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nd
SOLID MENSURATION PRELIMS REVIEWER (2 Sem; 2011)
VI. Rhombus
FORMULAS:
22. A = eh
I. Triangles in General
23. A =
1.
P=a+b+c
2.
bh A= 2
3.
1 A = rP 2
4.
A=
5.
s=
2
a+b+c 2
2
c = a + b
7.
A=
9.
h=
P = perimeter A = Area r = apothem b = base h = height a, b, c, e = sides
d2 2
2
2
1 26. A = (b1 + b2)h 2 27. A = mh
VIII. General Polygon: n-gon
2
6.
h=
d1 2
( ) + ( )
VII. Trapezoids
II. Right Triangles
8.
24. P = 4e 25. e =
s(s-a)(s-b)(s-c)
2
d1d2 2
28. A =
ab 2
e
29. r =
θ
2tan2
bc·ac ab c
30. R =
III. Equilateral Triangles
rP rne = 2 2
e θ
2sin2
31. sum of interior angles = 180(n – 2)
10. P = 3e
32. one interior angle =
3 11. h = e 2
180(n - 2) n
12. A =
3 2 e 4
33. one exterior angle =
13. r =
3 e 6
34. central angle =
360 n
360 n
***n = number of sides
2 3 14. R = h = e 3 3
Samples: IV. Parallelograms/Rectangles 1. 15. P = 2a + 2b
2
The area area of an equila equilateral teral triangle triangle is is 16 3 cm . Find the altitude of the triangle.
16. A = bh = ab
A= V. Squares 17. r =
16 3 =
e 2
2
e=
2e 2
2
20. A = e = 21. R =
3 2 e 4
e = 16 3 x
18. P = 4e 19. d =
3 2 1 e = bh = 16 3 4 2
d 2
2 e 2
4
64 = 8cm
1 16 3 = bh 2 1 16 3 = 8h 2
h = 4 3 cm
3
2.
Find the area of the largest circle that can be cut from
We subtract the area of the 2 semi-circles from the area
an octagon of edge 12cm.
of the square to get the unshaded a rea of the last figure.
360 = 45° 8
θ=
100 – 25∏ = 21.46cm
2
e
r=
r=
Let’s represent the 2 unshaded areas by x. Sooo…
θ
Then we subtract twice the value of the 2 unshaded
2tan2
2
regions (21.46cm ) from the area of the square.
12
2
100 – (21.46x2) = 57.08cm
45 2tan 2
r = 14.49cm A of the circle = ∏r
5.
Find <1, Arc ED, <2 and Arc BD.
6.
A rectangle measures 12cm x 20cm. If the shorter side is
2
2
A = (14.49) ∏
A = 659.18cm 3.
2
The sides of a triangle are 14 cm, 28cm, and 27cm. Find the base of a rectangle whose altitude is 20cm and whose area is equal to the area of the triangle. s=
a+b+c 2
s=
14 + 27 + 28 = 34.5cm 2
parallel to the plane and the longer side forms an angle
A=
s(s-a)(s-b)(s-c)
of 35° with the plane, what is the area of the projection
A=
34.5(34.5-14)( 34.5-27)( 34.5-28)
on the plane?
A=
34.5(20.5)( 7.5)( 6.5)
A = 185.68cm
The sketch would turn out this way:
2
A = ab 185.86 = 20x
x = 9.28cm
Solve for x: 4.
Find the area of the shaded portion. ABCD is a square.
cos35° =
AB = 10cm, AOB, AOD, DOC and BOC are semicircles. Area of the Square = 10
semi-circles):
() A=∏ ( )
A = lw = 12x = (12)(16.38)
2
Area of one circle (or two
A=∏
x = 16. 38cm
2
= 100cm
d 2
2
A = 196.5965cm 7.
What is the area of the largest polygon that can be cut from a circle of diameter 12cm?
2
10 2
A = 25∏ cm
72
cos 2 = 2
2
72 sin 2
So the area of the shaded region here is 25∏cm
r 6
r = 6cos36 = 4.85cm
2
x 20
=
e 2
6
e = 2(6sin36) = 7.05cm 1 1 A = rP = (4.85)(7.05x5) 2 2
A = 85.28cm
2
8.
Three circles of radii 14cm, 20cm and 18cm are tangent to each other. Find the area of the triangle formed by
12. If GC = 183cm, BH = 3.05cm, AB = 7.32cm , DF = 3.66cm
and DE = 4.88cm, find the length of BD.
joining their centers. a = 20 + 14 = 34 b = 20 + 18 = 38
Use ratio and
c = 14 + 18 = 32
proportion to get the values of x1 and x 2
s=
a+b+c 2
s=
34 + 38 + 32 = 52 2
A= =
s(s-a)(s-b)(s-c) 52(52 -34)( 52-38)( 52-32)
A=
The base angles of the isosceles trapezoid are 70° each. If its altitude is 8cm and the upper base is also 8cm, what is the length of the other base? 8 x
x = 2.91cm Length of other base: 8 + 2x 8 + 2(2.91) = 13.8235cm 10. The diameter of the circle is 30cm. Find the area of the
shaded portion. *Yung dalawang semi circle na shaded na nasa ibang quadrant, pwede mong ilipat sa missing na semi-circles dun sa 1
st
quadrant. So in other words, ¼ of the whole circle’s shaded. Get the
area of the whole circle and divide it by 4. 2
2
2
2
A = ∏r = ∏(15) = 706.8583cm / 4 = 176.7146cm
11. The centers of two circles of radii 25cm are 48cm apart.
