cevap

28
The correct answer is 28. According to the properties of triangles, for any point PP strictly inside ABC\triangle ABC, the sum of the interior segments is strictly less than the sum of the other two sides: BP+CP<AB+AC=12+18=30BP + CP < AB + AC = 12 + 18 = 30. Since BPBP and CPCP are integers, we check the boundary where PP lies on the side ACAC at an integer distance CP=zCP = z from CC. Applying Stewart's Theorem, the boundary length BP=yboundBP = y_{bound} is z222.5z+225\sqrt{z^2 - 22.5z + 225}. For the largest possible integer value z=17z = 17, the boundary value is approximately 11.4711.47. Because the point must lie strictly inside the triangle, BPBP must be strictly less than this boundary, so the maximum integer value for BPBP is 11. This yields a maximum sum of 11+17=2811 + 17 = 28. Lower integer values of zz yield smaller maximum sums (for example, if z=16z = 16, the boundary is exactly 11, so BPBP can be at most 10, giving a sum of 26).

Adım Adım Çözüm

1
Apply the interior point triangle inequality theorem.
For any point PP strictly inside ABC\triangle ABC, the sum of the distances to two vertices is strictly less than the sum of the other two sides: BP+CP<AB+ACBP + CP < AB + AC.
This establishes the theoretical upper bound for the sum of the two segment lengths.
2
Calculate the theoretical upper bound.
Since AB=12AB = 12 and AC=18AC = 18, we have BP+CP<12+18=30BP + CP < 12 + 18 = 30. Since BPBP and CPCP must be integers, the sum BP+CPBP + CP can be at most 29.
This sets the initial integer limit before evaluating if it is geometrically possible.
3
Analyze the boundary conditions for integer lengths using Stewart's Theorem.
Let CP=zCP = z and BP=yBP = y, where yy and zz are integers. As PP approaches the side ACAC, the boundary value yboundy_{bound} represents the distance from BB to a point on ACAC at distance zz from CC. Using Stewart's Theorem, this boundary satisfies: ybound2=z222.5z+225y_{bound}^2 = z^2 - 22.5z + 225. Since PP is strictly inside the triangle, yy must be strictly less than yboundy_{bound}.
This provides the mathematical relationship determining whether a point is inside the triangle for any given integer length of one segment.
4
Test the maximum possible integer value for zz to maximize the sum y+zy + z.
Since PP is strictly inside, zz must be strictly less than AC=18AC = 18, so the maximum integer for zz is 17. For z=17z = 17, the boundary value is ybound=17222.5(17)+225=131.511.47y_{bound} = \sqrt{17^2 - 22.5(17) + 225} = \sqrt{131.5} \approx 11.47. Since y<yboundy < y_{bound}, the maximum integer value for yy is 11. This yields a maximum sum of 11+17=2811 + 17 = 28.
This determines the actual maximum integer sum that can be geometrically realized within the triangle.

Anahtar Kavram

Triangle Inequality Theorem and Interior Point Properties
Soru 27Soru

In ABC\triangle ABC, the lengths of sides ABAB and ACAC are both 1313. A point DD lies on side BCBC such that ADAD is an integer. If the perimeter of ABD\triangle ABD is equal to the perimeter of ACD\triangle ACD, what is the sum of all possible integer values for the length of BCBC?

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Cevap: 34

Cevap

34
The perimeters of ABD\triangle ABD and ACD\triangle ACD are equal, and since AB=AC=13AB = AC = 13, this forces BD=CDBD = CD, making DD the midpoint of BCBC. In the isosceles triangle ABC\triangle ABC, the median ADAD is perpendicular to BCBC, making ABD\triangle ABD a right triangle. By the Pythagorean theorem, BD2+AD2=169BD^2 + AD^2 = 169. Since ADAD is an integer, BDBD must also be an integer (a half-integer would result in AD2AD^2 ending in .25.25, which cannot be a perfect square of an integer). The only positive integer solutions for (BD,AD)(BD, AD) are (5,12)(5, 12) and (12,5)(12, 5). This results in BC=2BDBC = 2 \cdot BD being either 1010 or 2424. The sum of these possible values is 10+24=3410 + 24 = 34.

Adım Adım Çözüm

1
Set the perimeters of ABD\triangle ABD and ACD\triangle ACD equal to each other.
BD=CDBD = CD
Since AB=AC=13AB = AC = 13, equating AB+BD+AD=AC+CD+ADAB + BD + AD = AC + CD + AD simplifies directly to BD=CDBD = CD.
2
Determine the relationship between ADAD and BCBC.
ABD\triangle ABD is a right triangle with hypotenuse 1313.
In an isosceles triangle, the median to the base is also the altitude, so ADBCAD \perp BC.
3
Apply the Pythagorean theorem to ABD\triangle ABD.
BD2+AD2=169BD^2 + AD^2 = 169
The sum of the squares of the legs in right triangle ABD\triangle ABD must equal the square of the hypotenuse AB=13AB = 13.
4
Analyze the parity and integer constraints of BDBD and ADAD.
BDBD must be a positive integer.
If BDBD were a half-integer, BD2BD^2 would end in .25.25, preventing AD2AD^2 from being an integer, which contradicts the given condition that ADAD is an integer.
5
Identify the Pythagorean triples with a hypotenuse of 1313.
(BD,AD){(5,12),(12,5)}(BD, AD) \in \{(5, 12), (12, 5)\}
The only positive integer solutions to x2+y2=132x^2 + y^2 = 13^2 are (5,12)(5, 12) and (12,5)(12, 5).
6
Calculate the possible lengths of BCBC and sum them.
BC{10,24}BC \in \{10, 24\}, and their sum is 3434.
Since DD is the midpoint of BCBC, the length of BCBC is 2BD2 \cdot BD, yielding 25=102 \cdot 5 = 10 and 212=242 \cdot 12 = 24. Both satisfy the triangle inequality because BC<AB+AC=26BC < AB + AC = 26.

