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SSAT Upper Level Math : Properties of Triangles

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Example Questions

Example Question #41 : Right Triangles

A right triangle has leg lengths of $10\text{ and }12$ . What is the area of this triangle?

Possible Answers:

120

Correct answer:

Explanation:

Since the legs of a right triangle form a right angle, you can use these as the base and the height of the triangle.

$\text{Area}=\frac{1}{2}(\text{base})(\text{height})$

$\text{Area}=\frac{1}{2}(10)(12)=\frac{1}{2}(120)=60$

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Example Question #42 : Right Triangles

A right triangle has leg lengths of $6\:cm$ and $9\:cm$ . Find the area of the right triangle.

Possible Answers:

$15\:cm^2$

$27\:cm^2$

$54\:cm^2$

$\frac{15}{2}\:cm^2$

Correct answer:

$27\:cm^2$

Explanation:

The legs of a right triangle also make up its base and its height.

$\text{Area}=\frac{1}{2}\text{base}\times\text{height}$

$\text{Area}=\frac{1}{2}\times6\:cm\times9\:cm=27\:cm^2$

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Example Question #43 : Right Triangles

A right triangle has leg lengths of $9\:cm$ and $12\:cm$ . Find the area of this triangle.

Possible Answers:

$54\:cm^2$

$15\sqrt2\:cm^2$

$15\:cm^2$

$108\:cm^2$

Correct answer:

$54\:cm^2$

Explanation:

The legs of a right triangle are also its height and its base.

$\text{Area}=\frac{1}{2}\text{base}\times\text{height}$

$\text{Area}=\frac{1}{2}(9\:cm)(12\:cm)=54\:cm^2$

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

A right triangle has two legs of lengths $9\:cm$ and $12\:cm$ , respectively. What is the area of the right triangle?

Possible Answers:

$108\:cm^{2}$

$54\:cm^{2}$

$72\:cm^{2}$

$36\:cm^2$

$124\:cm^2$

Correct answer:

$54\:cm^{2}$

Explanation:

The area of a right triangle with a base and a height can be found with the formula $A=\frac{1}{2}bh$ . Since the two legs of a right triangle are perpendicular to each other, we can use these as the base and height in the formula. Therefore:

$A=\frac{1}{2}(9\:cm)(12\:cm)$

$A=\frac{1}{2}(108\:cm^{2})$

$A=54\:cm^{2}$

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Example Question #44 : Properties Of Triangles

A given right triangle has two legs of lengths $6\:cm$ and $8\:cm$ , respectively. What is the area of the triangle?

Possible Answers:

$7\:cm^2$

Not enough information to solve

$48\:cm^2$

$24\:cm^2$

$14\:cm^2$

Correct answer:

$24\:cm^2$

Explanation:

$A=\frac{1}{2}bh$

$A=\frac{1}{2}(6\:cm)(8\:cm)$

$A=\frac{1}{2}(48\:cm^2)$

$A=24\:cm^2$

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Example Question #6 : How To Find The Area Of A Right Triangle

A given right triangle has legs of lengths $14\:cm$ and $7\:cm$ , respectively. What is the area of the right triangle?

Possible Answers:

$21\:cm^2$

Not enough information available

$42\:cm^2$

$98\:cm^2$

$49\:cm^2$

Correct answer:

$49\:cm^2$

Explanation:

$A=\frac{1}{2}(7\:cm)(14\:cm)$

$A=\frac{1}{2}(98\:cm^2)$

$A=49\:cm$

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

The lengths of the hypotenuses of ten similar right triangles form an arithmetic sequence. The smallest triangle has legs of lengths 3 and 4 inches; the second-smallest triangle has a hypotenuse of length one foot.

Which of the following responses comes closest to the area of the largest triangle?

Possible Answers:

7 square feet

6 square feet

8 square feet

9 square feet

5 square feet

Correct answer:

8 square feet

Explanation:

The hypotenuse of the smallest triangle can be calculated using the Pythagorean Theorem:

$c = \sqrt{a^{2}+b^{2}} = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25}=5$ inches.

Let $c_{n}$ be the lengths of the hypotenuses of the triangles in inches. $c_{1} = 5$ and $c_{2} = 12$ , so their common difference is

$d = c_{2} - c_{1} = 12 - 5= 7$

The arithmetic sequence formula is

$c_{n} = c_{1} + (n-1)d$

The length of the hypotenuse of the largest triangle - the tenth triangle - can be found by substituting $c_{1} = 5, n= 10, d = 7$ :

$c_{10} =5+ (10-1) 7 = 5 + 9 \cdot 7 = 5 + 63 = 68$ inches.

The largest triangle has hypotenuse of length 68 inches. Since the triangles are similar, corresponding sides are in proportion. If we let and be the lengths of the legs of the largest triangle, then

$\frac{a}{3} = \frac{68}{5}$

$\frac{a}{3} \cdot 3 = \frac{68}{5} \cdot 3$

a = 40.8

Similarly,

$\frac{b}{4} = \frac{68}{5}$

$\frac{b}{4} \cdot 4 = \frac{68}{5} \cdot 4$

b = 54.4

The area of a right triangle is half the product of its legs:

$A = \frac{ab}{2}= \frac{40.8 \cdot 54.4}{2} = 1,109.76$ square inches.

Divide this by 144 to convert to square feet:

$1,109.76 \div 144 \approx 7.7$

Of the given responses, 8 square feet is the closest, and is the correct choice.

