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# Solving Geometry Problems Involving System of Equations In Two Variables

Solving Geometry Problems Involving System of Equations In Two Variables

Algebra has also found wide applications in Geometry. Among its common problems are problems involving perimeters and angles. In this hub I included five sample problems involving perimeter and angles.

Sample Problem Number One :

A rectangular garden has a perimeter of 100 meters. The length is four times its width. Find the dimensions of the garden.

Solution :

Let X = width

Let Y = length

P = 2L + 2w

Equation one : 2X + 2Y = 100

Equation Two : Y = 4X

Substitute equation two in equation one :

2X + 2 (4X) = 100

2X + 8X = 100

10X = 100

(1/10) 10X = 100 (1/10)

X = 10 ==è width

4X = 4 (10) = 40 ==è lenth

Sample Problem Two :

The perimeter of a rectangular picture frame is 80 centimeter. Two times the width is equal to the length increased by five. Find its dimensions.

Solution :

Let X = width

Let Y = length

P = 2L + 2W

Equation one : 2X + 2Y = 80

Equation two: 2X = Y + 5

We use elimination method by subtraction to solve for the unknown variables.

2X + 2Y = 80

(-)2X (+) -Y = (-) 5

3Y = 75

(1/3) 3Y = 75 (1/3)

Y = 25 ==è length

To solve for width substitute in equation two

2X = 25 + 5

2X = 30

X = 15 ====è width

Sample Problem Three:

The perimeter of a rectangular flower garden is 120 meters. The length is ten meters greater than its width. Find its dimensions.

Solution :

Let X = width

Let Y = length

P = 2L +2W

Equation one : 2X + 2Y = 120

Equation two : Y = X + 10

Substitute equation two in equation one :

2X + 2 (X + 10 ) = 120

2X + 2X + 20 = 120

4X = 120 – 20

4X = 100

(1/4) 4X = 100 (1/4)

X = 25 ==è width

Y = 25 + 10 = 35 ===è lenth

Sample Problem Number Four :

The sum of the two nonright angles in a right triangle is of course 90 degrees. If twice the first is 40 degrees more than three times the second. Find the measurement of the angles of the right triangle.

Solution :

Let X = first angle

Let Y = second angle

Equation one : X + Y = 90

Equation two : 2X = 3Y + 40

From equation one we derive Y = 90 – X the substitute this in equation two.

2X = 3(90-X) + 40

2X = 270 – 3X + 40

2X + 3X = 270 + 40

5X = 310

(1/5) 5X = 310 (1/5)

X = 62 degrees ===è first angle

Y = 90 – 62 = 28 ===è second angle

Sample Problem Number Five :

Two angles are supplementary. The bigger angle is thrice as large as the smaller angle. Find the measurement of the angles.

Solution :

Let X = smaller angle

Let Y = bigger angle

The two angles are supplementary it means that their sum is equal to 180 degrees.

Equation one : X + Y = 180

Equation two : Y = 3X

Substitute equation two in equation one :

X +3X = 180

4X = 180

(1/4) 4X = 180 (1/4)

X = 45 ==èsmaller angle

Y = 3(45) = 135 ====. Bigger angle