Plane Geometry

218 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

A square tabletop has a diagonal length of 88 feet. What is the area of the tabletop, in square feet?

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Cevap: 32

Cevap

32
The correct answer is 32. The diagonal of a square divides it into two congruent 45-45-90 right triangles, where the diagonal is the hypotenuse. The ratio of the diagonal to the side length in a 45-45-90 triangle is 2\sqrt{2} to 11. Therefore, a square with a diagonal of 88 feet has a side length of s=82s = \frac{8}{\sqrt{2}} feet. The area of the square is s2=(82)2=642=32s^2 = \left(\frac{8}{\sqrt{2}}\right)^2 = \frac{64}{2} = 32 square feet.

Adım Adım Çözüm

1
Relate the diagonal of a square to its side length using special right triangles.
The diagonal of a square splits the square into two 45-45-90 right triangles. The hypotenuse of these triangles is the diagonal, 88 feet, and the legs are the sides of the square, ss. The relationship is s2=8s\sqrt{2} = 8.
In a 45-45-90 triangle, the hypotenuse is 2\sqrt{2} times the length of a leg.
2
Solve for the side length ss of the square tabletop.
s=82s = \frac{8}{\sqrt{2}} feet.
Divide both sides of the equation by 2\sqrt{2} to isolate the side length ss.
3
Calculate the area of the square tabletop.
Area =s2=(82)2=642=32= s^2 = \left(\frac{8}{\sqrt{2}}\right)^2 = \frac{64}{2} = 32 square feet.
The area of a square is calculated by squaring its side length.

Anahtar Kavram

Using the properties of 45-45-90 special right triangles to find side lengths and area from the diagonal of a square.
Soru 28Soru

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 29Soru

A right triangle has two legs of equal length. If the hypotenuse of the triangle is 10210\sqrt{2} centimeters, what is the length, in centimeters, of one of the legs?

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Cevap: 10

Cevap

The length of one of the legs is 10 centimeters.
An isosceles right triangle possesses acute angles of 4545^\circ and side ratios of x:x:x2x : x : x\sqrt{2}, where xx represents the leg length. Given a hypotenuse of 10210\sqrt{2} centimeters, we equate x2=102x\sqrt{2} = 10\sqrt{2}. Dividing both sides of the equation by 2\sqrt{2} isolates the leg length, giving x=10x = 10 centimeters.

Adım Adım Çözüm

1
Determine the triangle type from the given properties.
The triangle is a 4545^\circ-4545^\circ-9090^\circ special right triangle (isosceles right triangle).
A right triangle with two legs of equal length must have acute angles measuring 4545^\circ each, making it an isosceles right triangle.
2
Set up an equation utilizing the ratios of the side lengths.
Let xx be the leg length. The hypotenuse length is represented by x2=102x\sqrt{2} = 10\sqrt{2} centimeters.
The hypotenuse of a 4545^\circ-4545^\circ-9090^\circ special right triangle is always 2\sqrt{2} times the length of one of its legs.
3
Solve the equation for the variable xx.
x=10x = 10
Dividing both sides of the equation by 2\sqrt{2} isolates the variable xx representing the leg length.

Anahtar Kavram

Properties of 4545^\circ-4545^\circ-9090^\circ special right triangles.

Alternatif Yöntem

Alternatively, you can apply the Pythagorean Theorem: a2+b2=c2a^2 + b^2 = c^2. Since both legs are equal in length, we can set a=b=xa = b = x. This yields the equation x2+x2=(102)2x^2 + x^2 = (10\sqrt{2})^2. Simplifying both sides gives 2x2=100×2=2002x^2 = 100 \times 2 = 200. Dividing by 2 yields x2=100x^2 = 100, and taking the square root of both sides gives x=10x = 10 centimeters.
Tahmini Süre:45s
Soru 30Soru

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 31Soru

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 32Soru

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?

