AMC 8 Daily Practice - Triangle Properties

Complete problem set with solutions and individual problem pages

Problem 1 Easy

As shown in the figure, \angle A = 30^\circ, \angle B = 60^\circ, and \angle C = 20^\circ. What is the value of x?

  • A.

    99^\circ

  • B.

    105^\circ

  • C.

    110^\circ

  • D.

    120^\circ

  • E.

    130^\circ

Answer:C

Connect BD.

In \triangle ABD, \angle ADB = 180^\circ - 30^\circ - \angle ABD = 150^\circ - \angle ABD.

In \triangle BCD, \angle BDC = 180^\circ - 20^\circ - \angle CBD = 160^\circ - \angle CBD.

Then, \angle ADC = 360^\circ - \angle ADB - \angle BDC

= 360^\circ - (150^\circ - \angle ABD) - (160^\circ - \angle CBD)

= 360^\circ - 150^\circ + \angle ABD - 160^\circ + \angle CBD

= 110^\circ

Thus, x = 110^\circ.

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Problem 2 Easy

As shown in the figure, in triangle ABC, \angle ACB = 90^\circ, BC = 1, point D is the midpoint of side AB, and CD bisects the perimeter of triangle ABC. What is the length of BD?

  • A.

    \frac{\sqrt{2}}{4}

  • B.

    \frac{\sqrt{2}}{2}

  • C.

    1

  • D.

    \sqrt{2}

  • E.

    2

Answer:B

Since CD bisects the perimeter of triangle ABC, the perimeter of triangle ACD is equal to the perimeter of triangle BCD.

That is: AC + AD + CD = BC + BD + CD.

Simplifying, we get: AC + AD = BC + BD.

Since D is the midpoint of AB, AD = BD.

Substituting this into the equation above, we have: AC = BC

Given BC = 1, it follows that AC = 1.

In the right triangle ABC with \angle ACB = 90^\circ, by the Pythagorean theorem: AB = \sqrt{AC^2 + BC^2} = \sqrt{1^2 + 1^2} = \sqrt{2}.

Since D is the midpoint of AB, the length of BD is: BD = \frac{AB}{2} = \frac{\sqrt{2}}{2}.

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Problem 3 Easy

In a triangle, the largest angle is 5 times the smallest angle, and the other angle is 3 times the smallest angle. What is the degree measure of the largest angle?

  • A.

    90

  • B.

    92.5

  • C.

    100

  • D.

    102.5

  • E.

    112.5

Answer:C

Let the three angles of the triangle be \angle a, \angle b, and \angle c, where \angle a < \angle b < \angle c.

According to the problem, we know that \angle c = 5\angle a and \angle b = 3\angle a.

Thus, the ratio of the angles is \angle a : \angle b : \angle c = 1 : 3 : 5.

Since the sum of the interior angles of a triangle is 180^\circ, the measure of the largest angle \angle c is: \angle c = 180^\circ \div (1 + 3 + 5) \times 5 = 100^\circ.

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Problem 4 Easy

In triangle ABC, \angle B = 65^\circ. If \angle B is cut off along the dashed line, what is the sum of \angle ADE and \angle CED?

  • A.

    220^\circ

  • B.

    235^\circ

  • C.

    245^\circ

  • D.

    255^\circ

  • E.

    260^\circ

Answer:C

We know that: \angle ADE = \angle BDE + \angle B, \angle CED = \angle DEB + \angle B.

Then, the sum of \angle ADE and \angle CED is: \angle ADE + \angle CED = (\angle BDE + \angle B) + (\angle DEB + \angle B) = \angle BDE + \angle DEB + \angle B + \angle B.

In \triangle BDE, \angle BDE + \angle DEB + \angle B = 180^\circ.

Substituting this in, we get: \angle ADE + \angle CED = 180^\circ + \angle B

Since \angle B = 65^\circ: \angle ADE + \angle CED = 180^\circ + 65^\circ = 245^\circ.

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Problem 5 Easy

Two equilateral triangles ABC and DCE lie on the same horizontal plane, with areas 4 and 9 respectively. What is the area of triangle ACD?

  • A.

    5

  • B.

    6

  • C.

    6.5

  • D.

