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ISEE Upper Level Quantitative : Right Triangles

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Example Questions

Example Question #62 : Triangles

Consider a triangle, $\bigtriangleup ABC$ , in which AB = 36 , BC = 48 , and AC = 60 . Which is the greater number?

(a) The measure of $\angle B$ in degrees

(b)

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:

(a) and (b) are equal

Explanation:

By the Converse of the Pythagorean Theorem, a triangle is right if and only if the sum of the squares of the lengths of the smallest two sides is equal to the square of the longest side. Compare the quantities $(AB)^{2}+ ( BC )^{2}$ and $(AC)^{2}$

$(AB)^{2}+ ( BC )^{2} = 36 ^{2} + 48^{2} = 1,296 + 2,304 = 3,600$

$(AC)^{2} = 60^{2} = 3,600$

$(AB)^{2}+ ( BC )^{2} =(AC)^{2}$ , so $\bigtriangleup ABC$ is right, with the right angle opposite longest side $\overline{AC}$ . Thus, $\angle B$ is right and has degree measure 90.

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Example Question #31 : Right Triangles

The length of a side of a square is two-thirds the length of a leg of an isosceles right triangle. Which is the greater quantity?

(a) The area of the square

(b) The area of the triangle

Possible Answers:

It is impossible to tell from the information given

(a) and (b) are equal

(b) is greater

(a) is greater

Correct answer:

(b) is greater

Explanation:

Let be the length of a leg of the right triangle. Then the sidelength of the square is $\frac{2}{3}s$ .

(a) The square has area $A = \left ( \frac{2}{3} s \right ) ^{2} = \frac{4}{9} s ^{2}$

(b) The isosceles right triangle has base and height area $A = \frac{1}{2}s^{2}$

$\frac{1}{2} > \frac{4}{9}$ , so (b) is greater.

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Example Question #64 : Triangles

Two triangles are on the coordinate plane. Each has a vertex at the origin.

Triangle A has its other two vertices at $\left (18,0 \right )$ and (0,b) .

Triangle B has its other two vertices at $\left (2b,0 \right )$ and (0,9) .

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

It is impossible to tell from the information given

(b) is greater

Correct answer:

(a) and (b) are equal

Explanation:

Each triangle is a right triangle with legs along the - and -axes, so the area of each can be calculated by taking one-half the product of the two legs.

(a) The horizontal and vertical legs have measures 18 and , respectively, so the triangle has area $A = \frac{1}{2} \cdot 18 \cdot b = 9b$ .

(b) The horizontal and vertical legs have measures and 9, respectively, so the triangle has area $A = \frac{1}{2} \cdot 2b \cdot 9= 9b$ .

The areas are equal.

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Example Question #66 : Geometry

Construct rectangle RECT , and locate midpoint of side $\overline{RE}$ . Now construct segment $\overline{MC}$ .

Which is the greater quantity?

(a) The area of Quadrilateral RMCT

(b) Three times the area of $\Delta MEC$

Possible Answers:

It is impossible to tell from the information given

(a) and (b) are equal

(b) is greater

(a) is greater

Correct answer:

(a) and (b) are equal

Explanation:

$\Delta MEC$ is a right triangle with right angle $\angle E$ , so its legs measure and ; its area is $\frac{1}{2}\left ( ME \right ) \left (EC \right )$ . Since is the midpoint of $\overline{RE}$ , $ME = \frac{1}{2} \left ( RE \right )$ , making the area of the triangle

$\frac{1}{2}\left ( ME \right ) \left (EC \right ) = \frac{1}{2} \cdot \frac{1}{2} \left ( RE \right ) \left (EC \right ) = \frac{1}{4} \left ( RE \right ) \left (EC \right )$

Rectangle RECT has area $\left ( RE \right ) \left (EC \right )$ .

Quadrilateral RMCT , which is the portion of RECT not in $\Delta MEC$ , has as its area

$\left ( RE \right ) \left (EC \right ) - \frac{1}{4} \left ( RE \right ) \left (EC \right ) = \frac{3}{4} \left ( RE \right ) \left (EC \right )$

Therefore, the area of Quadrilateral RMCT is three times that of $\Delta MEC$ , making (a) and (b) equal.

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Example Question #67 : Geometry

Construct rectangle RECT . Let and be the midpoints of $\overline{RE}$ and $\overline{EC}$ , respectively, and draw the segments $\overline{MC}$ and $\overline{RN }$ . Which is the greater quantity?

(a) The area of $\Delta MEC$

(b) The area of $\Delta NER$

Possible Answers:

It is impossible to tell from the information given.

