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ISEE Upper Level Math : Plane Geometry

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Example Questions

Example Question #61 : Plane Geometry

In a right triangle, the legs are 7 feet long and 12 feet long. How long is the hypotenuse?

Possible Answers:

$3\sqrt{19$

193

$\sqrt{193}$

Correct answer:

$\sqrt{193}$

Explanation:

The pythagorean theory should be used to solve this problem.

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

$7^{2}+12^{2}+=c^{2}$

$49+144=c^{2}$

$193=c^{2}$

$\sqrt{193}=c$

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Example Question #62 : Plane Geometry

The legs of a right triangle are equal to 4 and 5. What is the length of the hypotenuse?

Possible Answers:

$\sqrt{41}$

Correct answer:

$\sqrt{41}$

Explanation:

If the legs of a right triangle are 4 and 5, to find the hypotenuse, the following equation must be used to find the hypotenuse, in which c is equal to the hypotenuse:

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

$4^2{+5^{2}}=c^2$

16+25=c^2

$41=c^{2}$

$c={\sqrt{41}}$

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Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

Right_triangle

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

Possible Answers:

$10\sqrt{2}$

$9\sqrt{3}$

$3 \sqrt{ 13}$

Correct answer:

$3 \sqrt{ 13}$

Explanation:

By the Pythagorean Theorem, which says a^2+b^2=c^2 . being the hypotenuse, or in this problem.

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

$x^{2} = 36 + 81$

$x^{2} = 117$

$x =\sqrt{ 117}$

Simply

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

$3 \sqrt{ 13}$

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Example Question #3 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

At your local university, there is a triangular walking path around a grassy area. If the two legs of the triangle form a ninety degree angle, and are each 30 meters long, what is the length of the hypotenuse of the triangle?

Possible Answers:

35m

45.6m

42m

30m

Correct answer:

42m

Explanation:

This problem asks us to find the hypotenuse of a right triangle. To do so, we can use Pythagorean Theorem.

a^2+b^2=c^2

Where a and b are the short sides of the triangle, and c is our hypotenuse.

We know a and b are both 30 meters, so plug that in and begin!

(30m)^2+(30m)^2=c^2

c^2=1800m^2

$c=\sqrt{1800m^2}=\sqrt{2*900m^2}=\sqrt{2}\sqrt{900m^2}=30\sqrt{2}m\approx42m$

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Example Question #62 : Geometry

In a right triangle, if the base of the triangle is 3, and the height is 5, what must be the length of the hypotenuse?

Possible Answers:

$5\sqrt3$

$2\sqrt{17}$

$\sqrt{34}$

Correct answer:

$\sqrt{34}$

Explanation:

Write the formula of the Pythagorean Theorem.

a^2+b^2 =c^2

The value of is what we will be solving for.

Substitute the side lengths.

3^2+5^2 =c^2

9+25 = c^2

34 =c^2

Square root both sides.

$c=\sqrt{34}$

The answer is: $\sqrt{34}$

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Example Question #61 : Geometry

One angle of a right triangle measures 45, and the hypotenuse length is 6 centimeters. Give the perimeter of the triangle.

Possible Answers:

$12\ cm$

$6\ cm$

$6(\sqrt{2}+1)\ cm$

$4(1+\sqrt{2})\ cm$

$6(2+\sqrt{2})\ cm$

Correct answer:

$6(\sqrt{2}+1)\ cm$

Explanation:

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.

$c^2=s^2+s^2\Rightarrow c^2=2s^2\Rightarrow 6^2=2s^2\Rightarrow s^2=18$

$\Rightarrow s=\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2}\ cm$

We can now find the perimeter since we have all three sides:

$Perimeter=s+s+c=2s+c=2\times 3\sqrt{2}+6=6\sqrt{2}+6=6(\sqrt{2}+1)\ cm$

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Example Question #62 : Geometry

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:

$19 \frac{1}{3}$

$20 \frac{2}{3}$

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:

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

$8 \sqrt{2}$

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

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

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 #2 : 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 $\overline{ED}$ ?

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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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Plane Geometry

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #61 : Plane Geometry

Example Question #62 : Plane Geometry

Example Question #1 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

Example Question #3 : How To Find The Length Of The Hypotenuse Of A Right Triangle : Pythagorean Theorem

Example Question #62 : Geometry

Example Question #61 : Geometry

Example Question #62 : Geometry

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

All ISEE Upper Level Math Resources

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