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Example Question #8 : How To Find The Area Of A Right Triangle

Right triangle 3

Figure NOT drawn to scale

In the above figure, $\bigtriangleup ABC$ is a right triangle. BC = 12 , AC = 15 , AB = AD . What fraction of $\bigtriangleup ABC$ has been shaded in?

Possible Answers:

$\frac{1}{2}$

$\frac{4}{5}$

$\frac{3}{5}$

$\frac{2}{3}$

Correct answer:

$\frac{3}{5}$

Explanation:

The length of the leg $\overline{AB}$ , which we will call , can be calculated by setting c = 15 , the length of hypotenuse $\overline{AC}$ , and b= 12 , the length of leg $\overline{BC}$ , and applying the Pythagorean Theorem:

$a = \sqrt{c^{2} - b^{2}} = \sqrt{15^{2} - 12^{2}} =\sqrt{225- 144} = \sqrt{81} = 9$

AD = AB = 9 .

Construct the altitude of $\bigtriangleup ABC$ - which is also that of $\bigtriangleup ABD$ , the shaded region - from to $\overline{AC}$ . We call the length of this altitude (height). The figure is seen below.

Untitled 3

The area of $\bigtriangleup ABC$ is one half this height multiplied by the corresponding base length :

$\frac{1}{2} \cdot AC \cdot h$

The area of $\bigtriangleup ABD$ , the shaded region - is, similarly,

$\frac{1}{2} \cdot AD \cdot h$

Therefore, the fraction that $\bigtriangleup ABD$ is of $\bigtriangleup ABC$ is the fraction of their areas:

$\frac{\frac{1}{2} \cdot AD \cdot h}{\frac{1}{2} \cdot AC \cdot h} = \frac{AD }{AC}$

Substituting the measures of the two segments:

$\frac{AD}{AC} = \frac{9}{15} = \frac{3}{5}$

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Example Question #11 : How To Find The Area Of A Right Triangle

What is the area of a right triangle whose hypotenuse is 13 inches and whose legs each measure a number of inches equal to an integer?

Possible Answers:

$32.5 \;in^{2}$

$25 \;in^{2}$

It cannot be determined from the information given.

$28 \;in^{2}$

$30 \;in^{2}$

Correct answer:

$30 \;in^{2}$

Explanation:

We are looking for a Pythagorean triple - that is, three integers that satisfy the relationship $a^{2}+b^{2} = c^{2}$ . We know that c=13 , and the only Pythagorean triple with c=13 is (5,12,13) . The legs of the triangle are therefore 5 and 12, and the area of the right triangle is

$A = \frac{1}{2} bh= \frac{1}{2} \cdot 5\cdot 12 = 30$

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Example Question #1 : How To Find If Right Triangles Are Congruent

Given:

$\bigtriangleup ABC$ , where $\angle B$ is a right angle; AB = 10; BC = 24 ;

$\bigtriangleup DEF$ , where $\angle E$ is a right angle and DF = 28 ;

$\bigtriangleup GHI$ , where $\angle H$ is a right angle and $\bigtriangleup GHI$ has perimeter 60;

$\bigtriangleup JKL$ , where $\angle K$ is a right angle and $\bigtriangleup JKL$ has area 120;

$\bigtriangleup MNO$ , where $\angle N$ is a right triangle and $\angle M \cong \angle A$

Which of the following must be a false statement?

Possible Answers:

$\bigtriangleup JKL \cong \bigtriangleup ABC$

All of the statements given in the other responses are possible

$\bigtriangleup DEF \cong \bigtriangleup ABC$

$\bigtriangleup GHI \cong \bigtriangleup ABC$

$\bigtriangleup MNO \cong \bigtriangleup ABC$

Correct answer:

$\bigtriangleup DEF \cong \bigtriangleup ABC$

Explanation:

$\bigtriangleup ABC$ has as its leg lengths 10 and 24, so the length of its hypotenuse, $\overline{AC}$ , is

$AC = \sqrt{(AB)^{2}+(BC)^{2}} = \sqrt{10^{2}+24^{2}} = \sqrt{100+576} = \sqrt{676 } = 26$

Its perimeter is the sum of its sidelengths:

P= 10+24+26 = 60

Its area is half the product of the lengths of its legs:

$A = \frac{1}{2} \cdot 10 \cdot 24 = 120$

$\bigtriangleup GHI$ and $\bigtriangleup JKL$ have the same perimeter and area, respectively, as $\bigtriangleup ABC$ ; also, between $\bigtriangleup MNO$ and $\bigtriangleup ABC$ , corresponding angles are congruent. In the absence of other information, none of these three triangles can be eliminated as being congruent to $\bigtriangleup ABC$ .

However, DF = 28 and AC = 26 . Therefore, $\overline{AC} \ncong \overline{DF}$ . Since a pair of corresponding sides is noncongruent, it follows that $\bigtriangleup DEF \ncong \bigtriangleup ABC$ .

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SSAT Upper Level Math : Properties of Triangles

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #41 : Right Triangles

Example Question #42 : Right Triangles

Example Question #43 : Right Triangles

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

Example Question #44 : Properties Of Triangles

Example Question #6 : How To Find The Area Of A Right Triangle

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

Example Question #8 : How To Find The Area Of A Right Triangle

Example Question #11 : How To Find The Area Of A Right Triangle

Example Question #1 : How To Find If Right Triangles Are Congruent

All SSAT Upper Level Math Resources

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