SAT Mc Graw Hill 2011

Concept Review 3

1. Your diagram should include one each of a
   3 x- 4 x- 5 x,a 5x- 12 x- 13 x, a 30°- 60 °- 90 °, and a
   45 °- 45 °- 90 °triangle.


2. Obtuse: 5^2 + 62 =61 < 9^2 = 81


3. Acute: 2^2 + 122 =148 > 12^2 = 144


4. Obtuse: 6^2 + 82 =100 < 11^2 = 121


5. Impossible: 2 +2 isn’t greater than 12


6. Impossible: 3 +4 isn’t greater than 7


7. Right: 1.5^2 + 22 =6.25=2.5^2


8. Since the area of a circle is
   πr^2 = 16 π,r=4. Put the infor-
   mation into the diagram.
   Use the Pythagorean
   theorem or notice that,
   since the hypotenuse is twice
   the shorter side, it is a 30°- 60 °- 90 °triangle.
   Either way, , so the area of the triangle

is ()bh^2443283 =()()=.

CB= 43

9. At first, consider the shorter
   leg as the base. In this case,
   the other leg is the
   height. Since the
   area is
   (bh)/2=30, the
   other leg must be 12. This is a 5-12-13 triangle,
   so the hypotenuse is 13. Now consider the
   hypotenuse as the base. Since 13h/2=30,
   h=60/13=4.615.


10. Your diagram should look like
    this: The height is
    .


11. Sketch the diagram. Use the
    Pythagorean theo-
    rem or distance for-
    mula to find the
    lengths. The
    perimeter is
    4 + 13 + 15 =32.

()()33 3=^9

Answer Key 3: The Pythagorean Theorem

C

A

P B

4

4 4

43

5

h

12

13

(^6363)

3 3

9

P(0, 0) M(4, 0)

4

N(9, 12)

12

5

13

15

y

x

SAT Practice 3

1. A Draw the rectangle. If the length and width are
   in the ratio of 5:12, then they can be expressed as
   5 xand 12x.The area, then, is (5x)(12x)= 60 x^2 =240.
   Sox=2, and the length and width are 10 and 24.
   Find the diagonal with the Pythagorean theorem:
   102 + 242 =a^2 , so 100 + 576 = 676 =a^2 andd=26.
   (Notice that this is a 5-12-13 triangle times 2!)


2. B Draw a diagram like this.
   The distance from the starting
   point to the finishing point is
   the hypotenuse of a right triangle
   withlegs of 6 and 8. Therefore, the
   distance is found with Pythagoras:
   62 + 82 = 36 + 64 = 100 =d^2 , so d=10.
   (Notice that this is a 3-4-5 triangle times 2!)


3. A Draw in the angle measures.
   All angles in a square are 90°and
   all angles in an equilateral tri-
   angle are 60°. Since all of the
   angles around point Fadd up
   to 360°,∠EFA=360 – 60 − 60 −
   90 = 150 °.SinceEF=AF,∆EFA
   is isosceles, so ∠AEF=(180−
   150)/2= 15 °.
   4. B If the perimeter of
   the triangle is 36, each
   side must have a length
   of 12. Since the altitude
   forms two 30°-60°-90°
   triangles, the altitude
   must have length.
   This is also the diameter
   of the circle. The circum-
   ference of a circle is πtimes the diameter:.
   5. C Draw in AEandAC.Since
   all radii of a circle are equal,
   their measures are both 10
   as well. Therefore ∆ACEis
   equilateral, and ADdivides it
   into two 30°-60°-90°triangles.
   You can use the Pythagorean
   theorem, or just use the 30°-60°-90°relationships
   to see that AD=^53.

63 π

63

10

6

4

2

6

8

Start

Finish

A B

C

E D

F

60 °

60 °

60 °

60 °

60 ° 60 °

90 ° 90 °

150 ° 90 °

12 12

6 6

60 ° 60 °

63

C

A

B

E

D

10

10 10

5

5

376 McGRAW-HILL’S SAT