Anahtar Kavram

Properties of Isosceles Triangles and the Pythagorean Theorem
Tahmini Süre:3m 0s
Soru 28Soru

In ABC\triangle ABC, the measure of B\angle B is 8080^\circ and the measure of C\angle C is 4040^\circ. A point DD lies on side BCBC such that ADAD bisects BAC\angle BAC, and a point EE lies on side ACAC such that AD=AEAD = AE. What is the measure, in degrees, of CDE\angle CDE?

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Cevap: 35

Cevap

35
The correct answer is 3535. By first finding that BAC=60\angle BAC = 60^\circ, we use the angle bisector ADAD to find CAD=30\angle CAD = 30^\circ. In ADC\triangle ADC, we find the interior angle ADC=110\angle ADC = 110^\circ. In the isosceles triangle ADE\triangle ADE with AD=AEAD=AE, the base angles are ADE=AED=75\angle ADE = \angle AED = 75^\circ. Finally, subtracting ADE\angle ADE from ADC\angle ADC gives CDE=35\angle CDE = 35^\circ.

Adım Adım Çözüm

1
Find the measure of the third angle of the main triangle, BAC\angle BAC.
BAC=60\angle BAC = 60^\circ
The sum of the interior angles of any triangle is 180180^\circ. Therefore, BAC=180BC=1808040=60\angle BAC = 180^\circ - \angle B - \angle C = 180^\circ - 80^\circ - 40^\circ = 60^\circ.
2
Determine the measure of the bisected angle CAD\angle CAD.
CAD=30\angle CAD = 30^\circ
Since ADAD bisects BAC\angle BAC, it divides the angle into two equal parts: BAD=CAD=602=30\angle BAD = \angle CAD = \frac{60^\circ}{2} = 30^\circ.
3
Calculate the interior angle ADC\angle ADC in ADC\triangle ADC.
ADC=110\angle ADC = 110^\circ
In ADC\triangle ADC, the sum of angles is 180180^\circ. Therefore, ADC=180CADC=1803040=110\angle ADC = 180^\circ - \angle CAD - \angle C = 180^\circ - 30^\circ - 40^\circ = 110^\circ.
4
Find the base angles of the isosceles triangle ADEADE.
ADE=75\angle ADE = 75^\circ
Since AD=AEAD = AE, ADE\triangle ADE is an isosceles triangle with vertex angle DAE=30\angle DAE = 30^\circ. The two base angles, ADE\angle ADE and AED\angle AED, are equal. Thus, ADE=180302=75\angle ADE = \frac{180^\circ - 30^\circ}{2} = 75^\circ.
5
Determine the final angle CDE\angle CDE by subtraction.
CDE=35\angle CDE = 35^\circ
Since point EE lies on side ACAC, ray DEDE lies between rays DADA and DCDC. Therefore, ADC=ADE+CDE\angle ADC = \angle ADE + \angle CDE. Rearranging gives CDE=ADCADE=11075=35\angle CDE = \angle ADC - \angle ADE = 110^\circ - 75^\circ = 35^\circ.

Anahtar Kavram

Applying triangle angle sum theorem, angle bisector properties, and isosceles triangle base angle properties to perform multi-step angle tracing.

Alternatif Yöntem

Use the exterior angle theorem on ADC\triangle ADC at vertex DD: ADB=CAD+C=30+40=70\angle ADB = \angle CAD + \angle C = 30^\circ + 40^\circ = 70^\circ. Then, since EE is on ACAC, AA, EE, and CC are collinear. In ADE\triangle ADE, the exterior angle at EE is DEC=DAE+ADE=30+75=105\angle DEC = \angle DAE + \angle ADE = 30^\circ + 75^\circ = 105^\circ. In DEC\triangle DEC, the sum of angles is 180180^\circ, so CDE=18010540=35\angle CDE = 180^\circ - 105^\circ - 40^\circ = 35^\circ.
Tahmini Süre:2m 30s
Soru 29Soru

In a certain triangle, the ratio of the measure of the first angle to the measure of the second angle is 1:21:2. The measure of the third angle is 2020^\circ less than the measure of the second angle. What is the measure of the largest angle in the triangle?

