Triangles - ISEE Upper Level: Quantitative Reasoning

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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 area of $\Delta ABC$

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

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Answer

If segments are constructed in which the endpoints form the midpoints of the sides of a triangle, then four triangles, congruent to each other and similar to the larger triangle, are formed. Therefore, one of these triangles - specifically, $\Delta DEF$ - would have one-fourth the area of $\Delta ABC$ . This means $\Delta ABC$ has more than twice the area of $\Delta DEF$ .

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

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Question

Which of the following could be the three sidelengths of an equilateral triangle?

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Answer

By definition, an equilateral triangle has three sides of equal length. We can identify the equilateral triangle by converting the given sidelengths to the same units and comparing them.

We can eliminate the following by showing that at least two sidelengths differ.

$2 \textrm{ yd, } 84 \textrm{ in, }7\textrm{ ft}$

2 yards = $2 \times 3 = 6$ feet.

Two sides have lengths 6 feet and 7 feet, so we can eliminate this choice.

$1\frac{1}{2} \textrm{ yd, } 50 \textrm{ in, }4\textrm{ ft}$

4 feet = $4 \times 12 = 48$ inches

Two sides have lengths 48 inches and 50 inches, so we can eliminate this choice.

$1\frac{1}{3} \textrm{ yd, } 48 \textrm{ in, }5\textrm{ ft}$

5 feet = $5 \times 12 = 60$ inches

Two sides have lengths 48 inches and 60 inches, so we can eliminate this choice.

$1\frac{2}{3} \textrm{ yd, } 48 \textrm{ in, }4\textrm{ ft}$

$1\frac{2}{3}$ yards = $1\frac{2}{3} \times 3 = 5$ feet

Two sides have lengths 4 feet and 5 feet, so we can eliminate this choice.

$\frac{2}{3} \textrm{ yd, } 24 \textrm{ in, }2\textrm{ ft}$

$\frac{2}{3}$ yards = $\frac{2}{3} \times3 = 2$ feet = $2 \times 12 = 24$ inches

All three sides have the same length, making this the triangle equilateral. This choice is correct.

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Question

What is the hypotenuse of a right triangle with sides 5 and 8?

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Answer

Because this is a right triangle, we can use the Pythagorean Theorem which says _a_2 + _b_2 = _c_2, or the squares of the two sides of a right triangle must equal the square of the hypotenuse. Here we have a = 5 and b = 8.

_a_2 + _b_2 = _c_2

52 + 82 = _c_2

25 + 64 = _c_2

89 = _c_2

c = √89

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Question

Which is the greater quantity?

(a) The hypotenuse of a right triangle with legs and .

(b) The hypotenuse of a right triangle with legs and .

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Answer

The hypotenuses of the triangles measure as follows:

(a) $c = \sqrt {10^{2} + 15^{2} } = \sqrt {100 + 225 } = \sqrt {325}$

(b) $c = \sqrt {12^{2} + 13^{2} } = \sqrt {144 + 169 } = \sqrt {313}$

, so $\sqrt {325} > \sqrt {313}$ , making (a) the greater quantity.

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Question

Which is the greater quantity?

(a) The hypotenuse of a $45^{\circ}-45^{\circ}-90^{\circ}$ right triangle with a leg of length 20

(b) The hypotenuse of a right triangle with legs of length 19 and 21

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Answer

The hypotenuses of the triangles measure as follows:

(a) $c = \sqrt {20^{2} + 20^{2} } = \sqrt {400 + 400 } = \sqrt {800}$

(b) $c = \sqrt {19^{2} + 21^{2} } = \sqrt {361 + 441} = \sqrt {802}$

, so $\sqrt {800} < \sqrt {802}$ , making (b) the greater quantity

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Question

A right triangle has a leg $4 \frac{1}{2}$ feet long and a hypotenuse $7 \frac{1}{2}$ feet long. Which is the greater quantity?

(a) The length of the second leg of the triangle

(b) 60 inches

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Answer

The length of the second leg can be calculated using the Pythagorean Theorem. Set $c = 7 \frac{1}{2} = \frac{15}{2} , a =4 \frac{1}{2} = \frac{9}{2}$ :

$b ^{2} = c ^{2}-a ^{2}$

$b ^{2} = \left ( \frac{15}{2} \right ) ^{2}- \left ( \frac{9}{2} \right )^{2}\textup{}$

$b ^{2} = \frac{225}{4} -\frac{81}{4}$

$b ^{2} = \frac{144}{4}$

$b ^{2} = 36$

$b = \sqrt{36} = 6$

The second leg therefore measures $6 \times 12 = 72$ inches.

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Question

What is the hypotenuse of a right triangle with sides 9 inches and 12 inches?

