(MATH13- B10)

Members: C06 Wrenbria Ngo

C07 Julie – Ann Parañal

C08 Dani Patalinghog

C09 Marino Penuliar

C10 Michael Sadsad

CPR

(MATH13- B10)

Members: C06 Wrenbria Ngo

C07 Julie – Ann Parañal

C08 Dani Patalinghog

C09 Marino Penuliar

C10 Michael Sadsad

Prof. Charity Hope Gayatin

Prof. Charity Hope Gayatin

Homework 1.1

#15. Find the sides of each of the two polygons if the total number of sides of the polygons is 13, and the sum of the number of diagonals of the polygons is 25. Assume: 8 and 5 D1= n2(n-3)

n=13D1= 82(8-3)

D=25D1= 20

D2= n2(n-3)

D2= 52(5-3)

D2= 5 Answer: The number of sides in each polygon is 8 and 5 #17. What is the name of a regular polygon that has 90 diagonals? Given: D = 90

Required: Name of regular polygon

D = n2 (n-3) -n2 + 3n + 180 = 0

90 = n2 (n-3) n-15 = 0 n+ 12= 0 180 = n2 – 3n n = 15 n= -12 Answer: PENTADECAGON

#19. Find the number of diagonals of a regular polygon whose interior angle measures 144°. Given: I.A = 144°

Required: D =? For D I.A = 180(n-2)n 144n = 180n – 3 D = n2 (n-3) D = 702 144 = 180(n-2)n -36 n = -360 D= 102(10-3) D = 35 144n = 180 (n-2) n = 10 2D = 10(7) Answer: D = 35

#21. The ratio of areas between two similar triangles is 1:4. If one side of the smaller triangle is 2 units, find the measure of the corresponding side of the other triangle. Given:

2 x2

A1 = 1unit2 A2 = 4unit2

Solution: A1A2 = (X1X2)2 X22 = 4(4)

14 = (2X2)2 X22 = 16

X2 = 4

Answer: X2 = 4 units

A3#33

A3#33

A2

A2

A1

A1

#23. A regular hexagon A has the midpoints of its edges joined to form a smaller hexagon B. This process is repeated by joining the midpoints of the edges of hexagon B to get a third hexagon C. What is the ratio of the area of hexagon C to the area of hexagon A? Given:

ABC

Required: Ratio of A1 to A3

Solution: A1A2 = (X1X2)2 A (X3216)2 = A3 X24

A1As4 = (X1X324)2 A1 X3216= A3 X244

4 (A1 X32) A1X2

Answer: Ratio of A1 to A3 = 4:1

#25. If ABCDE is a regular pentagon and diagonals EB and AC intersect at O, then what is the degree measure of < EOC?

O

O

Given: DSolution: θ=360n= 3605 = 75°

E C <EOC = 2θ

= 2(75°)

A B Answer: <EOC = 144° Exercise 1.2

#9. In a right triangle, the bisector of the right triangle divides the hypotenuse in the ratio of 3 is to 5. Determine the measures of the acute angles of the triangle.

Given: Answer: The Acute Triangle ∝ and β measures 52.24°

A=? 3 c = 8 and 37.76° respectively.

P= 5

Required: ∝= ? β=?

Solution: a=cp sin θ 2108 cos θ 2108 a=85 θ=52.24° θ=37.76° a=210 ∝ =52.24°...