2984 Words12 Pages

CPR

(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:*…show more content…*

Find the lengths of the three sides if the area of the triangle is 576cm2 .

Soltion : c/17=9/10=b/9

A= ss-as-b(s-c) s= a+b+c2 s= 10a+9a+17a20 b= 9(40)10 s= 95a b=36

576= 18a250; c= 17(40)10

9a2= 14400 c= 68 a2= 1600 a=40 Answer: 40cm, 36cm, 68cm

#15. Given triangle ABC whose sides are AB=15in., AC=25 in., and BC= 30in. From a point D on side AB, a line DE is drawn to a point E on side AC such that angle ADE is equal to angle ABC. If the perimeter of triangle ADE is 28 in., find the lengths of the line segments BD and CE. Given: ? A D 30in 15in

B E C

Required: BD =? ; CE =?

Solution: For BD P ADEP ABC = ADAB P ADE=28in Answer: The length of segments BD and CE is 9in and 10in AD = 15in( 28in)70in P ABC=70in AD = 6in P ADEP ABC= AEAC BD = 9in AE = 25in (28in)70in AE = 10 in

#17. What is the sum of the areas of the two triangles formed in number 16? Given: 3

(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:

Find the lengths of the three sides if the area of the triangle is 576cm2 .

Soltion : c/17=9/10=b/9

A= ss-as-b(s-c) s= a+b+c2 s= 10a+9a+17a20 b= 9(40)10 s= 95a b=36

576= 18a250; c= 17(40)10

9a2= 14400 c= 68 a2= 1600 a=40 Answer: 40cm, 36cm, 68cm

#15. Given triangle ABC whose sides are AB=15in., AC=25 in., and BC= 30in. From a point D on side AB, a line DE is drawn to a point E on side AC such that angle ADE is equal to angle ABC. If the perimeter of triangle ADE is 28 in., find the lengths of the line segments BD and CE. Given: ? A D 30in 15in

B E C

Required: BD =? ; CE =?

Solution: For BD P ADEP ABC = ADAB P ADE=28in Answer: The length of segments BD and CE is 9in and 10in AD = 15in( 28in)70in P ABC=70in AD = 6in P ADEP ABC= AEAC BD = 9in AE = 25in (28in)70in AE = 10 in

#17. What is the sum of the areas of the two triangles formed in number 16? Given: 3

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