How to find the length of the side of a right triangle - ISEE Upper Level Quantitative Reasoning

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Question

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

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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 $\overline{BC}$ .

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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 $\overline{AB}$ .

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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 .

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Answer

By the Pythagorean Theorem,

$x= 3$\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?

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Answer

By the Pythagorean Theorem,

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Question

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

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Answer

By the Pythagorean Theorem,

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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 $\overline{BC}$ .

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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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Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

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

Tap to reveal answer

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 $\overline{AB}$ .

Tap to reveal answer

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 .

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Answer

By the Pythagorean Theorem,

$x= 3$\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?

Tap to reveal answer

Answer

By the Pythagorean Theorem,

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Question

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

Tap to reveal answer

Answer

By the Pythagorean Theorem,

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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 $\overline{BC}$ .

Tap to reveal answer

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$

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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.

← Didn't Know|Knew It →

Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

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

Tap to reveal answer

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}$$

← Didn't Know|Knew It →

Question

Note: Figure NOT drawn to scale.

Refer to the above diagram.

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

Tap to reveal answer

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 .

Tap to reveal answer

Answer

By the Pythagorean Theorem,

$x= 3$\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?

Tap to reveal answer

Answer

By the Pythagorean Theorem,

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Question

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

Tap to reveal answer

Answer

By the Pythagorean Theorem,

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