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ISEE Upper Level Quantitative : Acute / Obtuse Triangles

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Example Questions

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

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

(a)

(b)

Possible Answers:

It is impossible to tell from the information given.

(a) is greater.

(b) is greater.

(a) and (b) are equal.

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 #5 : 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) is the greater quantity

(a) and (b) are equal

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:

(a) and (b) are equal

(b) is greater

It is impossible to tell from the information given

(a) is greater

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:

It is impossible to tell from the information given.

(a) is greater.

(a) and (b) are equal.

(b) 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 #1 : How To Find The Area Of A Triangle

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.

(a) and (b) are equal.

(b) is greater.

It is impossible to tell from the information given.

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:

It is impossible to tell from the information given

(b) is greater

(a) and (b) are equal

(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 #2 : How To Find The Area Of A Triangle

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

Possible Answers:

600

1,600

$75 \sqrt{105}$

1,200

None of the other responses is correct

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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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Acute / Obtuse Triangles

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

All ISEE Upper Level Quantitative Resources

Example Questions

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

Example Question #5 : 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 #1 : How To Find The Area Of A Triangle

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

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

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

All ISEE Upper Level Quantitative Resources

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