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Cevap: 8080^\circ

Cevap

8080^\circ
The correct answer is 8080^\circ. By representing the first angle as xx, the second angle is 2x2x, and the third angle is 2x202x - 20. The sum of the interior angles in a triangle is always 180180^\circ, which gives the equation x+2x+(2x20)=180x + 2x + (2x - 20) = 180. Simplifying this results in 5x20=1805x - 20 = 180, which yields 5x=2005x = 200 and x=40x = 40. Substituting this value back into the expressions for the three angles gives measures of 4040^\circ, 8080^\circ, and 6060^\circ. Comparing these values shows that the largest angle is 8080^\circ.

Adım Adım Çözüm

1
Define the measures of the first and second angles using a single variable based on their ratio.
Let the first angle be xx and the second angle be 2x2x.
Since the ratio of the first angle to the second angle is 1:21:2, we can represent them as xx and 2x2x respectively.
2
Express the measure of the third angle in terms of the same variable.
The third angle is 2x202x - 20.
The problem states the third angle is 2020^\circ less than the second angle, which has a measure of 2x2x.
3
Set up an equation using the triangle angle sum theorem and solve for xx.
x+2x+(2x20)=180    5x20=180    5x=200    x=40x + 2x + (2x - 20) = 180 \implies 5x - 20 = 180 \implies 5x = 200 \implies x = 40.
The sum of the measures of the interior angles of any triangle is always 180180^\circ.
4
Calculate the measures of all three angles to determine which is the largest.
First angle = 4040^\circ, Second angle = 2(40)=802(40^\circ) = 80^\circ, Third angle = 2(40)20=602(40^\circ) - 20^\circ = 60^\circ. The largest angle is 8080^\circ.
We must substitute x=40x = 40 back into our expressions for each angle to find their actual degree measures and identify the largest one.

Anahtar Kavram

The interior angles of a triangle always sum to 180180^\circ. Ratios and word problems can be modeled algebraically to determine unknown angle measures.
Tahmini Süre:1m 15s
Soru 30Soru

In ABC\triangle ABC, point DD lies on side BCBC. The segment ADAD divides the interior angle BAC\angle BAC into two angles, BAD\angle BAD and DAC\angle DAC, whose measures are in the ratio 3:23:2, respectively. The measures of the interior angles B\angle B and C\angle C are in the ratio 5:45:4, respectively. If the measure of ADC\angle ADC is 104104^\circ, what is the measure of BAC\angle BAC?

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Cevap: 9090^\circ

Cevap

9090^\circ
The correct answer is 9090^\circ. By expressing the angles in terms of variables using their ratios, we set up two independent linear equations: 3x+5y=1043x + 5y = 104 (from the exterior angle theorem on ABD\triangle ABD) and 2x+4y=762x + 4y = 76 (from the sum of angles in ADC\triangle ADC). Solving this system yields x=18x = 18. Since BAC\angle BAC is composed of BAD\angle BAD and DAC\angle DAC, its measure is 3x+2x=5x=5(18)=903x + 2x = 5x = 5(18^\circ) = 90^\circ.

Adım Adım Çözüm

1
Define variables for the partitioned angles and the base angles using the given ratios.
Let the measures of BAD\angle BAD and DAC\angle DAC be 3x3x and 2x2x respectively, so that BAC=5x\angle BAC = 5x. Let the measures of B\angle B and C\angle C be 5y5y and 4y4y respectively.
Ratios express quantities as multiples of a common variable, which simplifies setting up equations.
2
Apply the exterior angle theorem to ABD\triangle ABD at vertex DD.
The exterior angle ADC=BAD+B104=3x+5y\angle ADC = \angle BAD + \angle B \Rightarrow 104^\circ = 3x + 5y.
The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.
3
Apply the angle sum theorem to ADC\triangle ADC.
DAC+C+ADC=1802x+4y+104=1802x+4y=76\angle DAC + \angle C + \angle ADC = 180^\circ \Rightarrow 2x + 4y + 104^\circ = 180^\circ \Rightarrow 2x + 4y = 76^\circ.
The sum of the measures of the interior angles of any triangle is always 180180^\circ.
4
Solve the system of linear equations: (1) 3x+5y=1043x + 5y = 104 and (2) 2x+4y=762x + 4y = 76.
Multiply equation (1) by 2 and equation (2) by 3 to align the coefficients of xx:
6x+10y=2086x + 10y = 208
6x+12y=2286x + 12y = 228
Subtract the first aligned equation from the second:
2y=20y=102y = 20 \Rightarrow y = 10.
Substitute y=10y = 10 back into equation (2):
2x+4(10)=762x=36x=182x + 4(10) = 76 \Rightarrow 2x = 36 \Rightarrow x = 18.
Solving the system of linear equations determines the values of the variables xx and yy.
5
Calculate the measure of BAC\angle BAC.
BAC=5x=5(18)=90\angle BAC = 5x = 5(18^\circ) = 90^\circ.
We defined the total measure of BAC\angle BAC as the sum of its two partitioned parts, 3x+2x=5x3x + 2x = 5x.

