Isosceles Triangles - ISEE Upper Level Quantitative Reasoning

Card 0 of 216

Question

Given $\Delta ABC$ with . Which is the greater quantity?

(a)

(b)

Answer

Use the Triangle Inequality:

This makes (b) the greater quantity.

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Question

Which of the following could be the lengths of the three sides of a scalene triangle?

Answer

A scalene triangle, by definition, has sides all of different lengths. Since all of the given choices fit that criterion, the correct choice is that all can be scalene.

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Question

Given $\Delta ABC$ with right angle $\angle B$ , $m \angle C = 50 ^{\circ }$ .

Which is the greater quantity?

(a)

(b)

Answer

$m \angle B = 90^{\circ}$

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

The sum of the measures of the angles of a triangle is 180, so

$m \angle A + m \angle B + m \angle C = 180$

$m \angle A +90 + 50 = 180$

$m \angle A +140= 180$

$m \angle A +140-140= 180-140$

$m \angle A = 40^{\circ }$

$m \angle C > m \angle A$ , so the side opposite $\angle C$ , which is $\overline{AB}$ , is longer than the side opposite $\angle A$ , which is $\overline{BC}$ . This makes (a) the greater quantity.

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Question

$\Delta ABC$ is acute; . Which is the greater quantity?

(a)

(b)

Answer

Since $\Delta ABC$ is an acute triangle, $\angle B$ is an acute angle, and

$\left ( AC \right )^{2} < \left ( AB \right )^{2} + \left ( BC \right )^{2} = 5^{2} + 12^{2} = 25 + 144 = 169$

$\left ( AC \right )^{2} < 169$ ,

$AC< \sqrt{169}$

(b) is the greater quantity.

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Question

Given $\Delta ABC$ with obtuse angle $\angle A$ , which is the greater quantity?

(a)

(b)

Answer

To compare the lengths of $\overline{AB}$ and $\overline{BC}$ from the angle measures, it is necessary to know which of their opposite angles - $\angleC$ $\angle C$ and $\angle A$ , respectively - is the greater angle. Since $\angle A$ is the obtuse angle, it has the greater measure, and $\overline{BC}$ is the longer side. This makes (b) greater.

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Question

$\Delta ABC$ has obtuse angle $\angle B$ ; . Which is the greater quantity?

(a)

(b)

Answer

Since $\angle B$ is the obtuse angle of $\Delta ABC$ ,

$\left ( AC \right )^{2} >\left ( AB \right )^{2} + \left ( BC \right )^{2} = 7^{2} + 24^{2} = 49 + 576 = 625$ .

$\left ( AC \right )^{2} > 625$ ,

$AC > \sqrt{625}$

,

so (a) is the greater quantity.

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Question

Given $\Delta ABC$ with . Which is the greater quantity?

(a)

(b)

Answer

By the Converse of the Pythagorean Theorem,

$AC = \sqrt{(AB)^{2}+(BC)^{2}}= \sqrt{5^{2}+12^{2}}= \sqrt{25+144}=\sqrt{169} = 13$

if and only if $\angle B$ is a right angle.

However, if $\angle B$ is acute, then ; if $\angle B$ is obtuse, then .

Since we do not know whether $\angle B$ is acute, right, or obtuse, we cannot determine whether (a) or (b) is greater.

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Question

Given: $\bigtriangleup ABC$ . $AB= AC = 12, m \angle A = 80 ^{\circ }$ . Which is the greater quantity?

(a) 18

(b)

Answer

Suppose there exists a second triangle $\bigtriangleup DEF$ such that and . Whether $\angle D$ , the angle opposite the longest side, is acute, right, or obtuse can be determined by comparing the sum of the squares of the lengths of the shortest sides to the square of the length of the longest:

$(DE)^{2}+ (DF)^{2} = 12^{2}+ 12^{2} = 144 + 144 = 288$

$(EF) ^{2}= 18 ^{2} = 324$

$(EF) ^{2}> (DE)^{2}+ (DF)^{2}$ , making $\angle D$ obtuse, so $m \angle D > 90 ^{\circ }$ .

We know that

$AB= AC = 12, m \angle A = 80 ^{\circ }$

and

$DE = DF = 12, m \angle D > m \angle A$ .

