The ratio of the perimeter of triangle PQR to the perimeter of rectangle ABCD is 5 : 9. All in centimeters

Question

Triangle (sides) PQ: (x-3) PR: (3x-5) QR: (4x+3)
Rectangle (sides) AB: (2x+2) BC: (4x - 1/2) [AB is longer than BC]

a. Write algebraic expressions for the perimeters of triangle PQR and rectangle ABCD.

b. Write a linear equation using the algebraic expressions written in part a. Then solve for x.

c. Find the area of rectangle ABCD.

Answer 1

[ (x-3) + (3x-5) + (4x+3) ] : [2(2x+2)+ 2(4x - 1/2) ] = 4 : 9
(8x-5) / (12x+3) = 4/9

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Answer 2

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Answer 3

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Answer 4

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Answer 5

The ratio of the perimeter of triangle PQR to the perimeter of rectangle ABCD is 5 : 9. All in centimeters
Triangle (sides) PQ: (x-3) PR: (3x-5) QR: (4x+3)
Rectangle (sides) AB: (2x+2) BC: (4x - 1/2) [AB is longer than BC]

a. Write algebraic expressions for the perimeters of triangle PQR and rectangle ABCD.

b. Write a linear equation using the algebraic expressions written in part a. Then solve for x.

c. Find the area of rectangle ABCD.

Answer 6

a. Perimeter of triangle PQR = (x-3) + (3x-5) + (4x+3) = 8x-5 cm
Perimeter of rectangle ABCD = 2(2x+2) + 2(4x - 1/2) = 12x+3 cm

b. The linear equation using the given ratio of perimeters is:

(8x-5)/(12x+3) = 5/9

Cross-multiplying and simplifying, we get:

72x - 45 = 60x + 15

12x = 60

x = 5

Therefore, the value of x is 5.

c. The area of rectangle ABCD is:

Area = AB x BC

Area = (2x+2) x (4x - 1/2)

Area = 8x^2 + 3x - 1 cm^2

Substituting the value of x as 5, we get:

Area = 203 cm^2

Therefore, the area of rectangle ABCD is 203 square centimeters.