Anahtar Kavram

Using triangle angle properties and exterior angle theorems to set up and solve systems of linear equations.
Soru 31Soru

In a triangle, the lengths of the sides are xx, yy, and zz, where xx, yy, and zz are integers such that x<y<zx < y < z. If x=7x = 7 and the perimeter of the triangle is 3232, what is the number of possible integer values for zz?

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Cevap: 3

Cevap

There are exactly 3 possible integer values for the side length z.
The correct answer is 3. By expressing the second side as y=25zy = 25 - z and applying the ordering constraint 7<25z<z7 < 25 - z < z, we determine that 12.5<z<1812.5 < z < 18. Applying the Triangle Inequality Theorem (7+y>z7 + y > z) yields the restriction z<16z < 16. Combining these conditions restricts the integer values of zz to {13,14,15}\{13, 14, 15\}, which counts to exactly 3 possible values.

Adım Adım Çözüm

1
Express the side length yy in terms of zz.
y=25zy = 25 - z
The perimeter of the triangle is the sum of the side lengths: x+y+z=32x + y + z = 32. Substituting x=7x = 7 gives 7+y+z=327 + y + z = 32, which simplifies to y=25zy = 25 - z.
2
Apply the given inequality constraint x<y<zx < y < z to find initial bounds for zz.
12.5<z<1812.5 < z < 18
Substituting x=7x = 7 and y=25zy = 25 - z into x<y<zx < y < z yields 7<25z<z7 < 25 - z < z. The left inequality 7<25z7 < 25 - z simplifies to z<18z < 18. The right inequality 25z<z25 - z < z simplifies to 25<2z25 < 2z, or z>12.5z > 12.5.
3
Apply the Triangle Inequality Theorem to establish the final constraint on zz.
z<16z < 16
Since zz is the longest side, the sum of the two shorter sides must be strictly greater than zz: x+y>zx + y > z. Substituting x=7x = 7 and y=25zy = 25 - z gives 7+25z>z7 + 25 - z > z, which simplifies to 32>2z32 > 2z, or z<16z < 16.
4
Combine all constraints and count the valid integer values for zz.
3 possible values (13,14,1513, 14, 15)
Combining the bounds from the steps gives 12.5<z<1612.5 < z < 16. The integers satisfying this inequality are 1313, 1414, and 1515, which gives a total of 3 possible integer values.

Anahtar Kavram

Triangle Inequality Theorem and algebraic constraints on side lengths
Soru 32Soru

A triangle has two sides of length 5 and 12. The third side has a length of xx, where xx is an integer. If the perimeter of the triangle is a multiple of 5, what is the sum of all possible values of xx?

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Cevap: 21

Cevap

The sum of all possible values of the third side length is 21.
To find the sum of all possible values of xx, we first apply the Triangle Inequality Theorem. For a triangle with side lengths 5, 12, and xx, the third side must satisfy 125<x<12+512 - 5 < x < 12 + 5, which simplifies to 7<x<177 < x < 17. The perimeter PP of the triangle is given by P=5+12+x=17+xP = 5 + 12 + x = 17 + x. Given that 7<x<177 < x < 17, the perimeter must be between 17+7=2417 + 7 = 24 and 17+17=3417 + 17 = 34. The only multiples of 5 within this range are 25 and 30. Setting the perimeter equal to these values gives 17+x=25    x=817 + x = 25 \implies x = 8, and 17+x=30    x=1317 + x = 30 \implies x = 13. Both values are integers and satisfy the triangle inequality. The sum of these values is 8+13=218 + 13 = 21.

Adım Adım Çözüm

1
Apply the Triangle Inequality Theorem to find the range of possible values for the third side, xx.
125<x<12+512 - 5 < x < 12 + 5, which simplifies to 7<x<177 < x < 17.
The length of any side of a triangle must be strictly greater than the difference between the other two sides and strictly less than their sum.
2
Determine the expression for the perimeter of the triangle and find the bounds for the perimeter.
Perimeter P=5+12+x=17+xP = 5 + 12 + x = 17 + x. Since 7<x<177 < x < 17, the perimeter must satisfy 17+7<P<17+1717 + 7 < P < 17 + 17, which means 24<P<3424 < P < 34.
The perimeter of a triangle is the sum of its three side lengths.
3
Identify which values of the perimeter in this range are multiples of 5, and find the corresponding values of xx.
The multiples of 5 between 24 and 34 are 25 and 30. If P=25P = 25, then 17+x=25    x=817 + x = 25 \implies x = 8. If P=30P = 30, then 17+x=30    x=1317 + x = 30 \implies x = 13. Both x=8x = 8 and x=13x = 13 are integers that satisfy the initial inequality.
We must find the integer values of xx that make the perimeter a multiple of 5.
4
Calculate the sum of all possible values of xx.
8+13=218 + 13 = 21.
The question asks for the sum of all valid integer values of xx.

Anahtar Kavram

Triangle Inequality Theorem and Perimeter Calculations
Tahmini Süre:1m 30s
Soru 33Soru

An isosceles triangle has two sides of length 55 and 1111. A second triangle has side lengths of 1212, 1818, and dd, where dd is an integer. If dd is equal to the perimeter of the first triangle, what is the perimeter of the second triangle?