Cevabı ve açıklamayı göster

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 33Soru

A rectangular garden has a straight walking path that connects two opposite corners. The length of the path is 170170 meters, and the width of the garden is 8080 meters. What is the length, in meters, of the garden?

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Cevap: 150

Cevap

The length of the garden is 150150 meters.
The diagonal path, width, and length of the rectangular garden form a right triangle where the path is the hypotenuse. According to the Pythagorean theorem, the square of the length plus the square of the width equals the square of the path: length2+802=1702\text{length}^2 + 80^2 = 170^2. This simplifies to length2+6,400=28,900\text{length}^2 + 6,400 = 28,900. Subtracting 6,4006,400 from both sides gives length2=22,500\text{length}^2 = 22,500. Taking the square root of 22,50022,500 yields 150150 meters.

Adım Adım Çözüm

1
Identify the right triangle formed by the length, width, and diagonal path of the garden.
The width (8080 meters) and the unknown length are the legs, while the diagonal path (170170 meters) is the hypotenuse.
The diagonal of a rectangle forms two congruent right triangles with the rectangle's sides.
2
Set up the Pythagorean equation to solve for the unknown leg.
length2+802=1702\text{length}^2 + 80^2 = 170^2
The Pythagorean theorem states that a2+b2=c2a^2 + b^2 = c^2, where cc is the hypotenuse.
3
Calculate the squares of the known lengths.
802=6,40080^2 = 6,400 and 1702=28,900170^2 = 28,900
Evaluate the exponents to simplify the equation.
4
Isolate the squared unknown variable.
length2=28,9006,400=22,500\text{length}^2 = 28,900 - 6,400 = 22,500
Subtract 6,4006,400 from both sides of the equation.
5
Take the square root of both sides to find the length.
length=22,500=150\text{length} = \sqrt{22,500} = 150
The square root operation reverses the squaring of the variable.

Anahtar Kavram

Applying the Pythagorean theorem to find the length of an unknown leg in a right triangle.
Tahmini Süre:1m 0s
Soru 34Soru

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?

Cevabı ve açıklamayı göster

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 35Soru

In a right triangle, the length of the side opposite the 6060^\circ angle is 636\sqrt{3} centimeters. What is the length, in centimeters, of the hypotenuse of this triangle?

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Cevap: 12

Cevap

The length of the hypotenuse is 1212 centimeters.
In a 30609030^\circ-60^\circ-90^\circ special right triangle, the sides opposite the 3030^\circ angle, the 6060^\circ angle, and the 9090^\circ (hypotenuse) angle are in the ratio x:x3:2xx : x\sqrt{3} : 2x. Given that the side opposite the 6060^\circ angle is 636\sqrt{3} centimeters, we have x3=63x\sqrt{3} = 6\sqrt{3}, which means x=6x = 6. The hypotenuse is 2x=2(6)=122x = 2(6) = 12 centimeters.

Adım Adım Çözüm

1
Determine the type of special right triangle.
A 30609030^\circ-60^\circ-90^\circ right triangle.
Since the triangle is a right triangle and has a 6060^\circ angle, the remaining angle must be 1809060=30180^\circ - 90^\circ - 60^\circ = 30^\circ.
2
Set up the relation for the side lengths using the ratio of a 30609030^\circ-60^\circ-90^\circ triangle.
The side opposite the 6060^\circ angle is x3x\sqrt{3} centimeters, where xx is the length of the side opposite the 3030^\circ angle.
In any 30609030^\circ-60^\circ-90^\circ triangle, the side lengths are in the ratio 1:3:21 : \sqrt{3} : 2.
3
Solve for the base variable xx.
x=6x = 6
We are given that the side opposite the 6060^\circ angle is 636\sqrt{3} centimeters, so x3=63x\sqrt{3} = 6\sqrt{3}.
4
Calculate the length of the hypotenuse.
The hypotenuse is 2x=2(6)=122x = 2(6) = 12 centimeters.
The hypotenuse of a 30609030^\circ-60^\circ-90^\circ triangle is twice the length of the shorter leg, which is 2x2x.

