Quadrilaterals - ISEE Upper Level Quantitative Reasoning

Card 0 of 1040

Question

$\Delta ABC$ is an equilateral triangle. Points are the midpoints of $\overline{AB},\overline{BC},\overline{AC}$ , respectively. $\Delta DEF$ is constructed.

Which is the greater quantity?

(a) The perimeter of $\Delta ABC$

(b) Twice the perimeter of $\Delta DEF$

Answer

If segments are constructed in which the endpoints form the midpoints of the sides of a triangle, then each of the sides of the smaller triangle is half as long as the side of the larger triangle that it does not touch. Therefore:

$AB = 2 \cdot EF$

$BC = 2 \cdot DF$

$AC = 2 \cdot DE$

The perimeter of $\Delta ABC$ is:

$P = 2 \cdot EF+ 2 \cdot DF + 2 \cdot DE$

,

which is twice the perimeter of $\Delta DEF$ .

Note that the fact that the triangle is equilateral is irrelevant.

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Question

Column A Column B

The perimeter The perimeter

of a square with of an equilateral

sides of 4 cm. triangle with a side

of 9 cm.

Answer

Perimeter involves adding up all of the sides of the shape. Therefore, the square's perimeter is or 16. An equialteral shape means that all of the sides are equal. Therefore, the perimeter of the triangle is or 27. Therefore, Column B is greater.

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure.

Which is the greater quantity?

(a)

(b) $30 ^{\circ }$

Answer

Since the shorter leg of the right triangle is half the hypotenuse, the triangle is a $30 ^{\circ } -60^{\circ } -90^{\circ }$ triangle, with the $30 ^{\circ }$ angle opposite the shorter leg. That makes .

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Question

Right triangle $\Delta ABC$ has right angle $\angle B$ .

$m\angle A = \left ( x + 30 \right )^ {\circ} ,m\angle C = \left ( 2x- 57\right ) ^ {\circ}$

Which is the greater quantity?

(a) $m\angle A$

(b) $m\angle C$

Answer

The degree measures of the acute angles of a right triangle total 90, so we solve for in the following equation:

$\left ( x + 30 \right )+ \left ( 2x- 57\right ) = 90$

$3x \div 3 = 117 \div 3$

(a) $m\angle A = \left ( x + 30 \right )^ {\circ} = \left ( 39 + 30 \right )^ {\circ} = 69^ {\circ}$

(b) $m\angle C =\left ( 90 - 69 \right ) ^ {\circ} = 21^ {\circ}$

$m\angle A > m\angle C$

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Question

$\bigtriangleup ABC \sim \bigtriangleup DEF$

$\angle A$ is a right angle.

Which is the greater quantity?

(a) $m \angle D$

(b) $m \angle E$

Answer

Corresponding angles of similar triangles are congruent, so, since $\bigtriangleup ABC \sim \bigtriangleup DEF$ , and $\angle A$ is right, it follows that

$m \angle D = m \angle A = 90 ^{\circ }$

$\angle D$ is a right angle of a right triangle $\bigtriangleup DEF$ . The other two angles must be acute - that is, with measure less than $90 ^{\circ }$ - so $m \angle D > m \angle E$ .

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Question

$\bigtriangleup ABC$ is inscribed in a circle. $\angle B$ is a right angle, and $m \overarc{ACB} = 270 ^{\circ }$ .

Which is the greater quantity?

(a) $m \angle A$

(b) $m \angle C$

Answer

The figure referenced is below:

$\overarc{ACB}$ has measure $270 ^{\circ }$ , so its corresponding minor arc, $\overarc{A B}$ , has measure $360 ^{\circ } - 270 ^{\circ } = 90^{\circ }$ . The inscribed angle that intercepts this arc, which is $\angle C$ , has measure half this, or $45 ^{\circ }$ . Since $\angle B$ is a right angle, the other acute angle, $\angle A$ , has measure

$m \angle A = 90 ^{\circ } - m \angle C = 90 ^{\circ } - 45^{\circ } = 45 ^{\circ }$

Therefore, $m \angle A = 45 ^{\circ } = m \angle C$ .

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Question

Consider a triangle, $\bigtriangleup ABC$ , in which , , and . Which is the greater number?

(a) The measure of $\angle B$ in degrees

(b)

Answer

By the Converse of the Pythagorean Theorem, a triangle is right if and only if the sum of the squares of the lengths of the smallest two sides is equal to the square of the longest side. Compare the quantities $(AB)^{2}+ ( BC )^{2}$ and $(AC)^{2}$

$(AB)^{2}+ ( BC )^{2} = 36 ^{2} + 48^{2} = 1,296 + 2,304 = 3,600$

$(AC)^{2} = 60^{2} = 3,600$

$(AB)^{2}+ ( BC )^{2} =(AC)^{2}$ , so $\bigtriangleup ABC$ is right, with the right angle opposite longest side $\overline{AC}$ . Thus, $\angle B$ is right and has degree measure 90.