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Cevap: 57

Cevap

57
To find the perimeter of the second triangle, we must first determine the value of dd, which is the perimeter of the first triangle. The first triangle is isosceles with two sides of length 55 and 1111. By the Triangle Inequality Theorem, the sum of any two side lengths must exceed the third. A triangle with sides 5,5,115, 5, 11 is impossible because 5+5=10<115 + 5 = 10 < 11. Thus, the sides of the first triangle must be 11,11,511, 11, 5, giving a perimeter of 11+11+5=2711 + 11 + 5 = 27. This means d=27d = 27. The second triangle has sides of length 1212, 1818, and 2727. Since 12+18=30>2712 + 18 = 30 > 27, this is a valid triangle. Its perimeter is 12+18+27=5712 + 18 + 27 = 57.

Adım Adım Çözüm

1
Determine the possible side lengths of the first isosceles triangle.
The sides must be 1111, 1111, and 55.
An isosceles triangle has two equal sides. The side lengths must be either 5,5,115, 5, 11 or 11,11,511, 11, 5. According to the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side. If the sides were 5,5,115, 5, 11, then 5+5=10<115 + 5 = 10 < 11, which violates this theorem. Thus, the only valid side lengths are 1111, 1111, and 55 (since 5+11=16>115 + 11 = 16 > 11).
2
Calculate the perimeter of the first triangle to find the value of dd.
d=27d = 27
The perimeter of the first triangle is the sum of its three sides: 11+11+5=2711 + 11 + 5 = 27. Since dd is equal to this perimeter, d=27d = 27.
3
Verify that a triangle with side lengths 1212, 1818, and 2727 is valid.
The triangle is valid.
We check the Triangle Inequality Theorem: 12+18=30>2712 + 18 = 30 > 27, 12+27=39>1812 + 27 = 39 > 18, and 18+27=45>1218 + 27 = 45 > 12. Since all inequalities hold, the second triangle is valid.
4
Calculate the perimeter of the second triangle.
Perimeter = 5757
The perimeter of the second triangle is the sum of its side lengths: 12+18+27=5712 + 18 + 27 = 57.

Anahtar Kavram

Triangle Inequality Theorem and Isosceles Triangle Properties
Tahmini Süre:2m 0s
Soru 34Soru

In ABC\triangle ABC, the measures of the interior angles A\angle A, B\angle B, and C\angle C are in the ratio 3:4:53:4:5, respectively. What is the measure of the largest exterior angle of ABC\triangle ABC?

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Cevap: 135135^\circ

Cevap

135 degrees
The interior angles of a triangle sum to 180180^\circ. Since the angles are in the ratio 3:4:53:4:5, they can be represented as 3x3x, 4x4x, and 5x5x. Adding these gives 12x=18012x = 180^\circ, which simplifies to x=15x = 15^\circ. The interior angles are therefore 4545^\circ, 6060^\circ, and 7575^\circ. Because an exterior angle is supplementary to its adjacent interior angle, the largest exterior angle is paired with the smallest interior angle: 18045=135180^\circ - 45^\circ = 135^\circ. Alternatively, the exterior angle is equal to the sum of the two non-adjacent interior angles: 60+75=13560^\circ + 75^\circ = 135^\circ.

Adım Adım Çözüm

1
Represent the interior angles algebraically using the given ratio.
Let the measures of the interior angles be 3x3x, 4x4x, and 5x5x.
The ratio of the angles is 3:4:53:4:5, so their measures must be multiples of these ratio numbers by the same factor xx.
2
Set up and solve an equation for xx using the triangle angle sum theorem.
3x+4x+5x=180    12x=180    x=153x + 4x + 5x = 180^\circ \implies 12x = 180^\circ \implies x = 15^\circ.
The sum of the interior angles of any triangle is always 180180^\circ.
3
Determine the measures of the three interior angles.
The angles measure 3(15)=453(15^\circ) = 45^\circ, 4(15)=604(15^\circ) = 60^\circ, and 5(15)=755(15^\circ) = 75^\circ.
Multiplying the value of xx by each term of the ratio gives the individual interior angle measures.
4
Find the largest exterior angle of the triangle.
The largest exterior angle is supplementary to the smallest interior angle: 18045=135180^\circ - 45^\circ = 135^\circ.
An exterior angle and its adjacent interior angle form a linear pair and sum to 180180^\circ. The smallest interior angle will yield the largest exterior angle.

Anahtar Kavram

Triangle Angle Sum Theorem and Exterior Angle Relationships

Alternatif Yöntem

The exterior angle at any vertex of a triangle is equal to the sum of the measures of the two opposite interior angles. The two largest interior angles are 6060^\circ and 7575^\circ. Therefore, the largest exterior angle is the sum of these two angles: 60+75=13560^\circ + 75^\circ = 135^\circ.
Tahmini Süre:1m 0s
Soru 35Soru

In ABC\triangle ABC, point DD lies on side BCBC such that AD=BDAD = BD. If the measure of ADC\angle ADC is 112112^\circ and the measure of BAC\angle BAC is 8585^\circ, what is the measure of C\angle C, in degrees?