    7

  • E.

    8

Answer:B

Since triangles ABC and DCE are equilateral, AB \parallel DC and AC \parallel DE.

Given that the area of a triangle is \frac{1}{2} \times \text{base} \times \text{height}, we have:

\text{Area of } \triangle ABC : \text{Area of } \triangle ACD=AB : DC

\text{Area of } \triangle ACD : \text{Area of } \triangle CDE=BC : DE

Since AB = BC and DC = DE , it follows that: \text{Area of } \triangle ABC : \text{Area of } \triangle ACD = \text{Area of } \triangle ACD : \text{Area of } \triangle CDE.

Let the area of \triangle ACD be A.

Then: A^2 = \text{Area of } \triangle ABC \times \text{Area of } \triangle CDE = 4 \times 9.

Solving for A: A = \sqrt{4 \times 9} = 6.

Thus, the area of \triangle ACD is 6.

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Problem 6 Easy

The perimeter of a triangle is 60cm, and its shortest side is 13cm. If all three sides of the triangle are integers, what is the maximum possible length of the longest side in centimeters?

  • A.

    24

  • B.

    25

  • C.

    28

  • D.

    29

  • E.

    30

Answer:D

Assume the three sides of the triangle, arranged in ascending order, are 13, a, and b.

According to the triangle inequality theorem, the sum of any two sides of a triangle must be greater than the third side, so: 13 + a > b.

Since the perimeter of the triangle is 60 cm, we have: 13 + a + b = 60, which simplifies to: a = 47 - b.

Substituting a = 47 - b into the inequality 13 + a > b, we get: 13 + (47 - b) > b, 60 - b > b, 60 > 2b, b < 30.

Since b is an integer, the maximum possible value of b is 29.

Thus, the longest side can be at most 29 centimeters.

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Problem 7 Easy

In rectangle ABCD, AB = 18cm and BC = 12cm. The two line segments AE and AF divide the area of the rectangle into three equal parts. What is the area of triangle AEF?

  • A.

    50cm^{2}

  • B.

    54cm^{2}

  • C.

    60cm^{2}

  • D.

    66cm^{2}

  • E.

    72cm^{2}

Answer:C

The area of rectangle ABCD is AB \times BC.

The area of \triangle ABE is \frac{1}{2} \times AB \times BE = \frac{1}{3} \times AB \times BC, so BE = \frac{2}{3}BC.

The area of \triangle ADF is \frac{1}{2} \times AD \times DF = \frac{1}{3} \times AB \times BC, so DF = \frac{2}{3}AB.

Connect EF.

​

 

The area of \triangle CEF is \frac{1}{2} \times CF \times CE = \frac{1}{2} \times \left(\frac{1}{3}AB\right) \times \left(\frac{1}{3}BC\right) = \frac{1}{18} \times \text{Area of rectangle } ABCD.

Therefore, the area of \triangle AEF is: \left(\frac{1}{3} - \frac{1}{18}\right) \times \text{Area of rectangle } ABCD = \frac{5}{18} \times AB \times BC

Substituting the values AB = 18cm and BC = 12cm: \text{Area of } \triangle AEF = \frac{5}{18} \times 18 \times 12 = 60cm^{2}.

Thus, the area of \triangle AEF is 60cm^{2}.

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Problem 8 Easy

What is the largest whole number smaller than the perimeter of any triangle with a side of length 4 and a side of length 91?

  • A.

    182

  • B.

    184

  • C.

    186

  • D.

    188

  • E.

    190

Answer:A

From the Triangle Inequality Theorem, we know that for the third side s of the triangle, the following must hold: s < 4 + 91 = 95 and s > 91 - 4 = 87.

Let P be the perimeter of the triangle, so P = 4 + 91 + s.

Adding 4 + 91 to all parts of the inequality for s, we get: 87 + 4 + 91 < P < 95 + 4 + 91

182 < P < 190

Thus, the largest whole number smaller than the perimeter of any such triangle is 182.

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Related Paper
AMC 8 Daily Practice - Circles AMC 8 Daily Practice - Solid Figures AMC 8 Daily Practice - Area Tricks AMC 8 Daily Practice - Perimeter Tricks
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