Concept Review 3
- 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. - Obtuse: 5^2 + 62 =61 < 9^2 = 81
- Acute: 2^2 + 122 =148 > 12^2 = 144
- Obtuse: 6^2 + 82 =100 < 11^2 = 121
- Impossible: 2 +2 isn’t greater than 12
- Impossible: 3 +4 isn’t greater than 7
- Right: 1.5^2 + 22 =6.25=2.5^2
- 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
- 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. - Your diagram should look like
this: The height is
. - 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
- 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!) - 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!) - 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