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ISEE Upper Level Math : Right Triangles

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Example Questions

Example Question #4 : Triangles

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

Possible Answers:

Correct answer:

Explanation:

To find the missing side, use the Pythagorean Theorem (a^2+b^2=c^2) . Plug in (remember c is always the hypotenuse!) so that 6^2+b^2=10^2 . Simplify and you get 36+b^2=100. Subtract 36 from both sides so that you get b^2=64. Take the square root of both sides. B is 8.

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Example Question #44 : Geometry

Right_triangle

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

Possible Answers:

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

$x^{2}+6x+409 = 0$

$2x^{2}+6x+409 = 0$

$2x^{2}+6x-11 = 0$

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

Correct answer:

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

Explanation:

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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Example Question #1 : How To Find The Length Of The Side Of A Right Triangle

Right_triangle

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?

Possible Answers:

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

$x^{2} -4x+74= 0$

$x^{2} +4x+74= 0$

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

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

Correct answer:

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

Explanation:

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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Example Question #1 : Right Triangles

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram.

AF = 6 , FD = 8

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

Possible Answers:

$22\frac{2}{9}$

$16\frac{2}{3}$

$33\frac{1}{3}$

Correct answer:

$22\frac{2}{9}$

Explanation:

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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Example Question #51 : Geometry

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram.

DE = 6, EC = 12

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

Possible Answers:

$7\frac{1}{2}$

$7\frac{1}{4}$

$7\frac{3}{4}$

Correct answer:

$7\frac{1}{2}$

Explanation:

First, find .

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

$\Delta DEC \sim \Delta BED$

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

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

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

CB =BE + EC = 3+ 12 = 15

$\Delta DEC \sim \Delta ABC$ 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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Example Question #4 : Triangles

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

Possible Answers:

$30\frac{3}{4}$

$3\frac{3}{8}$

$15\frac{3}{8}$

$6\frac{3}{4}$

Correct answer:

$3\frac{3}{8}$

Explanation:

By the Pythagorean Theorem,

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

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

225 = 24x+144

24x= 81

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

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Example Question #52 : Geometry

A right triangle $\Delta ABC$ with hypotenuse $\overline{AC}$ is inscribed in $\odot O$ , a circle with radius 26. If AB = 20 , evaluate the length of $\overline{BC}$ .

Possible Answers:

Insufficient information is given to answer the question.

Correct answer:

Explanation:

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

$\Delta ABC \sim \Delta DEF$

$\angle B$ is a right angle; DE = 21 , DF = 42 .

Which is the greater quantity?

(a) $m \angle A$

(b) $60 ^{\circ }$

Possible Answers:

(b) is greater.

(a) and (b) are equal.

(a) is greater.

It is impossible to tell from the information given.

Correct answer:

(a) and (b) are equal.

Explanation:

$\Delta ABC \sim \Delta DEF$ . Corresponding angles of similar triangles are congruent, so since $\angle B$ is a right angle, so is $\angle E$ .

The hypotenuse $\overline{DF}$ of $\Delta DEF$ is twice as long as leg $\overline{DE}$ ; by the $30 ^{\circ } - 60^{\circ } -90^{\circ }$ Theorem, $m \angle D = 60 ^{\circ }$ . Again, by similiarity,

$m \angle A = m \angle D = 60 ^{\circ }$ .

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Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

If a right triangle has a base of and a height of , what is the length of the hypotenuse?

Possible Answers:

$\sqrt{14}$

$\sqrt{11}$

$\sqrt{65}$

Correct answer:

$\sqrt{65}$

Explanation:

To solve this problem, we must utilize the Pythagorean Theorom, which states that:

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

We know that the base is , so we can substitute in for . We also know that the height is , so we can substitute in for .

$7^{2}+4^{2}=c^{2}$

Next we evaluate the exponents:

$7^{2}=49$

$4^{2}=16$

Now we add them together:

49+16=65

Then, $65=c^{2}$ .

is not a perfect square, so we simply write the square root as $\sqrt{65}$ .

$c=\sqrt{65}$

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Example Question #51 : Geometry

If a right triangle has a base of and a height of , what is the length of the hypotenuse?

Possible Answers:

$\sqrt{90}$

144

Correct answer:

$\sqrt{90}$

Explanation:

To solve this problem, we are going to use the Pythagorean Theorom, which states that $a^{2}+b^{2}=c^{2}$ .

We know that this particular right triangle has a base of , which can be substituted for , and a height of , which can be substituted for . If we rewrite the theorom using these numbers, we get:

$3^{2}+9^{2}=c^{2}$

Next, we evaluate the expoenents:

$3^{2}=9$

$9^{2}=81$

$9+81=c^{2}$

9+81=90

Then, $90=c^{2}$ .

To solve for , we must find the square root of . Since this is not a perfect square, our answer is simply $c=\sqrt{90}$ .

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Right Triangles

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #4 : Triangles

Example Question #44 : Geometry

Example Question #1 : How To Find The Length Of The Side Of A Right Triangle

Example Question #1 : Right Triangles

Example Question #51 : Geometry

Example Question #4 : Triangles

Example Question #52 : Geometry

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

Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

Example Question #51 : Geometry

All ISEE Upper Level Math Resources

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