Anahtar Kavram

Using the side length ratios of a 30609030^\circ-60^\circ-90^\circ special right triangle to find missing lengths.
Soru 36Soru

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 37Soru

In right triangle ABCABC, the measure of angle BB is 9090^\circ and the measure of angle AA is 4545^\circ. If the length of leg ABAB is 88 inches, what is the length, in inches, of the hypotenuse ACAC?

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Cevap: 828\sqrt{2}

Cevap

The length of the hypotenuse is 828\sqrt{2} inches.
In right triangle ABCABC, the angle measures are 4545^\circ, 4545^\circ, and 9090^\circ. The lengths of the sides of a 4545^\circ-4545^\circ-9090^\circ triangle are in the ratio 1:1:21 : 1 : \sqrt{2}. Since the leg is 88 inches, the hypotenuse is 828\sqrt{2} inches.

Adım Adım Çözüm

1
Identify the type of right triangle.
Since angle B=90B = 90^\circ and angle A=45A = 45^\circ, angle CC must also be 4545^\circ. This is a 4545^\circ-4545^\circ-9090^\circ special right triangle.
The sum of the angles in a triangle is always 180180^\circ.
2
Recall the ratio of the side lengths of a 4545^\circ-4545^\circ-9090^\circ triangle.
The ratio of the sides opposite the angles 45:45:9045^\circ : 45^\circ : 90^\circ is 1:1:21 : 1 : \sqrt{2}. Thus, the hypotenuse is equal to leg×2\text{leg} \times \sqrt{2}.
This is a standard geometric property of isosceles right triangles.
3
Calculate the length of the hypotenuse.
Multiply the leg length of 88 inches by 2\sqrt{2} to get 828\sqrt{2} inches.
The leg adjacent to the 4545^\circ angle is given as 88 inches.

Anahtar Kavram

Hypotenuse of a 4545^\circ-4545^\circ-9090^\circ special right triangle

Alternatif Yöntem

Alternatively, use the Pythagorean theorem: AB2+BC2=AC2AB^2 + BC^2 = AC^2. Since it is an isosceles right triangle, BC=AB=8BC = AB = 8. Thus, 82+82=AC2    64+64=AC2    AC=128=828^2 + 8^2 = AC^2 \implies 64 + 64 = AC^2 \implies AC = \sqrt{128} = 8\sqrt{2}.
Tahmini Süre:45s
Soru 38Soru

A regular hexagon ABCDEFABCDEF has a side length of 88 centimeters. A point PP lies on the side CDCD such that the ratio of the length of CPCP to the length of PDPD is 1:31:3. What is the length, in centimeters, of the segment APAP?

Cevabı ve açıklamayı göster

Cevap: 14

Cevap

The length of the segment APAP is 1414 centimeters.
The correct answer is 1414. Dropping a perpendicular from PP to the main diagonal ADAD creates a 30609030^\circ-60^\circ-90^\circ triangle PHD\triangle PHD with hypotenuse PD=6PD = 6. The legs are DH=3DH = 3 and PH=33PH = 3\sqrt{3}. This leaves AH=13AH = 13. Applying the Pythagorean Theorem to the right triangle AHP\triangle AHP with legs 1313 and 333\sqrt{3} yields AP=132+(33)2=14AP = \sqrt{13^2 + (3\sqrt{3})^2} = 14.