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Cevap: 39

Cevap

The measure of C\angle C is 3939^\circ.
The measure of C\angle C is found by first calculating the interior angle ADB=180112=68\angle ADB = 180^\circ - 112^\circ = 68^\circ since BDCBDC forms a straight line. Because AD=BDAD = BD, ABD\triangle ABD is isosceles with B=BAD\angle B = \angle BAD. Using the angle sum of ABD\triangle ABD, we have 2(B)+68=1802(\angle B) + 68^\circ = 180^\circ, which yields B=56\angle B = 56^\circ. Finally, using the angle sum of ABC\triangle ABC, we calculate C=180(85+56)=39\angle C = 180^\circ - (85^\circ + 56^\circ) = 39^\circ.

Adım Adım Çözüm

1
Find the measure of ADB\angle ADB using the linear pair relationship with ADC\angle ADC.
ADB=68\angle ADB = 68^\circ
Angles on a straight line add up to 180180^\circ. Since DD lies on BCBC, ADB+ADC=180\angle ADB + \angle ADC = 180^\circ.
2
Calculate the measure of B\angle B using the properties of the isosceles triangle ABDABD.
B=56\angle B = 56^\circ
Since AD=BDAD = BD, the base angles opposite to these sides are equal: BAD=B\angle BAD = \angle B. The sum of angles in ABD\triangle ABD is 180180^\circ, so 2(B)+68=1802(\angle B) + 68^\circ = 180^\circ.
3
Find the measure of C\angle C using the triangle angle sum theorem on the large triangle ABCABC.
C=39\angle C = 39^\circ
The sum of the angles in ABC\triangle ABC is 180180^\circ, meaning BAC+B+C=180\angle BAC + \angle B + \angle C = 180^\circ. Substituting the known values gives 85+56+C=18085^\circ + 56^\circ + \angle C = 180^\circ.

Anahtar Kavram

Using the Isosceles Triangle Theorem, the Triangle Angle Sum Theorem, and linear pairs to trace unknown angles in a geometric figure.

Daha Fazla Pratik

Try finding the missing angles when a transversal cuts two parallel lines that form a triangle with a third intersecting line.

Alternatif Yöntem

Instead of finding B\angle B first and then solving for C\angle C in ABC\triangle ABC, we can find the angle DAC\angle DAC first. Since ADC=112\angle ADC = 112^\circ is an exterior angle to ABD\triangle ABD, we have ADC=B+BAD\angle ADC = \angle B + \angle BAD. Since B=BAD\angle B = \angle BAD, we get 2(BAD)=112    BAD=562(\angle BAD) = 112^\circ \implies \angle BAD = 56^\circ. Because BAC=85\angle BAC = 85^\circ, we have DAC=8556=29\angle DAC = 85^\circ - 56^\circ = 29^\circ. Now looking at ADC\triangle ADC, we can solve for C\angle C directly: C=180(112+29)=39\angle C = 180^\circ - (112^\circ + 29^\circ) = 39^\circ.
Tahmini Süre:1m 30s
Soru 36Soru

The lengths of the three sides of a triangle are in the ratio 3:4:x3:4:x, where xx is an integer. If the perimeter of the triangle is 3636 centimeters, how many different possible values can xx have?

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Cevap: 2

Cevap

There are 2 possible integer values for xx.
The correct answer is the value of 2. By expressing the side lengths as 3k3k, 4k4k, and xkxk where kk must be a positive integer, the perimeter equation becomes k(7+x)=36k(7 + x) = 36. Solving for 7+x7 + x yields factors of 3636 greater than 77, which are 9,12,18,9, 12, 18, and 3636. These correspond to xx values of 2,5,11,2, 5, 11, and 2929. Checking each set of side lengths against the Triangle Inequality Theorem shows that only the sets corresponding to x=2x = 2 (sides 12,16,812, 16, 8) and x=5x = 5 (sides 9,12,159, 12, 15) form valid triangles.

