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ISEE Upper Level Quantitative : Triangles

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

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All ISEE Upper Level Quantitative Resources

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Example Questions

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Example Question #21 : Triangles

$\Delta ABC$ is acute; AB = 5, BC = 12 . Which is the greater quantity?

(a)

(b)

Possible Answers:

(a) and (b) are equal.

(b) is greater.

(a) is greater.

It is impossible to tell from the information given.

Correct answer:

(b) is greater.

Explanation:

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

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

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

$AC< \sqrt{169}$

AC < 13

(b) is the greater quantity.

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Example Question #2 : How To Find The Length Of The Side Of A Triangle

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

(a) 18

(b)

Possible Answers:

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

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

Correct answer:

(a) is the greater quantity

Explanation:

Suppose there exists a second triangle $\bigtriangleup DEF$ such that DE = DF = 12 and EF = 18 . 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,

BC < EF = 18 .

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Example Question #1 : How To Find If Two Triangles Are Similar

$\Delta ABC \sim \Delta DEF$

AB = 12 ,AC = 16, EF = 27, DF = 24

Which is the greater quantity?

(a)

(b)

Possible Answers:

It is impossible to tell from the information given

(b) is greater

(a) is greater

(a) and (b) are equal

Correct answer:

(a) and (b) are equal

Explanation:

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

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

Substitute and solve for :

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

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

BC = 18

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

Substitute and solve for :

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

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

DE = 18

The two are equal.

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Example Question #2 : How To Find If Two Triangles Are Similar

$\Delta ABC \sim \Delta DEF$

AB = 18, BC = 17, AC = 16

Which is the greater quantity?

(a)

(b)

Possible Answers:

(b) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

(a) is greater.

Correct answer:

(a) is greater.

Explanation:

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

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

Substitute:

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

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

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

DE > EF , so (a) is greater.

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Example Question #21 : Triangles

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

Possible Answers:

(a) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

(b) is greater.

Correct answer:

(a) and (b) are equal.

Explanation:

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.

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Example Question #2 : How To Find The Area Of A Triangle

Two triangles on the coordinate plane have a vertex at the origin and a vertex at (c,d) , where c > d>0 .

Triangle A has its third vertex at (10,0) .

Triangle B has its third vertex at (0,10) .

Which is the greater quantity?

(a) The area of Triangle A

(b) The area of Triangle B

Possible Answers:

(b) is greater

(a) and (b) are equal

It is impossible to tell from the information given

(a) is greater

Correct answer:

(b) is greater

Explanation:

(a) Triangle A has as its base the horizontal segment connecting (0,0) and (10,0) , the length of which is 10. Its (vertical) altitude is the segment from (c,d) 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 (0,0) and (0,10) , the length of which is 10. Its (horizontal) altitude is the segment from (c,d) 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 $A = \frac{1}{2} \cdot 10 \cdot c = 5c$

c > d , so 5c > 5d . (b), the area of Triangle B, is greater.

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Example Question #22 : Triangles

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

Possible Answers:

1,600

$75 \sqrt{105}$

None of the other responses is correct

600

1,200

Correct answer:

None of the other responses is correct

Explanation:

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,

30 + 40 = 70 < 80 ;

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

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Example Question #1 : How To Find The Area Of A Triangle

Pentagon 2

The above depicts Square ABCD ; X, Y , 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 $\bigtriangleup ZXD$

Possible Answers:

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

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

Correct answer:

(a) and (b) are equal

Explanation:

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, AZ = BZ= XD = BY = 1 , as shown in this diagram.

Pentagon 3

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.

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Example Question #21 : Geometry

$\Delta ABC$ is an equilateral triangle. Points D,E,F 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 $\Delta DEF$

Possible Answers:

(a) is greater.

It is impossible to tell from the information given.

(a) and (b) are equal.

(b) is greater.

Correct answer:

(a) and (b) are equal.

Explanation:

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 = AB + BC + AC

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

P = 2 ( EF+ DF + DE) ,

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

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

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Example Question #21 : Triangles

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.

Possible Answers:

The quantity in Column B is greater.

There is not enough info to determine a relationship between the columns.

The quantity in Column A is greater.

The quantities in both columns are equal.

Correct answer:

The quantity in Column B is greater.

Explanation:

Perimeter involves adding up all of the sides of the shape. Therefore, the square's perimeter is 4+4+4+4, or 16. An equialteral shape means that all of the sides are equal. Therefore, the perimeter of the triangle is 9+9+9, or 27. Therefore, Column B is greater.

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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 #21 : Triangles

Example Question #2 : How To Find The Length Of The Side Of A Triangle

Example Question #1 : How To Find If Two Triangles Are Similar

Example Question #2 : How To Find If Two Triangles Are Similar

Example Question #21 : Triangles

Example Question #2 : How To Find The Area Of A Triangle

Example Question #22 : Triangles

Example Question #1 : How To Find The Area Of A Triangle

Example Question #21 : Geometry

Example Question #21 : Triangles

All ISEE Upper Level Quantitative Resources

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