Acute / Obtuse Triangles - ISEE Upper Level Quantitative Reasoning

Card 0 of 496

Question

In isosceles triangle ABC, the measure of angle A is 50 degrees. Which is NOT a possible measure for angle B?

Answer

If angle A is one of the base angles, then the other base angle must measure 50 degrees. Since 50 + 50 + x = 180 means x = 80, the vertex angle must measure 80 degrees.

If angle A is the vertex angle, the two base angles must be equal. Since 50 + x + x = 180 means x = 65, the two base angles must measure 65 degrees.

The only number given that is not possible is 95 degrees.

Compare your answer with the correct one above

Question

Let the three angles of a triangle measure , , and .

Which of the following expressions is equal to ?

Answer

The sum of the measures of the angles of a triangle is $180^{\circ }$ , so simplify and solve for in the equation:

Compare your answer with the correct one above

Question

The angles of a triangle measure $\left ( x +30 \right )^{\circ }$ , $\left ( 2x - 3\right ) ^{\circ }$ , and $y ^{\circ }$ . Give in terms of .

Answer

The sum of the measures of three angles of a triangle is $180 ^{\circ }$ , so we can set up the equation:

$\left ( x +30 \right )+\left ( 2x - 3\right ) + y = 180$

We can simplify and solve for :

Compare your answer with the correct one above

Question

Which of the following is true about a triangle with two angles that measure $44^{\circ }$ each?

Answer

The measures of the angles of a triangle total $180^{\circ }$ , so if two angles measure $44^{\circ }$ and we call the measure of the third, then

$x= 92^{\circ } > 90^{\circ }$

This makes the triangle obtuse.

Also, since the triangle has two congruent angles (the $44^{\circ }$ angles), the triangle is also isosceles.

Compare your answer with the correct one above

Question

You are given two triangles, $\Delta ABC$ and $\Delta DEF$ .

, $\angle B$ is an acute angle, and $\angle E$ is a right angle.

Which quantity is greater?

(a)

(b)

Answer

We invoke the SAS Inequality Theorem, which states that, given two triangles $\Delta ABC$ and $\Delta DEF$ , with , $m\angle B < m \angle E$ ( the included angles), then - that is, the side opposite the greater angle has the greater length. Since $\angle B$ is an acute angle, and $\angle E$ is a right angle, we have just this situation. This makes (b) the greater.

Compare your answer with the correct one above

Question

Refer to the above figure. Which is the greater quantity?

(a)

(b)

Answer

The Triangle Exterior-Angle Theorem states that the measure of an exterior angle is equal to the sum of its remote interior angles. Therefore,

,

making the quantities equal.

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Refer to the above figure. Which is the greater quantity?

(a)

(b)

Answer

(a) The measures of the angles of a linear pair total 180, so:

(b) The Triangle Exterior-Angle Theorem states that the measure of an exterior angle is equal to the sum of its remote interior angles. Therefore, .

Therefore (a) is the greater quantity.

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Refer to the above figure. Which is the greater quantity?

(a)

(b)

Answer

The two angles at bottom are marked as congruent. Each of these two angles forms a linear pair with a $140 ^{\circ }$ angle, so it is supplementary to that angle, making its measure $(180-140)^{\circ } = 40^{\circ }$ . Therefore, the other marked angle also measures $40^{\circ }$ .

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ , so:

The quantities are equal.

Compare your answer with the correct one above

Question

$\Delta ABC$ is equilateral; $\Delta CBD$ is isosceles

Which is the greater quantity?

(a) $m \angle BCD$

(b) $60 ^{\circ }$

Answer

$\Delta ABC$ is equilateral, so

.

In $\Delta CBD$ , we are given that

.

Since the triangles have two pair of congruent sides, the third side with the greater length is opposite the angle of greater measure. Therefore,

$m \angle BCD >m \angle BAC$ .

