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ISEE Upper Level Quantitative : Triangles

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

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Example Questions

Example Question #71 : Triangles

The lengths of the hypotenuses of ten similar right triangles form an arithmetic sequence. The smallest triangle has legs of lengths 5 and 12 inches; the second-smallest triangle has a hypotenuse of length one and one half feet.

Which of the following responses comes closest to the area of the largest triangle?

Possible Answers:

5 square feet

7 square feet

3 square feet

6 square feet

4 square feet

Correct answer:

4 square feet

Explanation:

The hypotenuse of the smallest triangle can be calculated using the Pythagorean Theorem:

$c = \sqrt{a^{2}+b^{2}} = \sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169}=13$ inches.

Let $a_{n}$ be the lengths of the hypotenuses of the triangles in inches. $a_{1} = 13$ and $a_{2} = 1 \frac{1}{2} \cdot 12 = 18$ , so their common difference is

$d = a_{2} - a_{1} = 18- 13= 5$

The arithmetic sequence formula is

$a_{n} = a_{1} + (n-1)d$

The length of the hypotenuse of the largest triangle - the tenth triangle - can be found by substituting $a_{1} = 13, n= 10, d = 5$ :

$a_{10} = a_{1} + (10-1)5 = 13+ 9 \cdot 5= 13 + 45 = 58$ inches.

The largest triangle has hypotenuse of length 58 inches. Since the triangles are similar, corresponding sides are in proportion. If we let and be the lengths of the legs of the largest triangle, then

$\frac{a}{5} = \frac{58}{13}$

$\frac{a}{5}\cdot 5= \frac{58}{13} \cdot 5$

$a \approx 22.3$

Similarly,

$\frac{b}{12} = \frac{58}{13}$

$\frac{b}{12} \cdot 12= \frac{58}{13} \cdot 12$

$b \approx 53.5$

The area of a right triangle is half the product of its legs:

$A = \frac{ab}{2}= \frac{22.3 \cdot 53.5}{2} \approx 596.5$ square inches.

Divide this by 144 to convert to square feet:

$596.5 \div 144 \approx 4.1$

Of the given responses, 4 square feet is the closest, and is the correct choice.

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Example Question #71 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

In Square SARE . is the midpoint of $\overline{SE}$ , is the midpoint of $\overline{SA}$ , and is the midpoint of $\overline{QA}$ . Draw the segments $\overline{QX}$ and $\overline{UR}$ .

Which is the greater quantity?

(a) The area of $\bigtriangleup SQX$

(b) The area of $\bigtriangleup UAR$

Possible Answers:

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

It cannot be determined which of (a) and (b) is greater

Correct answer:

(a) and (b) are equal

Explanation:

The figure referenced is below:
Square x

Let be the common sidelength of Square SARE .

Then $SX = SQ = \frac{1}{2}s$ .

The area of right triangle $\bigtriangleup SQX$ is half the product of its legs, so

$A _{\bigtriangleup SQX} = \frac{1}{2} \cdot SQ \cdot SX$

$= \frac{1}{2} \cdot \frac{1}{2}s \cdot \frac{1}{2}s = \frac{1}{8}s ^{2}$

$QA = \frac{1}{2}s$ , so

$UA = \frac{1}{2} \cdot QA = \frac{1}{2} \cdot \frac{1}{2}s = \frac{1}{4}s$

The area of right triangle $\bigtriangleup UAR$ is half the product of its legs, so

$A _{\bigtriangleup UAR } = \frac{1}{2} \cdot UA \cdot AR$

$= \frac{1}{2} \cdot \frac{1}{4}s \cdot s = \frac{1}{8}s ^{2}$

$\bigtriangleup SQX$ and $\bigtriangleup UAR$ have the same area.

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Example Question #72 : Geometry

Right triangle

Figure NOT drawn to scale

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 area of $\bigtriangleup AXB$

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

Possible Answers:

(a) and (b) are equal

(a) is the greater quantity

(b) is the greater quantity

It is impossible to determine which is greater from the information given

Correct answer:

(b) is the greater quantity

Explanation:

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 areas of two similar triangles is the square of their similarity ratio, which here is

$\left ( \frac{4}{3} \right )^{2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$ , or 16: 9 .