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Answer

Since we're dealing with right triangles, we can use the Pythagorean Theorem (). In this formula, a and b are the sides, while c is the hypotenuse. The hypotenuse of a right triangle is the longest side and the side that is opposite the right angle. Now, we can plug into our formula, which looks like this: We simplify and get . At this point, isolate c. This means taking the square root of both sides so that your answer is 15in.

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Question

The perimeter of a regular pentagon is 75% of that of the triangle in the above diagram. Which is the greater quantity?

(A) The length of one side of the pentagon

(B) One and one-half feet

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Answer

By the Pythagorean Theorem, the hypotenuse of the right triangle is

$\sqrt{14^{2}+48^{2}} = \sqrt{196+2,304} = \sqrt{2,500} = 50$ inches, making its perimeter

inches.

The pentagon in question has sides of length 75% of 112, or

$112 \times 0.75 = 84$ .

Since a pentagon has five sides of equal length, each side will have measure

$84 \div 5 = 16 \frac{4}{5}$ inches.

One and a half feet are equivalent to $12 \times 1 \frac{1}{2} = 18$ inches, so (B) is the greater quantity.

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Question

The track at Gauss High School is unusual in that it is shaped like a right triangle, as shown above.

Cary decides to get some exercise by running from point A to point B, then running half of the distance from point B to point C.

Which is the greater quantity?

(A) The distance Cary runs

(B) One-fourth of a mile

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Answer

By the Pythagorean Theorem, the distance from B to C is

$\sqrt{600^{2} + 800^{2} }$

$= \sqrt{360,000 + 640,000 }$

$= \sqrt{1,000,000} = 1,000$ feet

Cary runs

$800 + \frac{1}{2} \times1,000 = 800 + 500 = 1,300$ feet

Since 5,280 feet make a mile, one-fourth of a mile is equal to

$5,280 \div 4 = 1,320$ feet.

(B) is greater

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Question

Give the length of the hypotenuse of the above right triangle in terms of .

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Answer

If we let be the length of the hypotenuse, then by the Pythagorean theorem,

$c = \sqrt{(k+2)^{2}+(k-2)^{2}}$

$c = \sqrt{(k ^{2}+4k+4) +(k ^{2}-4k+4)}$

$c = \sqrt{2k ^{2} +8}$

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Question

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

Which is the greater quantity?

(a)

(b)

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Answer

The figure referenced is below:

For the sake of simplicity, assume that the square has sides of length 4. The following reasoning is independent of the actual lengths, and the reason for choosing 4 will become apparent in the explanation.

and are midpoints of their respective sides, so , making $\overline{QX}$ the hypotenuse of a triangle with legs of length 2 and 2. Therefore,

$(QX ) ^{2}= (SQ)^{2}+ (SX)^{2} = 2^{2}+ 2^{2} = 4+ 4 = 8$ .

Also, , and since is the midpoint of $\overline{AQ}$ , . , making $\overline{UR}$ the hypotenuse of a triangle with legs of length 1 and 4. Therefore,

$(UR)^{2} = (AU)^{2}+ (AR)^{2} = 1^{2}+ 4^{2} = 1+ 16 = 17$

$(UR)^{2} >(QX ) ^{2}$ , so

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Question

Figure NOT drawn to scale.

In the above figure, $\angle ABC$ is a right angle.

What is the length of $\overline{AC}$ ?

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Answer

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller triangles each similar to the larger triangle. In particular,

$\bigtriangleup BXC \sim \bigtriangleup ABC$ .

Their corresponding sides are in proportion, so, setting the ratios of the hypotenuses to the short legs equal to each other,

$\frac{AC}{BC} = \frac{BC}{CX}$

$\frac{AC}{13} = \frac{13}{5}$

$\frac{AC}{13} \cdot 13 = \frac{13}{5} \cdot 13$

$AC = \frac{169}{5} = 33 \frac{4}{5 }$

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Question

Figure NOT drawn to scale.

In the above figure, $\angle ABC$ is a right angle.

What is the length of $\overline{AB}$ ?

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Answer

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller triangles each similar to the larger triangle. In particular,

$\bigtriangleup BXC \sim \bigtriangleup ABC$ .

Their corresponding sides are in proportion, so, setting the ratios of the long legs to the short legs equal to each other,

$\frac{AB}{BC} = \frac{BX}{XC}$

By the Pythagorean Theorem.

$(BX)^{2} = (BC)^{2} - (CX)^{2}$

$(BX)^{2} = 13^{2} - 5^{2} = 169 - 25 = 144$

$BX = \sqrt{144} = 12$

The proportion statement becomes

$\frac{AB}{13} = \frac{12}{5}$

$\frac{AB}{13} \cdot 13 = \frac{12}{5} \cdot 13$

$AB = \frac{156}{5}= 31 \frac{1}{5}$

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Question

Given: $\bigtriangleup ABC$ with , , .

