Right Triangles - ISEE Upper Level Quantitative Reasoning

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Question

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

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 .

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

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

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?

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

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

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 .

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)

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 $\overline{AC}$ ?

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 $\overline{AB}$ ?

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) $90^{\circ }$

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 $\overline{AC}$ ?

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

A right triangle has a hypotenuse of 10 and a side of 6. What is the missing side?

Answer

To find the missing side, use the Pythagorean Theorem . Plug in (remember c is always the hypotenuse!) so that . Simplify and you get Subtract 36 from both sides so that you get Take the square root of both sides. B is 8.

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

Find the length of $\overline{BC}$ .

Answer

First, find .

Since $\overline{DF }$ is an altitude of right $\Delta ADB$ to its hypotenuse,

$\frac{FB}{FD}= \frac{FD}{AF}$

$\frac{FB}{8}= \frac{8}{6}$

$FB= \frac{8}{6} \cdot 8 = 10 \frac{2}{3}$

$AB = AF + FB = 6 + 10 \frac{2}{3} = 16 \frac{2}{3}$

$\Delta AFD \sim \Delta ABC$ by the Angle-Angle Postulate, so

$\frac{BC}{FD} = \frac{AB}{AF}$

$\frac{BC}{8} = \frac{16 \frac{2}{3}}{6}$

$BC = 16 \frac{2}{3}\cdot 8 \div 6 = 22\frac{2}{9}$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

Find the length of $\overline{AB}$ .

Answer

First, find .

Since $\overline{DE}$ is an altitude of $\Delta CDB$ from its right angle to its hypotenuse,

$\Delta DEC \sim \Delta BED$

$\frac{BE}{DE}= \frac{ED}{EC}$

$\frac{BE}{6}= \frac{6}{12}$

$BE= \frac{6}{12} \cdot 6 = 3$

$\Delta DEC \sim \Delta ABC$ by the Angle-Angle Postulate, so

$\frac{AB}{DE} = \frac{CB}{CE}$

$\frac{AB}{6} = \frac{15}{12}$

$AB= \frac{15}{12} \cdot 6$

$AB = 7\frac{1}{2}$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

Answer

By the Pythagorean Theorem,

$x^{2}+ 15^{2} = (x+12)^{2}$

$x^{2}+ 225 = x^{2}+24x+144$

$x= 3\frac{3}{8}$

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Which of the following quadratic equations would yield the value of as a solution?

Answer

By the Pythagorean Theorem,

$x^{2}+ (x+5)^{2} = (x+7)^{2}$

$x^{2}+ x^{2} +10x+25 = x^{2} +14x +49$

$2x^{2} +10x+25 = x^{2} +14x +49$

$x^{2} -4x-24= 0$

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Question

Refer to the above diagram. Which of the following quadratic equations would yield the value of as a solution?

Answer

By the Pythagorean Theorem,

$x^{2}+ (x+3)^{2} = 20^{2}$

$x^{2}+ x^{2}+6x+9= 400$

$2x^{2}+6x-391 = 0$

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Question

A right triangle $\Delta ABC$ with hypotenuse $\overline{AC}$ is inscribed in $\odot O$ , a circle with radius 26. If , evaluate the length of $\overline{BC}$ .

Answer

The arcs intercepted by a right angle are both semicircles, so hypotenuse $\overline{AC}$ shares its endpoints with two semicircles. This makes $\overline{AC}$ a diameter of the circle, and $AC = 2r = 2 \cdot 26 = 52$ .

By the Pythagorean Theorem,

$BC = \sqrt{(AC)^{2} - (AB)^{2}} = \sqrt{52^{2} -20^{2}} = \sqrt{2,704-400} = \sqrt{2,304} = 48$

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