Between $\bigtriangleup ABC$ and $\bigtriangleup DEF$ , we have two sets of congruent sides, with the included angle of the latter of greater measure than that of the former. It follows from the Side-Angle-Side Inequality (or Hinge) Theorem that between the third sides, $\overline{EF}$ is the longer. Therefore,

.

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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

$\Delta ABC$ 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

A square, a regular pentagon, a regular hexagon, and a regular octagon have the same sidelength. Which is the greater quantity?

(A) The mean of their perimeters

(B) The median of their perimeters

Answer

The answer is independent of the sidelength, so we can assume without loss of generality that the sidelength is 1. The square, the pentagon, the hexagon, and the octagon have 4, 5, 6, and 8 sides of equal length, respectively, so their perimeters are 4, 5, 6, and 8. The mean of these four perimeters is

$\left ( 4+5+6+8\right ) \div 4 = 23 \div 4 = 5.75$ units.

The median is the mean of the middle two perimeters, which are 5 and 6:

$\left ( 5 + 6\right ) \div 2 = 5.5$

The mean, (A), is greater.

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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

Two sides of a triangle have length 8 inches and 6 inches. Which of the following lengths of the third side would make the triangle isosceles?

Answer

An isosceles triangle, by definition, has two sides of equal length. Having the third side measure either 6 inches or 8 inches would make the triangle meet this criterion. Also, since 6 inches and 8 inches are equal to $\frac{1}{2} \textrm{ ft}$ and $\frac{2}{3} \textrm{ ft}$ , respectively, these also make the triangle isosceles. Therefore, the correct choice is that all four make the triangle isosceles.

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Question

$\Delta ABC$ is an isosceles triangle with obtuse angle $\angle A$ .

Which is the greater quantity?

(a)

(b)

Answer

A triangle must have at least two acute angles; if $\angle A$ is obtuse, then $\angle B$ and $\angle C$ are the acute angles of $\Delta ABC$ . Since $\Delta ABC$ is isosceles, the Isosceles Triangle Theorem requires two of the angles to be congruent; they must be the two acute angles $\angle B$ and $\angle C$ . Also, the sides opposite these two angles are the congruent sides; these sides are $\overline{AB}$ and $\overline{AC}$ , respectively. This makes the quantities (a) and (b) equal.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Which expression is equivalent to ?

Answer

This is an isosceles triangle, so the left and right sides are of equal length. Draw the altitude of this triangle, as follows:

The altitude is a perpendicular bisector of the base; it is one leg of a right triangle with half the base, which is 15 inches, as the other leg, and one side, which is inches, as the hypotenuse. By definition,

$\cos 40 ^{\circ } = \frac{15}{x}$ (adjacent side divided by hypotenuse), so

$\frac{x}{\cos 40 ^{\circ }} \cdot \cos 40 ^{\circ } =\frac{x}{\cos 40 ^{\circ }} \cdot \frac{15}{x}$

$x= \frac{15}{\cos 40 ^{\circ }}$

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Question

Which is the greater quantity?

(a) The perimeter of a regular pentagon with sidelength 1 foot

(b) The perimeter of a regular hexagon with sidelength 10 inches

Answer

The sides of a regular polygon are congruent, so in each case, multiply the sidelength by the number of sides to get the perimeter.

(a) Since one foot equals twelve inches, $5 \times 12 = 60$ inches.

(b) Multiply: $6 \times 10 = 60$ inches

The two polygons have the same perimeter.

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Question

An equilateral triangle, a square, a regular pentagon, a regular hexagon, and a regular octagon have the same sidelength. Which is the greater quantity?

(A) The median of their perimeters

(B) The midrange of their perimeters

Answer

The answer is independent of the sidelength, so we can assume without loss of generality that the sidelength is 1. The equilateral triangle, the square, the pentagon, the hexagon, and the octagon have 3, 4, 5, 6, and 8 sides of equal length, respectively, so their perimeters are 3, 4, 5, 6, and 8.

The median of these perimeters is the middle perimeter, 5. The midrange of these perimeters is the mean of the greatest and the least perimeters:

$\left ( 3+8\right ) \div 2 = 5.5$

The midrange, (B), is greater.

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