Since $\angle BAC$ is an angle of an equilateral triangle, its measure is $60^{\circ }$ , so $m \angle BCD >60^{\circ }$ .

Compare your answer with the correct one above

Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$

$m \angle A = 60^{\circ }$

$\angle E \cong \angle F$

Which is the greater quantity?

(a) $m \angle C$

(b) $m \angle D$

Answer

Corresponding angles of similar triangles are congruent, so, since $\bigtriangleup ABC \sim \bigtriangleup DEF$ , it follows that

$m \angle D = m \angle A = 60 ^{\circ }$

By similarity, $m \angle B = m \angle E$ and $m \angle C = m \angle F$ , and we are given that $m \angle E = m \angle F$ , so

$m \angle B = m \angle C$

Also,

$m \angle A + m \angle B + m \angle C = 180^{\circ }$

$60^{\circ }+ m \angle C + m \angle C = 180^{\circ }$

$60^{\circ }+2 \cdot m \angle C = 180^{\circ }$

$2 \cdot m \angle C = 120^{\circ }$

$m \angle C = 60^{\circ }$ ,

and $m \angle C =m \angle D$ .

Compare your answer with the correct one above

Question

Refer to the above figure. Which is the greater quantity?

(a)

(b)

Answer

Extend $\overline{AB}$ as seen in the figure below:

The measure of an exterior angle of a triangle is equal to the sum of the measures of its remote interior angles; specifically,

$m \angle CAD = m \angle B + m \angle C$ ,

and

$m \angle CAD =( x+ y) ^{\circ }$

However, $m \angle CAD > z ^{\circ }$ , so, by substitution,

Compare your answer with the correct one above

Question

Given: $\bigtriangleup ABC$ . . Which is the greater quantity?

(a) $m \angle B$

(b) $60^{\circ }$

Answer

Below is the referenced triangle along with $\bigtriangleup DEF$ , an equilateral triangle with sides of length 10:

As an angle of an equilateral triangle, $\angle D$ has measure $60 ^{\circ }$ . Applying the Side-Side-Side Inequality Theorem, since , , and , it follows that $m \angle A > m \angle D$ , so $m \angle A > 60^{\circ }$ .

Also, since , by the Isosceles Triangle Theorem, $m \angle B = m \angle C$ . Since $m \angle A > 60^{\circ }$ , and the sum of the measures of the angles of a triangle is $180 ^{\circ }$ , it follows that

$m \angle B + m \angle C < 120^{\circ }$

Substituting and solving:

$m \angle B + m \angle B < 120^{\circ }$

$2 m \angle B < 120^{\circ }$

$2 m \angle B \div 2 < 120^{\circ } \div 2$

$m \angle B < 60^{\circ }$ .

Compare your answer with the correct one above

Question

Consider $\Delta ABC$ and $\Delta DEF$ with .

Which is the greater quantity?

(a) $m \angle A$

(b) $m \angle D$

Answer

, so, by the Side-Side-Side Principle, since there are three pairs of congruent corresponding sides between the triangles, we can say they are congruent - that is,

$\Delta ABC \cong \Delta DEF$ .

Corresponding angles of congruent sides are congruent, so $m \angle A = m \angle D$ .

Compare your answer with the correct one above

Question

Given $\Delta ABC$ and $\Delta DEF$ with $AC = DF, AB = DE, m \angle B = m \angle E$

Which is the greater quantity?

(a)

(b)

Answer

Examine the diagram below, in which two triangles matching the given descriptions have been superimposed.

Note that and $m \angle B = m \angle E$ . The two question marks need to be replaced by and . No matter how you place these two points, . However, with one replacement, ; with the other replacement, . Therefore, the information given is insufficient to answer the question.

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

Which is the greater quantity?

(a)

(b)

Answer

$\Delta ABC \sim \Delta DEF$ , so by definition, the sides are in proportion. Therefore,

$\frac{DE}{AB} = \frac{EF}{BC}$ .