Which is the greater quantity?

(a) $m \angle B$

(b) $90^{\circ }$

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Answer

The measure of the angle formed by the two shorter sides of a triangle can be determined to be acute, right, or obtuse by comparing the sum of the squares of those lengths to the square of the length of the opposite side. We compare:

$(AB) ^{2} + (BC)^{2} = 6^{2} + 8 ^{2} = 36 + 64 = 100$

$(AC)^{2} = 12^{2} = 144$

$(AB) ^{2} + (BC)^{2} < (AC) ^{2}$ ; it follows that $\angle B$ is obtuse, and has measure greater than $90 ^{\circ }$

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Question

Figure NOT drawn to scale.

In the above figure, $\angle ABC$ is a right angle.

What is the length of $\overline{AC}$ ?

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Answer

The altitude of a right triangle from the vertex of its right angle divides the triangle into two smaller triangles each similar to the larger triangle. In particular,

$\bigtriangleup BXC \sim \bigtriangleup ABC$ .

Their corresponding sides are in proportion, so, setting the ratios of the hypotenuses to the short legs equal to each other,

$\frac{AC}{BC} = \frac{BC}{CX}$

$\frac{AC}{13} = \frac{13}{5}$

$\frac{AC}{13} \cdot 13 = \frac{13}{5} \cdot 13$

$AC = \frac{169}{5} = 33 \frac{4}{5 }$

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Question

Refer to the above right triangle. Which of the following is equal to ?

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Answer

By the Pythagorean Theorem,

$x^{2 } + 12 ^{2 } = 18 ^{2 }$

$x^{2 } + 144 = 324$

$x^{2 } + 144 -144 = 324-144$

$x^{2 } = 180$

$x = \sqrt{180} = \sqrt{36} \cdot \sqrt{5} = 6 \sqrt{5}$

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Question

Given $\Delta ABC$ with right angle $\angle B$ , $m \angle C = 45 ^{\circ }$

Which is the greater quantity?

(a)

(b)

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Answer

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

$m \angle C = 45 ^{\circ }$

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

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

$m \angle A +90 + 45 = 180$

$m \angle A +135= 180$

$m \angle A +135-135= 180-135$

$m \angle A = 45^{\circ }$

This is a $45^{\circ }-45^{\circ }-90^{\circ }$ triangle, so its legs $\overline{AB}$ and $\overline{BC}$ are congruent. The quantities are equal.

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Question

Give the length of one leg of an isosceles right triangle whose area is the same as the right triangle in the above diagram.

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Answer

The area of a triangle is half the product of its height and its base; in a right triangle, the legs, being perpendicular, can serve as these quantites.

The triangle in the diagram has area

$\frac{1}{2} \times 30 \times 18 = 270$ square inches.

An isosceles right triangle has two legs of the same length, which we will call . The area of that triangle, which is the same as that of the one in the diagram, is therefore

$\frac{1}{2} L ^{2} = 270$

$\frac{1}{2} L ^{2}\cdot 2 = 270 \cdot 2$

$L ^{2} = 540$

$L = \sqrt{540}$

$L = \sqrt{36}\cdot \sqrt{15} = 6 \sqrt{15}$ inches.

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Question

The perimeter of a regular octagon is 20% greater than that of the above right triangle. Which is the greater quantity?

(A) The length of one side of the octagon

(B) 3 yards

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Answer

By the Pythagorean Theorem, the shorter leg has length

$\sqrt{26^{2}-24^{2}} = \sqrt{676-576} = \sqrt{100} = 10$ feet.

The perimeter of the right triangle is therefore

feet.

The octagon has perimeter 20% greater than this, or

$60 + 0.20 \times 60 = 60 + 12 = 72$ feet.

A regular octagon has eight sides of equal length, so each side of this octagon has length

$72 \div 8 = 9$ feet, which is equal to 3 yards. This makes the quantities equal.

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Question

The area of a square is equal to that of the above right triangle. Which is the greater quantity?

(A) The sidelength of the square

(B) 4 yards

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Answer

By the Pythagorean Theorem, the shorter leg has length

$\sqrt{26^{2}-24^{2}} = \sqrt{676-576} = \sqrt{100} = 10$ feet.

The area of a triangle is equal to half the product of its base and height; for a right triangle, the legs can serve as these. The area of the above right triangle is

$\frac{1}{2} \times 10 \times 24 = 120$ square feet.

The sidelength is the square root of this; , so $\sqrt{120} < \sqrt{121} = 11$ . Therefore each sidelength of the square is just under 11 feet. 4 yards is 12 feet, so (B) is greater.

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