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

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Question

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

Answer

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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Question

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

Answer

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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Question

Note: Figure NOT drawn to scale.

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

Answer

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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Question

Note: Figure NOT drawn to scale.

In the above figure, .

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

Answer

and , so by the Pythagorean Theorem,

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

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

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

$= \sqrt{40,000}$

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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Question

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?

Answer

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:

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

$h = 30 \div 6.5 \approx 4.6$ 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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Question

Note: Figure NOT drawn to scale.

In the above figure, .

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

Answer

and , so by the Pythagorean Theorem,

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

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

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

$= \sqrt{40,000}$

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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Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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,

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

In the above figure, .

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

Answer

and , so by the Pythagorean Theorem,

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

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

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

$= \sqrt{40,000}$

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.

Compare your answer with the correct one above

Question

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?

Answer

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:

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

In the above figure, .

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

Answer

and , so by the Pythagorean Theorem,

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

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

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

$= \sqrt{40,000}$

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.

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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,

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

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

Answer

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

In the above figure, .

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

Answer

and , so by the Pythagorean Theorem,

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

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

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

$= \sqrt{40,000}$

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.

Compare your answer with the correct one above

Question

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?

Answer

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:

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

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

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

In the above figure, .

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

Answer

and , so by the Pythagorean Theorem,

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

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

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

$= \sqrt{40,000}$

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.

Compare your answer with the correct one above

Question

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

Answer

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

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

Compare your answer with the correct one above

Question

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

Answer

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,

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

Compare your answer with the correct one above

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