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Question

$\Delta ABC$ and $\Delta DEF$ are right triangles, with right angles $\angle B , \angle E$ , respectively.

Which is the greater quantity?

(a) The perimeter of $\Delta ABC$

(b) The perimeter of $\Delta DEF$

Answer

No information is given about the legs of either triangle; therefore, no information about their perimeters can be deduced.

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Question

Note: Figure NOT drawn to scale

Refer to the above triangle. Starting at point A, an insect walks clockwise along the sides of the triangle until he has walked 75% of the length of $\overline{CB}$ . What percent of the perimeter of the triangle has the insect walked?

Answer

By the Pythagorean Theorem, the distance from B to C, which we will call , is equal to

$D = \sqrt{13^{2}- 5^{2}} = \sqrt{169- 25 }= \sqrt{144} = 12$ .

The perimeter of the triangle is

.

The insect traveled the entirety of the hypotenuse, which is 13 units long, and 75% of the longer leg, which adds 75% of 12, or $0.75 \times 12 = 9$ units. Therefore, the insect has traveled 22 out of the 30 units perimeter, or

$\frac{22}{30} \times 100 = 73 \frac{1}{3} %$ of the perimeter.

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Question

Refer to the above diagram, in which $\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . Which is the greater quantity?

(a) Four times the perimeter of $\bigtriangleup AXB$

(b) Three times the perimeter of $\bigtriangleup CXB$

Answer

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other. The ratio of the hypotenuse of $\bigtriangleup CXB$ to that of $\bigtriangleup AXB$ (which are corresponding sides) is

$\frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}$ ,

making this the similarity ratio. The ratio of the perimeters of two similar triangles is the same as their similarity ratio; therefore, if is the perimeter of $\bigtriangleup AXB$ and is the perimeter of $\bigtriangleup CXB$ , it follows that

$\frac{Y}{X} = \frac{4}{3}$

$\frac{Y}{X} \cdot X = \frac{4}{3} \cdot X$

$Y = \frac{4}{3} X$

Multiply both sides by 3:

$3 \cdot Y =3 \cdot \frac{4}{3} X$

Three times the perimeter of $\bigtriangleup CXB$ is therefore equal to four times that of $\bigtriangleup AXB$ .

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Question

Note: Figure NOT drawn to scale.

Which of the following is the greater quantity?

(A) The perimeter of the triangle

(B) 90

Answer

The longest side of the triangle appears opposite the angle of greatest measure. The side of length 30 appears opposite an angle of measure $180 ^{\circ } - (65 ^{\circ } + 65 ^{\circ } ) = 50 ^{\circ }$ . Therefore, the sides opposite the $65^{\circ }$ angles must have lengths greater than 30.

If we let this common length be , then

The perimeter of the triangle is therefore greater than 90.

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Question

Examine the above diagram. If $l \parallel m$ , give in terms of .

Answer

The two marked angles are same-side exterior angles of parallel lines, which are supplementary - that is, their measures have sum 180. We can solve for in this equation:

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Question

Which is the greater quantity?

(a) The measure of an angle complementary to a $55 ^{\circ }$ angle

(b) The measure of an angle supplementary to a $135 ^{\circ }$ angle

Answer

Supplementary angles and complementary angles have measures totaling $180^{\circ }$ and $90^{\circ }$ , respectively.

(a) The measure of an angle complementary to a $55 ^{\circ }$ angle is $90^{\circ } - 55^{\circ } = 35^{\circ }$

(b) The measure of an angle supplementary to a $135 ^{\circ }$ angle is $180^{\circ } - 135 ^{\circ } = 45^{\circ }$

This makes (b) greater.

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Question

$\angle 1$ and $\angle 2$ are complementary; $m \angle 1 =2 m \angle 2 - 50$ .

Which is the greater quantity?

(A) $m \angle 1$

(B) $m \angle 2$

Answer

Two angles are complementary if their degree measures total 90. Therefore,

$m \angle 1 + m \angle 2 = 90$

Since $m \angle 1 =2 m \angle 2 - 50$ , we can substitute, and we can solve for $m\angle2$ :

$(2 m \angle 2 - 50)+ m \angle 2 = 90$

$3 m \angle 2 - 50 = 90$

$3 m \angle 2 - 50 + 50 = 90 + 50$

$3 m \angle 2 = 140$

$3 m \angle 2 \div 3 = 140 \div 3$

$m \angle 2 = 46 \frac{2}{3}$

$m \angle 1 = 90 - m \angle 2 = 90 - 46 \frac{2}{3} = 43 \frac{1}{3}$

$m \angle 2 > m \angle 1$ , making (B) the greater quantity.