Therefore, if is the area of $\bigtriangleup AXB$ and is the area of $\bigtriangleup CXB$ , it follows that

$\frac{Y}{X} = \frac{16}{9}$

$\frac{Y}{X} \cdot X = \frac{16}{9} \cdot X$

$Y = \frac{16}{9} X$

Four times the area of $\bigtriangleup AXB$ is ; three times the area of $\bigtriangleup CXB$ is

$3 Y = 3 \cdot \frac{16}{9} X = \frac{48}{9} X = 5 \frac{4}{9} \cdot X$

$5 \frac{4}{9} > 4$ , so three times the area of $\bigtriangleup CXB$ is the greater quantity.

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Example Question #72 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Right triangle 2

Figure NOT drawn to scale.

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

(a) Twice the area of $\bigtriangleup AXB$

(b) The area of $\bigtriangleup CXB$

Possible Answers:

It is impossible to determine which is greater from the information given

(a) is the greater quantity

(a) and (b) are equal

(b) is the greater quantity

Correct answer:

(b) is the greater quantity

Explanation:

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. Also, since $\angle ABC$ measures 90 degrees and $\angle C$ measure 30 degrees, $\angle A$ measures 60 degrees, making $\bigtriangleup ABC$ a 30-60-90 triangle.

Because of this, the ratio of the measures of the legs of $\bigtriangleup ABC$ is

$\frac{BC}{AB} = \frac{\sqrt{3}}{1}$ ,

Since these legs coincide with the hypotenuses of $\bigtriangleup CXB$ and $\bigtriangleup AXB$ , this is also the similarity ratio of the latter to the former. The ratio of the areas is the square of this, or

$\left ( \frac{\sqrt{3}}{1} \right )^{2} = \frac{3}{1}$

Therefore, the area of $\bigtriangleup CXB$ is three times that of $\bigtriangleup AXB$ . This makes (b) the greater quantity.

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Example Question #73 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Triangle 4

The above figure depicts Trapezoid TRAP . Which is the greater quantity?

(a) The area of $\bigtriangleup TRP$

(b) The area of $\bigtriangleup TRA$

Possible Answers:

(b) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

The area of a triangle is one half times the product of its height and the length of its base. As can be seen in the diagram below, both $\bigtriangleup TRP$ and $\bigtriangleup TRA$ have height and base of length :

Since both base length and height are the same between the two triangles, it follows that they have the same area.

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Example Question #1 : Isosceles Triangles

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?

Possible Answers:

$\frac{1}{2} \textrm{ ft}$

All of the other choices are correct.

$\frac{2}{3} \textrm{ ft}$

$8\textrm{ in}$

$6 \textrm{ in}$

Correct answer:

All of the other choices are correct.

Explanation:

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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Example Question #71 : Geometry

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

Which is the greater quantity?

(a)

(b)

Possible Answers:

(b) is greater.

(a) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

Correct answer:

(a) and (b) are equal.

Explanation:

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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Example Question #2 : Acute / Obtuse Isosceles Triangles

Isosceles

Note: Figure NOT drawn to scale.

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

Possible Answers:

The correct answer is not among the other choices.

$\frac{15}{\sin 40^{\circ }}$

$15 \cos 40^{\circ }$

$15 \sin 40^{\circ }$

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

Correct answer:

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

Explanation:

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

Isosceles

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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Example Question #3 : Isosceles Triangles

Isosceles

Note: Figure NOT drawn to scale.

Which of the following is the greater quantity?

(A) The perimeter of the triangle

(B) 90

Possible Answers:

(A) is greater

(B) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

Correct answer:

(A) is greater

Explanation:

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

s > 30

s + s > 30 + 30 = 60

P = s + s + 30 > 60 + 30 = 90

The perimeter of the triangle is therefore greater than 90.

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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Triangles

Study concepts, example questions & explanations for ISEE Upper Level Quantitative

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #71 : Triangles

Example Question #71 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #72 : Geometry

Example Question #72 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #73 : Isee Upper Level (Grades 9 12) Quantitative Reasoning

Example Question #1 : Isosceles Triangles

Example Question #71 : Geometry

Example Question #2 : Acute / Obtuse Isosceles Triangles

Example Question #3 : Isosceles Triangles

All ISEE Upper Level Quantitative Resources

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