Adım Adım Çözüm

1
Determine the length of the main diagonal ADAD of the regular hexagon.
AD=16AD = 16 cm
In a regular hexagon with side length ss, the main diagonal connecting opposite vertices has a length of 2s2s. Given s=8s = 8, we find AD=2×8=16AD = 2 \times 8 = 16.
2
Calculate the length of the segment PDPD on the side CDCD.
PD=6PD = 6 cm
The point PP divides the side CDCD of length 88 in the ratio CP:PD=1:3CP:PD = 1:3. Thus, PD=31+3×8=6PD = \frac{3}{1+3} \times 8 = 6.
3
Identify the angles and type of triangle formed by dropping a perpendicular from PP to diagonal ADAD.
PHD\triangle PHD is a 30609030^\circ-60^\circ-90^\circ right triangle.
The diagonal ADAD bisects the interior angle CDE=120\angle CDE = 120^\circ of the regular hexagon, making ADC=60\angle ADC = 60^\circ. Since PHADPH \perp AD, the triangle PHD\triangle PHD has angles 9090^\circ, 6060^\circ, and 3030^\circ.
4
Find the lengths of the legs DHDH and PHPH of the special right triangle PHD\triangle PHD.
DH=3DH = 3 cm and PH=33PH = 3\sqrt{3} cm
Using the ratios of a 30609030^\circ-60^\circ-90^\circ triangle with hypotenuse PD=6PD = 6, the leg adjacent to the 6060^\circ angle is DH=6cos(60)=3DH = 6 \cos(60^\circ) = 3, and the leg opposite to the 6060^\circ angle is PH=6sin(60)=33PH = 6 \sin(60^\circ) = 3\sqrt{3}.
5
Calculate the length of the segment AHAH.
AH=13AH = 13 cm
Since HH lies on the diagonal ADAD, we subtract the length of DHDH from the total length of the diagonal: AH=ADDH=163=13AH = AD - DH = 16 - 3 = 13.
6
Apply the Pythagorean Theorem to the right triangle AHP\triangle AHP to find the length of APAP.
AP=14AP = 14 cm
In the right triangle AHP\triangle AHP with legs AH=13AH = 13 and PH=33PH = 3\sqrt{3}, the hypotenuse is AP=AH2+PH2=132+(33)2=169+27=196=14AP = \sqrt{AH^2 + PH^2} = \sqrt{13^2 + (3\sqrt{3})^2} = \sqrt{169 + 27} = \sqrt{196} = 14.

Anahtar Kavram

Applying special right triangle ratios and the Pythagorean Theorem in multi-step geometric figures.
Soru 39Soru

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?

Cevabı ve açıklamayı göster

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 40Soru

In right triangle ABCABC, the hypotenuse ACAC has a length of 1515 centimeters, and leg ABAB has a length of 99 centimeters. What is the length, in centimeters, of leg BCBC?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The length of leg BCBC is 1212 centimeters.
The length of leg BCBC is found using the Pythagorean Theorem, AB2+BC2=AC2AB^2 + BC^2 = AC^2. Substituting the given values gives 92+BC2=1529^2 + BC^2 = 15^2, which simplifies to 81+BC2=22581 + BC^2 = 225. Subtracting 81 from both sides yields BC2=144BC^2 = 144. Taking the square root of 144 gives the correct length of 12 centimeters.

Adım Adım Çözüm

1
Identify the given dimensions and apply the Pythagorean Theorem.
AB2+BC2=AC2AB^2 + BC^2 = AC^2
For any right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
2
Substitute the known values AB=9AB = 9 and AC=15AC = 15 into the equation.
92+BC2=1529^2 + BC^2 = 15^2
The hypotenuse ACAC is the side opposite the right angle, and ABAB is one of the legs.
3
Simplify the squared terms.
81+BC2=22581 + BC^2 = 225
Squaring 9 yields 81, and squaring 15 yields 225.
4
Isolate the unknown term by subtracting 81 from both sides.
BC2=144BC^2 = 144
Subtracting 81 from both sides isolates BC2BC^2 on the left side of the equation.
5
Take the square root of both sides to solve for the leg length.
BC=12BC = 12
Taking the square root of 144 gives the side length, which must be positive.

Anahtar Kavram

Pythagorean Theorem
ÖncekiSayfa 2 / 11Sonraki
Plane Geometry Alıştırma Soruları — ACT — Sayfa 2 | Examkin