How to find the length of the side of a right triangle - ISEE Upper Level Mathematics Achievement

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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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