Right Triangles - ISEE Upper Level Mathematics Achievement

Card 0 of 20

Question

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

Answer

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.

Compare your answer with the correct one above

Question

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

Answer

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:

Compare your answer with the correct one above

Question

$\Delta ABC \sim \Delta DEF$

$\angle B$ is a right angle; , .

Which is the greater quantity?

(a) $m \angle A$

(b) $60 ^{\circ }$

Answer

$\Delta ABC \sim \Delta DEF$ . Corresponding angles of similar triangles are congruent, so since $\angle B$ is a right angle, so is $\angle E$ .

The hypotenuse $\overline{DF}$ of $\Delta DEF$ is twice as long as leg $\overline{DE}$ ; by the $30 ^{\circ } - 60^{\circ } -90^{\circ }$ Theorem, $m \angle D = 60 ^{\circ }$ . Again, by similiarity,

$m \angle A = m \angle D = 60 ^{\circ }$ .

Compare your answer with the correct one above

Question

One angle of a right triangle measures 45 degrees, and the hypotenuse measures 8 centimeters. Give the area of the triangle.

Answer

This triangle has two angles of 45 and 90 degrees, so the third angle must measure 45 degrees; this is therefore an isosceles right triangle.

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let hypotenuse and side length.

$c^2=s^2+s^2\Rightarrow c^2=2s^2\Rightarrow 8^2=2s^2\Rightarrow s^2=32$

$\Rightarrow s=\sqrt{32}=\sqrt{16\times 2}=4\sqrt{2}\ cm$

We can then plug this side length into the formula for area.

$area=\frac{s^2}{2}=\frac{(4\sqrt{2})^2}{2}=\frac{32}{2}=16\ cm^2$

Compare your answer with the correct one above

Question

The legs of a right triangle measure $2 \frac{2}{5}$ and $3 \frac{1}{5}$ . What is its perimeter?

Answer

The hypotenuse of the triangle can be calculated using the Pythagorean Theorem. Set $a = 2 \frac{2}{5} = \frac{12}{5} , b=3 \frac{1}{5}= \frac{16}{5}$ :

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

$c ^{2} = \left ( \frac{12}{5} \right ) ^{2} + \left ( \frac{16}{5} \right ) ^{2}$

$c ^{2} = \frac{144}{25} + \frac{256}{25}$

$c ^{2} = \frac{400}{25} = 16$

$c = \sqrt{16} = 4$

Add the three sidelengths:

$P =2 \frac{2}{5} + 3 \frac{1}{5} + 4 = 9 \frac{3}{5}$

Compare your answer with the correct one above

Question

Figure NOT drawn to scale

$\bigtriangleup ABC$ is a right triangle with altitude $\overline{BX}$ . Give the ratio of the area of $\bigtriangleup BXC$ to that of $\bigtriangleup AXB$ .

Answer

The altitude of a right triangle from the vertex of its right angle - which, here, is $\overline{BX}$ - divides the triangle into two triangles similar to each other. The ratio of the hypotenuse of the white triangle to that of the gray triangle (which are corresponding sides) is

$\frac{BC}{AB} = \frac{40}{30} =\frac{4}{3}$ ,

making this the similarity ratio. The ratio of the areas of two similar triangles is the square of their similarity ratio, which here is

$\left ( \frac{4}{3} \right )^{2} = \frac{4^{2}}{3^{2}} = \frac{16}{9}$ , or .

Compare your answer with the correct one above

Question

Find the area of a right triangle with a base of 7cm and a height of 20cm.

Answer

To find the area of a right triangle, we will use the following formula:

$A = \frac{1}{2} \cdot b \cdot h$

where b is the base and h is the height of the triangle.

We know the base of the triangle is 7cm. We know the height of the triangle is 20cm. Knowing this, we can substitute into the formula. We get

$A = \frac{1}{2} \cdot 7\text{cm} \cdot 20\text{cm}$

$A = 7\text{cm} \cdot 10\text{cm}$

$A = 70\text{cm}^2$

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 diagram, which depicts a right triangle. What is the value of ?