Adım Adım Çözüm

1
Define the side lengths using a multiplier kk.
Let the side lengths of the triangle be 3k3k, 4k4k, and xkxk for some positive multiplier kk. Since all three side lengths must be integers, the difference between the first two sides, 4k3k=k4k - 3k = k, must also be an integer. Thus, kk must be a positive integer.
This establishes that the scaling factor kk is a positive integer, allowing us to find discrete solutions.
2
Set up the perimeter equation and express xx in terms of kk.
The perimeter is the sum of the side lengths: 3k+4k+xk=36    k(7+x)=363k + 4k + xk = 36 \implies k(7 + x) = 36. Since kk and xx are positive integers, 7+x7 + x must be a factor of 3636 that is greater than 77.
This constrains the possible values of xx to the factors of 3636 that are larger than 77.
3
Find the potential values of xx and their corresponding side lengths.
The factors of 3636 greater than 77 are 9,12,18,9, 12, 18, and 3636. This yields four potential cases:
- If 7+x=9    x=27+x = 9 \implies x = 2, then k=4k = 4, and the sides are 12,16,812, 16, 8.
- If 7+x=12    x=57+x = 12 \implies x = 5, then k=3k = 3, and the sides are 9,12,159, 12, 15.
- If 7+x=18    x=117+x = 18 \implies x = 11, then k=2k = 2, and the sides are 6,8,226, 8, 22.
- If 7+x=36    x=297+x = 36 \implies x = 29, then k=1k = 1, and the sides are 3,4,293, 4, 29.
This identifies all mathematically possible configurations before checking if they can physically form a triangle.
4
Apply the Triangle Inequality Theorem to each case.
The sum of the lengths of any two sides must be strictly greater than the length of the remaining side:
- For sides 12,16,812, 16, 8: 12+8=20>1612 + 8 = 20 > 16 (Valid).
- For sides 9,12,159, 12, 15: 9+12=21>159 + 12 = 21 > 15 (Valid).
- For sides 6,8,226, 8, 22: 6+8=14<226 + 8 = 14 < 22 (Invalid).
- For sides 3,4,293, 4, 29: 3+4=7<293 + 4 = 7 < 29 (Invalid).
Only the cases where x=2x = 2 and x=5x = 5 form valid triangles.
This filters the candidate values of xx to only those that can form a valid geometric triangle.

Anahtar Kavram

Triangle Inequality Theorem and Integer Ratio Constraints
Tahmini Süre:1m 30s
Soru 37Soru

A triangle has side lengths of 77, x+2x+2, and 2x12x-1, where xx is an integer. What is the total number of possible values for xx?

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Cevap: 7

Cevap

The total number of possible integer values for xx is 77.
According to the Triangle Inequality Theorem, a triangle is formed if and only if the sum of any two side lengths is strictly greater than the third side length. Solving the three inequalities 7+(x+2)>2x17 + (x+2) > 2x-1, 7+(2x1)>x+27 + (2x-1) > x+2, and (x+2)+(2x1)>7(x+2) + (2x-1) > 7 gives x<10x < 10, x>4x > -4, and x>2x > 2. The overlapping interval is 2<x<102 < x < 10. The integers in this range are {3,4,5,6,7,8,9}\{3, 4, 5, 6, 7, 8, 9\}, which gives a total of 77 possible integer values.

Adım Adım Çözüm

1
Set up the three inequalities required by the Triangle Inequality Theorem.
1) 7+(x+2)>2x17 + (x+2) > 2x-1
2) 7+(2x1)>x+27 + (2x-1) > x+2
3) (x+2)+(2x1)>7(x+2) + (2x-1) > 7
The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the remaining side.
2
Solve the first inequality: 7+(x+2)>2x17 + (x+2) > 2x-1.
x<10x < 10
Simplifying the left side yields x+9>2x1x + 9 > 2x - 1. Subtracting xx and adding 11 to both sides results in 10>x10 > x, which means x<10x < 10.
3
Solve the second inequality: 7+(2x1)>x+27 + (2x-1) > x+2.
x>4x > -4
Simplifying the left side yields 2x+6>x+22x + 6 > x + 2. Subtracting xx and 66 from both sides results in x>4x > -4.
4
Solve the third inequality: (x+2)+(2x1)>7(x+2) + (2x-1) > 7.
x>2x > 2
Simplifying the left side yields 3x+1>73x + 1 > 7. Subtracting 11 and dividing by 33 results in x>2x > 2.
5
Determine the combined range for xx and identify the valid integers.
The combined range is 2<x<102 < x < 10. The valid integers are 3,4,5,6,7,8,3, 4, 5, 6, 7, 8, and 99.
To satisfy all three inequalities, xx must be greater than 22, greater than 4-4, and less than 1010, which simplifies to 2<x<102 < x < 10.
6
Count the total number of valid integer values for xx.
There are 77 integer values.
Counting the elements of the set {3,4,5,6,7,8,9}\{3, 4, 5, 6, 7, 8, 9\} yields a total of 77 values.

Anahtar Kavram

Triangle Inequality Theorem
Tahmini Süre:1m 30s
Soru 38Soru

In PQR\triangle PQR, the measure of P\angle P is 5050^\circ. The angle bisectors of PQR\angle PQR and PRQ\angle PRQ intersect at point II inside the triangle. What is the measure of QIR\angle QIR?

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Cevap: 115115^\circ

Cevap

115 degrees
To find the measure of the angle at the intersection, we first use the Triangle Angle-Sum Theorem on the larger triangle. The sum of the interior angles in the larger triangle is 180 degrees. Since the angle at one vertex is 50 degrees, the sum of the other two angles must be 130 degrees. The angle bisectors divide these two angles in half, meaning the sum of the two half-angles in the smaller triangle is half of 130 degrees, which is 65 degrees. Applying the Triangle Angle-Sum Theorem to the smaller triangle, the sum of its interior angles is also 180 degrees. Subtracting the sum of the two half-angles (65 degrees) from 180 degrees gives 115 degrees for the angle at the intersection.