(b) is greater.

(a) and (b) are equal.

(a) is greater.

Correct answer:

(a) and (b) are equal.

Explanation:

Each triangle is a right triangle, and each has its two legs as its base and height.

(a) is the midpoint of $\overline{RE}$ , so $ME = \frac{1}{2} RE$ .

The area of $\Delta MEC$ is $A = \frac{1}{2}bh= \frac{1}{2} \left ( ME \right ) \left ( EC\right )= \frac{1}{2} \left ( \frac{1}{2} RE \right ) \left ( EC\right ) =\frac{1}{4} \left ( RE \right ) \left ( EC\right )$ .

(b) is the midpoint of $\overline{EC}$ , so $NE = \frac{1}{2} EC$ .

The area of $\Delta NER$ is

$A = \frac{1}{2}bh= \frac{1}{2} \left ( RE \right ) \left ( NE \right )= \frac{1}{2} \left ( \frac{1}{2} RE \right ) \left ( \frac{1}{2} EC\right ) =\frac{1}{4} \left ( RE \right ) \left ( EC\right )$ .

The triangles have equal area.

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Example Question #65 : Triangles

The length of a side of a square is one-half the length of the hypotenuse of a $30 ^{\circ }- 60 ^{\circ }-90 ^{\circ }$ triangle. Which is the greater quantity?

(a) The area of the square

(b) The area of the triangle

Possible Answers:

(a) and (b) are equal.

It is impossible to tell from the information given.

(b) is greater.

(a) is greater.

Correct answer:

(a) is greater.

Explanation:

(a) Let be the sidelength of the square. Then its area is $A = s^{2}$ .

(b) In a $30 ^{\circ }- 60 ^{\circ }-90 ^{\circ }$ triangle, the shorter leg is one-half as long as the hypotenuse. The hypotenuse has length , so the shorter leg has length . The longer leg is $\sqrt{3}$ times as long as the shorter leg, so the longer leg will have length $s\sqrt{3}$ . The area of the triangle is

$A = \frac{1}{2} bh = \frac{1}{2} s \cdot s\sqrt{ 3} = \frac{\sqrt{ 3}}{2} s^{2}$ .

$\frac{\sqrt{3} }{2}\approx \frac{1.7 }{2} < 1$ , so $\frac{\sqrt{ 3}}{2} s^{2} < s^{2}$ ; the square has the greater area.

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Example Question #66 : Triangles

Right_triangle

Give the area of the above right triangle in terms of .

Possible Answers:

$k^{2}-2$

$k^{2}-4$

$\frac{1}{2} k^{2}-2$

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

$2k ^{2} +8$

Correct answer:

$\frac{1}{2} k^{2}-2$

Explanation:

The area of a triangle is half the product of its base and its height; for a right triangle, the legs, which are perpendicular, serve as base and height.

$A = \frac{1}{2} (k+2) (k-2)$

$A = \frac{1}{2} (k^{2}-2^{2})$

$A = \frac{1}{2} (k^{2}-4)$

$A = \frac{1}{2} k^{2}-2$

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Example Question #67 : Triangles

Triangle

Square

Note: Figures NOT drawn to scale.

Refer to the above figures - a right triangle and a square. The area of the triangle is what percent of the area of the square?

Possible Answers:

$79 \frac{1}{6} \%$

$93 \frac{3}{4} \%$

$50 \%$

$46 \frac{7}{8} \%$

$100 \%$

Correct answer:

$46 \frac{7}{8} \%$

Explanation:

The area of the triangle is

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

The sidelength of the square is $2 \times 12 = 24$ inches, so the area of the square is

$A = 24 ^{2} = 24 \times 24 = 576$ .

The question becomes "what percent of 576 is 270", which is answered as follows:

$P = \frac{270}{576} \times 100 = 46 \frac{7}{8}$

The correct answer is $46 \frac{7}{8} \%$ .

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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 : Geometry

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:

(a) is the greater quantity

(b) is the greater quantity

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

(a) and (b) are equal

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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All ISEE Upper Level Quantitative Resources

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ISEE Upper Level Quantitative : Right Triangles

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

All ISEE Upper Level Quantitative Resources

Example Questions

Example Question #62 : Triangles

Example Question #31 : Right Triangles

Example Question #64 : Triangles

Example Question #66 : Geometry

Example Question #67 : Geometry

Example Question #65 : Triangles

Example Question #66 : Triangles

Example Question #67 : Triangles

Example Question #71 : Triangles

Example Question #71 : Geometry

All ISEE Upper Level Quantitative Resources

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