Answer

By the Pythagorean Theorem, which says . being the hypotenuse, or in this problem.

$x^{2} = 6^{2} + 9^{2}$

$x^{2} = 36 + 81$

$x^{2} = 117$

$x =\sqrt{ 117}$

Simply

$\sqrt{ 9} \cdot \sqrt{ 13}$

$3 \sqrt{ 13}$

Compare your answer with the correct one above

Question

If a right triangle has a base of and a height of , what is the length of the hypotenuse?

Answer

To solve this problem, we must utilize the Pythagorean Theorom, which states that:

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

We know that the base is , so we can substitute in for . We also know that the height is , so we can substitute in for .

$7^{2}+4^{2}=c^{2}$

Next we evaluate the exponents:

$7^{2}=49$

$4^{2}=16$

Now we add them together:

Then, $65=c^{2}$ .

is not a perfect square, so we simply write the square root as $\sqrt{65}$ .

$c=\sqrt{65}$

Compare your answer with the correct one above

Question

If a right triangle has a base of and a height of , what is the length of the hypotenuse?

Answer

To solve this problem, we are going to use the Pythagorean Theorom, which states that $a^{2}+b^{2}=c^{2}$ .

We know that this particular right triangle has a base of , which can be substituted for , and a height of , which can be substituted for . If we rewrite the theorom using these numbers, we get:

$3^{2}+9^{2}=c^{2}$

Next, we evaluate the expoenents:

$3^{2}=9$

$9^{2}=81$

$9+81=c^{2}$

Then, $90=c^{2}$ .

To solve for , we must find the square root of . Since this is not a perfect square, our answer is simply $c=\sqrt{90}$ .

Compare your answer with the correct one above

Question

What is the hypotenuse of a right triangle with sides 5 and 8?

Answer

According to the Pythagorean Theorem, the equation for the hypotenuse of a right triangle is $a^{2} + b^{2}=c^{2}$ . Plugging in the sides, we get $5^{2}+8^{2}=c^{2}$ . Solving for , we find that the hypotenuse is $\sqrt{89}$ :

Compare your answer with the correct one above

Question

In a right triangle, two sides have length . Give the length of the hypotenuse in terms of .

Answer

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let hypotenuse and side length.

$c^2=s^2+s^2\Rightarrow c^2=(2t)^2+(2t)^2\Rightarrow c^2=8t^2\Rightarrow c=\sqrt{8t^2}=2\sqrt{2}t$

Compare your answer with the correct one above

Question

In a right triangle, two sides have lengths 5 centimeters and 12 centimeters. Give the length of the hypotenuse.

Answer

This triangle has two angles of 45 and 90 degrees, so the third angle must measure 45 degrees; this is therefore an isosceles right triangle.

By the Pythagorean Theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let hypotenuse and , lengths of the other two sides.

$c^2=a^2+b^2\Rightarrow c^2=5^2+12^2\Rightarrow c^2=25+144=169$

$\Rightarrow c=\sqrt{169}\Rightarrow c=13\ cm$

Compare your answer with the correct one above

Question

In a rectangle, the width is 6 feet long and the length is 8 feet long. If a diagonal is drawn through the rectangle, from one corner to the other, how many feet long is that diagonal?

Answer

Given that a rectangle has all right angles, drawing a diagonal will create a right triangle the legs are each 6 feet and 8 feet.

We know that in a 3-4-5 right triangle, when the legs are 3 feet and 4 feet, the hypotenuse will be 5 feet.

Given that the legs of this triangle are twice as long as those in the 3-4-5 triangle, it follows that the hypotense will also be twice as long.

Thus, the diagonal in through the rectangle creates a 6-8-10 triangle. 10 is therefore the length of the diagonal.

Compare your answer with the correct one above

Tap the card to reveal the answer