Substitute:

$\frac{DE}{18} = \frac{EF}{17}$

$\frac{DE}{18} \cdot 18 = \frac{EF}{17}\cdot 18$

$DE = \frac{18}{17}\cdot EF$

, so (a) is greater.

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

Which is the greater quantity?

(a)

(b)

Answer

$\Delta ABC \sim \Delta DEF$ , so by definition, the sides are in proportion.

(a) $\frac{BC}{EF} = \frac{AC} {DF}$

Substitute and solve for :

$\frac{BC}{27} = \frac{16} {24}$

$\frac{BC}{27} \cdot 27 = \frac{16} {24} \cdot 27$

(b) $\frac{DE}{AB} = \frac{DF}{AC}$

Substitute and solve for :

$\frac{DE}{12} = \frac{24}{16}$

$\frac{DE}{12} \cdot 12= \frac{24}{16}\cdot 12$

The two are equal.

Compare your answer with the correct one above

Question

Triangle B has a height that is twice that of Triangle A and a base that is one-half that of Triangle A. Which is the greater quantity?

(a) The area of Triangle A

(b) The area of Triangle B

Answer

Let and be the base and height of Triangle A. Then the base and height of Triangle B are $\frac{1}{2} b$ and , respectively.

(a) The area of Triangle A is $A = \frac{1}{2} bh$ .

(b) The area of Triangle B is $A = \frac{1}{2}\cdot \frac{1}{2} b \cdot 2h = \frac{1}{2} bh$ .

Therefore, (a) and (b) are equal.

Compare your answer with the correct one above

Question

Two triangles on the coordinate plane have a vertex at the origin and a vertex at , where .

Triangle A has its third vertex at .

Triangle B has its third vertex at .

Which is the greater quantity?

(a) The area of Triangle A

(b) The area of Triangle B

Answer

(a) Triangle A has as its base the horizontal segment connecting and , the length of which is 10. Its (vertical) altitude is the segment from to this horizontal segment, which is part of the -axis; its height is therefore the -coordinate of this point, or .

The area of Triangle A is therefore $A = \frac{1}{2} \cdot 10 \cdot d = 5d$

(b) Triangle B has as its base the vertical segment connecting and , the length of which is 10. Its (horizontal) altitude is the segment from to this vertical segment, which is part of the -axis; its height is therefore the -coordinate of this point, or .

The area of Triangle B is therefore $A = \frac{1}{2} \cdot 10 \cdot c = 5c$

, so . (b), the area of Triangle B, is greater.

Compare your answer with the correct one above

Question

A triangle has sides 30, 40, and 80. Give its area.

Answer

By the Triangle Inequality Theorem, the sum of the lengths of the two shorter sides of a triangle must exceed the length of its longest side. However,

;

Therefore, this triangle cannot exist, and the correct answer is "none of the other responses is correct".

Compare your answer with the correct one above

Question

The above depicts Square ; , and are the midpoints of $\overline{AD}$ , $\overline{BC}$ , and $\overline{AB}$ , respectively. Which is the greater quantity?

(a) The area of $\bigtriangleup BZY$

(b) The area of $\bigtriangleup ZXD$

Answer

For the sake of simplicity, assume that the square has sidelength 2; this reasoning is independent of the actual sidelength.

Since , , and are the midpoints of their respective sides, , as shown in this diagram.

The area of $\bigtriangleup BZY$ , it being a right triangle, is half the product of the lengths of its legs:

$\frac{1}{2} \cdot BZ \cdot BY = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$

The area of $\bigtriangleup ZXD$ is half the product of the length of a base and the height. Using $\overline{XD}$ as the base, and $\overline{AZ}$ as an altitude:

$\frac{1}{2} \cdot XD \cdot AZ= \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}$

The two triangles have the same area.

Compare your answer with the correct one above

Tap the card to reveal the answer