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ISEE Upper Level Math : Right Triangles

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Example Questions

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Example Question #2 : How To Find The Perimeter Of A Right Triangle

One leg of a right triangle measures $3\frac{1}{3}$ ; the hypotenuse measures $8 \frac{2}{3}$ . What is the perimeter of the triangle?

Possible Answers:

$20 \frac{2}{3}$

$19 \frac{1}{3}$

Correct answer:

Explanation:

The length of the second leg can be calculated using the Pythagorean Theorem. Set $c = 8 \frac{2}{3} = \frac{26}{3}, a =3\frac{1}{3} = \frac{10}{3}$ :

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

$b ^{2} = \left ( \frac{26}{3} \right ) ^{2}- \left ( \frac{10}{3} \right )^{2}$

$b ^{2} = \frac{676}{9}- \frac{100}{9}$

$b ^{2} = \frac{576}{9}$

$b ^{2} =64$

$b = \sqrt{64} = 8$

Add the three sidelengths to get the perimeter:

$P = 3\frac{1}{3} + 8 + 8 \frac{2}{3} = 20$

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Example Question #3 : How To Find The Perimeter Of A Right Triangle

Right triangle 2

$\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . If $BX = \sqrt{6}$ , give the perimeter of $\bigtriangleup ABC$ .

Possible Answers:

$8 \sqrt{2}$

$4 \sqrt{6} + 2\sqrt{2}$

$\sqrt{6} + 3\sqrt{2}$

$2 \sqrt{6} + 6\sqrt{2}$

Correct answer:

$2 \sqrt{6} + 6\sqrt{2}$

Explanation:

An altitude of a right triangle, which here is $\overline{BX}$ , divides the triangle into two triangles similar to each other and to the large triangle. Therefore, their corresponding angles are equal.

$\angle ABC$ is right - and has degree measure 90 - and $\angle C$ has degree measure 30, so $\angle A$ has degree measure 60, making the triangle a 30-60-90 triangle. Therefore, the smaller triangles are as well.

$\overline{BX}$ is the short leg of $\bigtriangleup BXC$ , so the hypotenuse, $\overline{BC}$ , has twice the length of that leg, or $2 \sqrt{6}$ .

$\overline{BC}$ is the long leg of $\bigtriangleup ABC$ ; short leg $\overline{AB}$ has as its length the length of $\overline{BC}$ divided by $\sqrt{3}$ , or $\frac{2 \sqrt{6}}{\sqrt{3}}= 2\sqrt{\frac{6}{3}} = 2\sqrt{2}$ .

The hypotenuse of $\bigtriangleup ABC$ , $\overline{AC}$ , has as its length twice the length of short leg $\overline{AB}$ , which is $2 \cdot 2\sqrt{2} = 4 \sqrt{2}$ .

Add the lengths of the sides of $\bigtriangleup ABC$ :

$2 \sqrt{6} + 2\sqrt{2} + 4\sqrt{2}= 2 \sqrt{6} + 6\sqrt{2}$

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

Right_triangle

Note: Figure NOT drawn to scale.

In the above figure, AB = 120, BC = 160 .

Which of the following comes closest to the length of $\overline{DF}$ ?

Possible Answers:

Correct answer:

Explanation:

AB = 120 and BC = 160 , so by the Pythagorean Theorem,

$AC = \sqrt{(AB)^{2}+(BC)^{2}}$

$= \sqrt{120+160^{2}}$

$= \sqrt{14,400+25,600 }$

$= \sqrt{40,000}$

= 200

Because $\overline{DB}$ is the altitude from the vertex of $\Delta ABC$ ,

$\Delta ADB \sim \Delta ABC$ .

Therefore,

$\frac{DB}{BC} = \frac{AB}{AC}$

$\frac{DB}{160} = \frac{120}{200}$

$DB = \frac{120}{200} \cdot 160 = 96$

Also,

$\frac{AD}{AB} = \frac{AB}{AC}$

$\frac{AD}{120} = \frac{120}{200}$

$AD = \frac{120}{200} \cdot 120 = 72$

For similar reasons,

$\Delta ADB \sim \Delta DFB$ .