Adım Adım Çözüm

1
Find the sum of the remaining interior angles of the triangle.
The sum of the angles at Q and R is 130 degrees.
The sum of all interior angles in any triangle is always 180 degrees, and the angle at P is given as 50 degrees.
2
Determine the sum of the bisected angles in the smaller triangle.
The sum of the half-angles is 65 degrees.
Since the lines QI and RI are angle bisectors, the sum of the interior angles of the smaller triangle at vertices Q and R is half the sum of the angles at Q and R of the larger triangle.
3
Calculate the measure of the angle at the intersection point.
The angle at the intersection point is 115 degrees.
The sum of the interior angles in the smaller triangle is also 180 degrees, so the unknown angle is found by subtracting the sum of the two half-angles from 180 degrees.

Anahtar Kavram

The Triangle Angle-Sum Theorem states that the sum of the measures of the interior angles of a triangle is always 180 degrees. Angle bisectors divide an angle into two equal parts.

Alternatif Yöntem

Alternatively, one can assign specific values to the angles that satisfy the given conditions. Since the sum of the angles in the large triangle must be 180 degrees and the angle at vertex P is 50 degrees, the sum of the other two angles must be 130 degrees. If we assume the triangle is isosceles with the two unknown angles being equal, each of those angles is 65 degrees. Their bisectors would each form an angle of 32.5 degrees with the base. In the smaller triangle, the sum of these two bisected angles is 65 degrees, which leaves 115 degrees for the angle at the intersection point.
Tahmini Süre:1m 15s
Soru 39Soru

In ABC\triangle ABC, the measure of exterior angle ACD\angle ACD is 135135^\circ, where DD lies on the extension of side BCBC past CC. If the measure of interior angle A\angle A is 2525^\circ greater than the measure of interior angle B\angle B, what is the measure, in degrees, of B\angle B?

Cevabı ve açıklamayı göster

Cevap: 55

Cevap

55
According to the Exterior Angle Theorem, the measure of exterior angle ACD\angle ACD is equal to the sum of the measures of its remote interior angles, A\angle A and B\angle B. This gives the equation mA+mB=135\text{m}\angle A + \text{m}\angle B = 135^\circ. Using the information that mA=mB+25\text{m}\angle A = \text{m}\angle B + 25^\circ, we substitute this expression into the equation to get (mB+25)+mB=135(\text{m}\angle B + 25^\circ) + \text{m}\angle B = 135^\circ. Simplifying this equation gives 2mB+25=1352\text{m}\angle B + 25 = 135, which simplifies to 2mB=1102\text{m}\angle B = 110, and dividing by 2 yields mB=55\text{m}\angle B = 55^\circ.

Adım Adım Çözüm

1
Apply the Exterior Angle Theorem to express the relation between the exterior angle and the two remote interior angles.
mA+mB=135\text{m}\angle A + \text{m}\angle B = 135^\circ
The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.
2
Substitute the relationship between the interior angles into the equation.
(mB+25)+mB=135(\text{m}\angle B + 25^\circ) + \text{m}\angle B = 135^\circ
The problem states that the measure of interior angle A\angle A is 2525^\circ greater than the measure of interior angle B\angle B.
3
Solve the algebraic equation for the measure of interior angle B\angle B.
mB=55\text{m}\angle B = 55^\circ
Combining like terms gives 2mB+25=1352\text{m}\angle B + 25 = 135. Subtracting 25 from both sides gives 2mB=1102\text{m}\angle B = 110. Dividing by 2 yields mB=55\text{m}\angle B = 55^\circ.

Anahtar Kavram

Exterior Angle Theorem and remote interior angles relation
Soru 40Soru

A triangle has two sides of length 99 and 1414. The length of the third side, ss, is a multiple of 44. What is the sum of all possible integer values for ss?

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Cevap: 56

Cevap

56
According to the Triangle Inequality Theorem, the length of the third side, ss, of a triangle with sides of length 9 and 14 must satisfy the inequality 149<s<14+914 - 9 < s < 14 + 9, which simplifies to 5<s<235 < s < 23. The integer values within this range that are multiples of 4 are 8, 12, 16, and 20. Adding these values together yields a sum of 56.

Adım Adım Çözüm

1
Apply the Triangle Inequality Theorem to determine the range of possible lengths for the third side.
The third side length, ss, must satisfy the inequality: 149<s<14+914 - 9 < s < 14 + 9, which simplifies to 5<s<235 < s < 23.
The Triangle Inequality Theorem states that the length of any side of a triangle must be strictly greater than the positive difference of the other two sides and strictly less than the sum of the other two sides.
2
Identify all integer values in the range (5,23)(5, 23) that are multiples of 4.
The multiples of 4 that are strictly greater than 5 and strictly less than 23 are: 8, 12, 16, and 20.
We must find the integers within the bounds that can be divided by 4 with a remainder of 0.
3
Calculate the sum of the identified multiples of 4.
8+12+16+20=568 + 12 + 16 + 20 = 56.
The question asks for the sum of all possible integer values for the third side length ss.

Anahtar Kavram

Triangle Inequality Theorem
Tahmini Süre:1m 0s
ÖncekiSayfa 2 / 3Sonraki