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Question

Note: diagram is not drawn to scale

Refer to the above diagram. If $m \angle1 = m \angle 2 + 30$ , what is $m \angle1$ ?

Answer

$\angle 1$ and $\angle 2$ form a linear pair, so

$m \angle1 + m \angle 2 = 180$

Since $m \angle1 = m \angle 2 + 30$ , this can be rewritten as

$m \angle1 - 30 = m \angle 2$ , and the first equation can be rewritten as:

$m \angle1 + m \angle1 - 30 = 180$

$2 m \angle1 - 30 = 180$

$2 m \angle1 - 30 + 30 = 180 + 30$

$2 m \angle1 = 210$

$2 m \angle1 \div 2 = 210 \div 2$

$m \angle1 = 105$

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Question

Note: Figure NOT drawn to scale.

Which of the following pairs of numbers could give the measures of $\angle 1$ and $\angle 2$ ?

Answer

The two angles form a linear pair and therefore their measures total $180^{\circ }$ . We check all of the pairs for this sum.

The correct pair is $112^{\circ }, 68^{\circ }$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Which of the following triples could refer to the measures of $\angle 1$ , $\angle 2$ , and $\angle 3$ ?

Answer

The measure of an exterior angle of a triangle, which here is $\angle 3$ , is the sum of the measures of its remote interior angles, which here are $\angle 1$ and $\angle 2$ . Therefore, we are looking for the sum of the first two angle measures to be equal to the third.

$42 ^{\circ }+ 72 ^{\circ } = 114 ^{\circ }$

$30 ^{\circ } + 80 ^{\circ } = 110 ^{\circ }$

$56 ^{\circ } + 86 ^{\circ } = 142 ^{\circ }$

$25 ^{\circ }+ 88 ^{\circ } = 113 ^{\circ }$

All four triples satisfy this condition.

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Question

Note: Figure NOT drawn to scale.

$m \angle 3 = 125 ^{\circ }$

$3 m \angle 1 = 2 m \angle 2$

Evaluate $m \angle 1$ .

Answer

$3 m \angle 1 = 2 m \angle 2$ , so

$3 \left (m \angle 1 \right ) \div 2 = 2\left ( m \angle 2 \right )\div 2$

$\frac{3}{2} m \angle 1 = m \angle 2$

The measure of an exterior angle of a triangle, which here is $\angle 3$ , is the sum of the measures of its remote interior angles, which here are $\angle 1$ and $\angle 2$ .

$m \angle 1 + m \angle 2 = m \angle 3$

$m \angle 1 + \frac{3}{2} m \angle 1 = 125 ^{\circ }$

$\frac{5}{2} m \angle 1 = 125 ^{\circ }$

$\frac{2}{5} \cdot \frac{5}{2} m \angle 1 = \frac{2}{5} \cdot 125 ^{\circ }$

$m \angle 1 =50 ^{\circ }$

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Question

Note: figure NOT drawn to scale

The degree measure of $\angle 1$ is five degrees greater than twice that of $\angle 2$ . Which is the greater quantity?

(A) $m \angle 2$

(B) $55^{\circ }$

Answer

The degree measure of $\angle 1$ is five degrees greater than twice that of $\angle 2$ - that is,

$m \angle 1 = 2 m \angle 2 + 5$

Since $\angle 1$ and $\angle 2$ form a linear pair,

$m\angle1 + m\angle 2 = 180$ , and by substitution,

$( 2 m \angle 2 + 5)+ m\angle 2 = 180$

$3 m \angle 2 + 5 = 180$

$3 m \angle 2 + 5 - 5 = 180 - 5$

$3 m \angle 2 = 175$

$3 m \angle 2 \div 3 = 175\div 3$

$m \angle 2 = 58 \frac{1}{3} > 55$

This makes (A) greater.

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Question

Note: Figure NOT drawn to scale

$m\angle 2 = 48 ^{\circ }$ . What is $m \angle 1$ ?

Answer

$\angle 1$ and $\angle 2$ form a linear pair and therfore, the the sum of their degree measures is $180^{\circ }$ .

$m\angle 1 + m\angle 2 = 180^{\circ }$

$m\angle 1 + 48 ^{\circ } = 180^{\circ }$

$m\angle 1 + 48 ^{\circ } - 48 ^{\circ } = 180^{\circ } - 48 ^{\circ }$

$m\angle 1 = 132 ^{\circ }$

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