Therefore,

$\frac{DF}{AD}= \frac{DB}{AB}$

$\frac{DF}{72}= \frac{96}{120}$

$DF= \frac{96}{120} \cdot 72 = 57.6$

Of the choices given, 60 comes closest.

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Example Question #61 : Geometry

Right_triangle

Note: Figure NOT drawn to scale.

In the above figure, AB = 120, BC = 160 .

Which of the following comes closest to the length of $\overline{ED}$ ?

Possible Answers:

Correct answer:

Explanation:

AB = 120 and BC = 160 , so by the Pythagorean Theorem,

$AC = \sqrt{(AB)^{2}+(BC)^{2}}$

$= \sqrt{120+160^{2}}$

$= \sqrt{14,400+25,600 }$

$= \sqrt{40,000}$

= 200

Because $\overline{DB}$ is the altitude from the vertex of $\Delta ABC$ ,

$\Delta BDC \sim \Delta ABC$ .

Therefore,

$\frac{BD}{AB} = \frac{BC}{AC}$

$\frac{DB}{120} = \frac{160}{200}$

$DB = \frac{160}{200} \cdot 120 = 96$

Also,

$\frac{CD}{CB} = \frac{BC}{AC}$

$\frac{CD}{160} = \frac{160}{200}$

$CD = \frac{160}{200} \cdot 160 = 128$

For similar reasons,

$\Delta BDC \sim \Delta DEC$ .

Therefore,

$\frac{ED}{DB}= \frac{CD}{CB}$

$\frac{ED}{96}= \frac{128}{160}$

$ED = \frac{128}{160} \cdot 96 = 76.8$ .

Of the choices given, 75 comes closest.

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

Right_triangle

Refer to the above figure. Evaluate the length of $\overline{BD}$ in terms of .

Possible Answers:

$\frac{\sqrt{n^{2}+36}}{6n}$

$\frac{3n}{\sqrt{n^{2}+36}}$

$\frac{\sqrt{n^{2}+36}}{3n}$

$\frac{6n}{\sqrt{n^{2}+36}}$

$\frac{\sqrt{n^{2}+36}}{12n}$

Correct answer:

$\frac{6n}{\sqrt{n^{2}+36}}$

Explanation:

The altitude of a right triangle to its hypotenuse divides the triangle into two smaller trangles similar to each other and to the large triangle.

Therefore,

$\Delta ABC \sim \Delta ADB$

and, consequently,

$\frac{DB}{BC} = \frac{AB}{AC}$ ,

or, equivalently,

$DB = \frac{AB \cdot BC}{AC}$

$AC = \sqrt{ n^{2}+36}$ by the Pythagorean Theorem, so

$DB = \frac{6n }{\sqrt{ n^{2}+36}}$ .

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Example Question #4 : How To Find The Height Of A Right Triangle

Right_triangle

Refer to the above figure. Evaluate the length of $\overline{BD}$ in terms of .

Possible Answers:

$\frac{n+6}{2}$

$\sqrt{n+6}$

$\sqrt{6n}$

$\sqrt{\frac{n+6}{2}}$

Correct answer:

$\sqrt{6n}$

Explanation:

The height of a right triangle from the vertex of its right angle is the geometric mean - in this case, the square root of the product - of the lengths of the two segments of the hypotenuse that it forms. Therefore,

$BD = \sqrt{(AD)(CD)} = \sqrt{6n}$

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Example Question #5 : How To Find The Height Of A Right Triangle

Right_triangle

Note: Figure NOT drawn to scale.

In the above right triangle, AC=10 . Give the length of $\overline{BD}$ .

