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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

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All SSAT Upper Level Math Resources

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Example Questions

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:

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:

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

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

$A=49\:cm$

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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 #491 : Geometry

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:

$32.5 \;in^{2}$

It cannot be determined from the information given.

$30 \;in^{2}$

$25 \;in^{2}$

$28 \;in^{2}$

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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Example Question #2 : How To Find If Right Triangles Are Congruent

Given: $\bigtriangleup ABC$ and $\bigtriangleup DEF$ with right angles $\angle B$ and $\angle E$ ; $\angle A \cong \angle D$ .

Which of the following statements alone, along with this given information, would prove that $\bigtriangleup ABC \cong \bigtriangleup DEF$ ?

I) $\overline{AB } \cong \overline{DE}$

II) $\overline{BC }\cong \overline{EF}$

III) $\overline{AC } \cong \overline{DF}$

Possible Answers:

I or III only

III only

II or III only

Any of I, II, or III

I or II only

Correct answer:

Any of I, II, or III

Explanation:

$\angle A \cong \angle D$ ; $\angle B \cong \angle E$ since both are right angles.

Given that two pairs of corresponding angles are congruent and any one side of corresponding sides is congruent, it follows that the triangles are congruent. In the case of Statement I, the included sides are congruent, so by the Angle-Side-Angle Congruence Postulate, $\bigtriangleup ABC \cong \bigtriangleup DEF$ . In the case of the other two statements, a pair of nonincluded sides are congruent, so by the Angle-Angle-Side Congruence Theorem, $\bigtriangleup ABC \cong \bigtriangleup DEF$ . Therefore, the correct choice is I, II, or III.

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

$\bigtriangleup ABC \cong \bigtriangleup DEF$ , where $\angle B$ is a right angle, AC = 20 , and AB = BC .

Which of the following is true?

Possible Answers:

$\bigtriangleup DEF$ has perimeter 40

$DE = 20\sqrt{2}$

$m \angle F = 60^{\circ }$

None of the statements given in the other choices is true.

$\bigtriangleup DEF$ has area 100

Correct answer:

$\bigtriangleup DEF$ has area 100

Explanation:

$\bigtriangleup ABC \cong \bigtriangleup DEF$ , and corresponding parts of congruent triangles are congruent.

Since $\angle B$ is a right angle, so is $\angle E$ . DE = AB and EF= BC ; since AB = BC , it follows that DE = EF . $\bigtriangleup DEF$ is an isosceles right triangle; consequently, $m \angle A = m \angle C = 45^{\circ }$ .

$\bigtriangleup DEF$ is a 45-45-90 triangle with hypotenuse of length DF= AC = 20 . By the 45-45-90 Triangle Theorem, the length of each leg is equal to that of the hypotenuse divided by $\sqrt{2}$ ; therefore,

$DE=EF = \frac{DF}{\sqrt{2}} = \frac{20}{\sqrt{2}} =\frac{20 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{20 \sqrt{2}}{2} = 10 \sqrt{2}$

$DE = 20\sqrt{2}$ is eliminated as the correct choice.

Also, the perimeter of $\bigtriangleup DEF$ is

$P = DE+EF+DF = 10 \sqrt{2} + 10 \sqrt{2} + 20 = 20+ 20 \sqrt{2} \ne 40$ .

This eliminates the perimeter of $\bigtriangleup DEF$ being 40 as the correct choice.

Also, $m \angle F = 60^{\circ }$ is eliminated as the correct choice, since the triangle is 45-45-90.

The area of $\bigtriangleup DEF$ is half the product of the lengths of its legs:

$A = \frac{1}{2} \cdot DE \cdot EF$

$= \frac{1}{2} \cdot 10 \sqrt{2}\cdot 10 \sqrt{2}$

$= \frac{1}{2} \cdot 10\cdot 10 \cdot \sqrt{2} \cdot \sqrt{2}$

$= \frac{1}{2} \cdot 10\cdot 10 \cdot 2$

= 100

The correct choice is the statement that $\bigtriangleup DEF$ has area 100.

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Example Question #711 : Ssat Upper Level Quantitative (Math)

One angle of a right triangle has measure $120^{\circ }$ . Give the measures of the other two angles.

Possible Answers:

This triangle cannot exist.

$30^{\circ }, 90^{\circ }$

$120^{\circ }, 120^{\circ }$

$90^{\circ }, 120^{\circ }$

$30^{\circ }, 30^{\circ }$

Correct answer:

This triangle cannot exist.

Explanation:

A right triangle must have one right angle and two acute angles; this means that no angle of a right triangle can be obtuse. But since $120^{\circ } > 90^{\circ }$ , it is obtuse. This makes it impossible for a right triangle to have a $120^{\circ }$ angle.

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Example Question #712 : Ssat Upper Level Quantitative (Math)

One angle of a right triangle has measure $68^{\circ }$ . Give the measures of the other two angles.

Possible Answers:

This triangle cannot exist.

$52^{\circ }, 60^{\circ }$

$44^{\circ },68^{\circ }$

$22^{\circ }, 90^{\circ }$

$56^{\circ },56^{\circ }$

Correct answer:

$22^{\circ }, 90^{\circ }$

Explanation:

One of the angles of a right triangle is by definition a right, or $90^{\circ }$ , angle, so this is the measure of one of the missing angles. Since the measures of the angles of a triangle total $180^{\circ }$ , if we let the measure of the third angle be , then:

x + 68 + 90 = 180

x + 158 = 180

x + 158 - 158= 180 - 158

x = 22

The other two angles measure $22 ^{\circ }, 90^{\circ }$ .

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Example Question #2 : How To Find An Angle In A Right Triangle

Find the degree measure of in the right triangle below.

Possible Answers:

$90^{\circ}$

$43^{\circ}$

$32^{\circ}$

$47^{\circ}$

Correct answer:

$43^{\circ}$

Explanation:

The total number of degrees in a triangle is 180 .

While $47^{\circ}$ is provided as the measure of one of the angles in the diagram, you are also told that the triangle is a right triangle, meaning that it must contain a $90^{\circ}$ angle as well. To find the value of , subtract the other two degree measures from 180 .

$x=180-90-47=43^{\circ}$

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SSAT Upper Level Math : SSAT Upper Level Quantitative (Math)

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

All SSAT Upper Level Math Resources

Example Questions

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 #491 : Geometry

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

Example Question #2 : How To Find If Right Triangles Are Congruent

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

Example Question #711 : Ssat Upper Level Quantitative (Math)

Example Question #712 : Ssat Upper Level Quantitative (Math)

Example Question #2 : How To Find An Angle In A Right Triangle

All SSAT Upper Level Math Resources

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