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SSAT Upper Level Math : Geometry

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Example Questions

Example Question #841 : Geometry

The base length of a parallelogram is t+2 . If the perimeter of the parallelogram is 24, give the side length in terms of .

Possible Answers:

10+t

20-t

20+t

10-t

22-2t

Correct answer:

10-t

Explanation:

Let:

Side length .

The perimeter of a parallelogram is:

2w+2h

where:

is the base length of the parallelogram and is the side length. The perimeter is known, so we can write:

Perimeter=24=2(t+2)+2x

Now we solve the equation for :

24=2(t+2)+2x

$\Rightarrow 24=2t+4+2x$

$\Rightarrow 24-2t-4=2x$

$\Rightarrow 20-2t=2x$

$\Rightarrow 10-t=x$

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Example Question #5 : How To Find The Perimeter Of A Parallelogram

The base length of a parallelogram is identical to its side length. If the perimeter of the parallelogram is 40, give the base length.

Possible Answers:

Correct answer:

Explanation:

The perimeter of a parallelogram is:

2(w+h)

Where:

is the base length of the parallelogram and is the side length. In this problem the base length and side length are identical, that means:

w=h

So we can write:

$Perimeter=40=2(w+w)\Rightarrow 20=2w\Rightarrow w=10$

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Example Question #6 : How To Find The Perimeter Of A Parallelogram

The base length of a parallelogram is 2t+3 and the side length is 2t-3 . Give the perimeter of the parallelogram in terms of and calculate it for t=1 .

Possible Answers:

6t, 6

6t+6, 12

4t, 4

8t, 8

8t+8, 16

Correct answer:

8t, 8

Explanation:

The perimeter of a parallelogram is:

2(w+h)

where:

is the base length of the parallelogram and is the side length. So we have:

$Perimeter=2(w+h)=2\left [ (2t+3)+(2t-3) \right ]=2(4t)=8t$

and:

$t=1\Rightarrow 8t=8\times 1=8$

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Example Question #1061 : Ssat Upper Level Quantitative (Math)

Parallelogram1

The above parallelogram has area 100. Give its perimeter.

Possible Answers:

$20+20\sqrt{2}$

$20+20\sqrt{3}$

$40\sqrt{2}$

$40\sqrt{3}$

Correct answer:

$20+20\sqrt{2}$

Explanation:

The height of the parallelogram is , and the base is . By the $45^{\circ}-45^{\circ}-90^{\circ}$ Theorem, BD=CD . Since the product of the height and the base of a parallelogram is its area,

$BD \cdot CD = A$

$\left (BD \right )^{2} = 100$

BD = 10

By the $45^{\circ}-45^{\circ}-90^{\circ}$ Theorem,

CD = AB = BD = 10 , and

$AD = BC = BD \cdot \sqrt{2} = 10\sqrt{2}$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 10 + 10 + 10\sqrt{2}+ 10\sqrt{2} = 20+20\sqrt{2}$

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Example Question #8 : How To Find The Perimeter Of A Parallelogram

Parallelogram2

Give the perimeter of the above parallelogram if BD = 10 .

Possible Answers:

$40+40\sqrt{3}$

$60\sqrt{3}$

$40+20\sqrt{2}$

$40+20\sqrt{3}$

$60\sqrt{2}$

Correct answer:

$40+20\sqrt{3}$

Explanation:

By the $30^{\circ}-60^{\circ}-90^{\circ}$ Theorem:

$AB = CD= BD\sqrt{3}= 10\sqrt{3}$ , and

$AD=BC= 2\cdot BD = 2\cdot 10 = 20$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 20+20+ 10\sqrt{3}+ 10\sqrt{3} = 40+20\sqrt{3}$

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SSAT Upper Level Math : Geometry

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

All SSAT Upper Level Math Resources

Example Questions

Example Question #841 : Geometry

Example Question #5 : How To Find The Perimeter Of A Parallelogram

Example Question #6 : How To Find The Perimeter Of A Parallelogram

Example Question #1061 : Ssat Upper Level Quantitative (Math)

Example Question #8 : How To Find The Perimeter Of A Parallelogram

All SSAT Upper Level Math Resources

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