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?

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,

$x= 3$\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,

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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,

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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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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.

Compare your answer with the correct one above

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}$$

Compare your answer with the correct one above

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}$$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

Answer

By the Pythagorean Theorem,

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

Compare your answer with the correct one above

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,

Compare your answer with the correct one above

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,

Compare your answer with the correct one above

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$

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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}$$

Compare your answer with the correct one above

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}$$

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

Answer

By the Pythagorean Theorem,

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

Compare your answer with the correct one above

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,

Compare your answer with the correct one above

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,

Compare your answer with the correct one above

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