Possible Answers:

$\frac{5\sqrt{3}}{3}$

$\frac{5\sqrt{2}}{3}$

$\frac{5\sqrt{3}}{2}$

$\frac{5\sqrt{2}}{2}$

Correct answer:

$\frac{5\sqrt{3}}{2}$

Explanation:

The two triangles formed by an altitude from the vertex of a right triangle are similar to each other and the large triangle, so all three are 30-60-90 triangles. Take advantage of this, applying it twice.

Looking at $\Delta ABC$ . By the 30-60-90 Theorem, the shorter leg of a hypotentuse measures half that of the hypotenuse. $AB = \frac{1}{2} AC = \frac{1}{2} \cdot 10 = 5$ .

Now, look at $\Delta ADB$ . By the same theorem,

$AD = \frac{1}{2} AB = \frac{1}{2} \cdot 5 = \frac{5}{2}$

and

$BD = AD \cdot \sqrt{3}= \frac{5}{2}\cdot \sqrt{3} = \frac{5\sqrt{3}}{2}$

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

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above map. A farmer owns a triangular plot of land flanked by the three highways as shown. The farmer uses Highway 2 frequently; however, he can only access it by driving five miles north on Highway 32 or twelve miles east on Highway 100.

He wants to build a dirt road that directly accesses Highway 2. He figures that it will cost $1,500 per tenth of a mile to construct the road. By his estimate, which answer will come closest to the total cost of the shortest possible road?

Possible Answers:

$\$50,000$

$\$80,000$

$\$70,000$

$\$60,000$

$\$40,000$

Correct answer:

$\$70,000$

Explanation:

The three roads form the legs and the hypotenuse of a right triangle with legs 5 and 12 miles; by the Pythagorean Theorem, the hypotenuse is

$\sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt {169}= 13$ miles.

To find the length of the shortest possible road, which must be perpendicular to Highway 2, it should be observed that this road serves as an altitude from the vertex of the triangle to the hypotenuse.

The area of a right triangle can be calculated two ways - by taking half the product of the lengths of the legs or by taking half the product of the length of the altitude (height) and that of the hypotenuse. Setting as the height, we can solve for in the equation:

$\frac{1}{2} \cdot 13 \cdot h = \frac{1}{2} \cdot 5 \cdot 12$

6.5 h = 30

$h = 30 \div 6.5 \approx 4.6$ miles.

This is 46 tenths of a mile, so the approximate cost of the road in dollars will be

$46 \times 1,500 = 69,000$

Among the five choices, $70,000 comes closest.

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Example Question #1 : How To Find An Angle In A Right Triangle

Which of the following is true about a triangle with two angles that measure $45^{\circ}$ and $45^{\circ}$ ?

Possible Answers:

This is an equilateral triangle.

This triangle is scalene and right.

This triangle is scalene.

This triangle is obtuse.

This is a right isosceles triangle.

Correct answer:

This is a right isosceles triangle.

Explanation:

The sum of the two given angles is 90 degrees, which means that the third angle should be a right angle (90 degrees). We also know that two of the angles are equal. Therefore, the triangle is right and isosceles.

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Example Question #71 : Isee Upper Level (Grades 9 12) Mathematics Achievement

What is the value of x in a right triangle if the two acute angles are equal to 5x and 25x?

Possible Answers:

Correct answer:

Explanation:

In a right triangle, there is one right angle of 90 degrees, while the two acute angles add up to 90 degrees.

Given that the two acute angles are equal to 5x and 25x, the value of x can be calculated with the equation below:

5x+25x=90

30x=90

x=3

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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 #2 : How To Find The Perimeter Of A Right Triangle

Example Question #3 : How To Find The Perimeter Of A Right Triangle

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

Example Question #61 : Geometry

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

Example Question #4 : How To Find The Height Of A Right Triangle

Example Question #5 : How To Find The Height Of A Right Triangle

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

Example Question #1 : How To Find An Angle In A Right Triangle

Example Question #71 : Isee Upper Level (Grades 9 12) Mathematics Achievement

All ISEE Upper Level Math Resources

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