SOLVING WORD PROBLEMS INVOLVING GEOMETRIC SHAPES

Solving Word Problems Involving Geometric Shapes :

In this section, we will learn how to solve word problems involving the shapes square, rectangle and triangle.

Example 1 :

The side of a square exceeds the side of another square by 4 cm and the sum of the area of two squares is 400 sq.cm. Find the dimensions of the squares.

Solution :

Let x be side length of one square

The side of a square exceeds the side of another square by 4 cm

So, the side length of another square  =  x + 4

Area of one square with side length x  =  x2

Area of one square with side length x + 4 is (x + 4)²

Sum of the area of two squares  =  400 sq.cm

x2 + (x + 4)2  =  400

x2 + x2 + 2 ⋅  4 + 42  =  400

2x2 + 8x + 16 - 400  =  0

2x2 + 8x - 384  =  0

By dividing the entire equation by 2, we get

x2 + 4x - 192  =  0

(x + 16) (x - 12) = 0

x + 16  =  0

x  =  -16

x - 12  =  0

x  =  12

Therefore sides of one square is 12 cm.

Side length of another square  =  (12 + 4)  =  16 cm.

Example 2 :

The length of the rectangle exceeds its width by 2 cm and the area of the rectangle is 195 sq.cm. Find the dimensions of the rectangle.

Solution :

Let x and y be the width and length of rectangle respectively

The length of the rectangle exceeds its width by 2 cm

So, length (y)  =  x + 2

Area of the rectangle  =  195 sq.cm

Length  width = 195

x(x + 2)  =  195

x2 + 2x - 195  =  0

(x + 15) (x - 13)  =  0

x + 15  =  0

x  =  -15

x - 13  =  0

x  =  13

Here x represents width of the rectangle. So, the negative value is not possible.

To find the value of y we have to apply the value of x in the equation y = x + 2

y  =  13 + 2

y  =  15 cm

Therefore length of rectangle is 15 cm and width of the rectangle is 13 cm.

Example 3 :

The footpath of uniform width runs all around a rectangular field 28 meters long and 22 meters wide. If the path occupies 600 m² area, find the width of the path.

Solution :

Let x be the width of the path

Length of the rectangular field  =  28 m

Width of the rectangular field  =  22 m

Area of the path  =  600 m2

Length of the larger rectangle :

  =  28 + x + x

  =  28 + 2x

Width of the larger rectangle :

  =  22 + x + x

  =  22 + 2x

Area of the path  =  Area of larger rectangle - Area of smaller rectangle

600  =  (28 + 2x)  (22 + 2x) - 28  22

600  =  616 + 56x + 44x + 4x2 - 616

600  =  56x + 44x + 4x2

600  =  4x2 + 100 x

4x² + 100x  =  600

Dividing the entire equation by 4, we get

x+ 25x  =  150

x² + 25x - 150  =  0

(x - 5) (x + 30)  =  0

x - 5  =  0

x  =  5

x + 30  =  0

x  =  -30

Negative value is not possible. Because x represents width of the path.

Therefore width of the path is 5 m.

Example 4  :

The area of the right angled triangle is 500 cm². If the base of the triangle exceeds the altitude by 15 cm,find the dimensions of the triangle.

Solution :

The area of the right angled triangle  =  500 cm²

Let x and y be the base and height of the triangle respectively.

base of the triangle exceeds the altitude by 15 cm

 y  =  x + 15

Area of the triangle  =  500

(1/2)  b  h  =  500

(1/2)  (x + 15)  x  =  500

  x (x + 15)  =  500  2

x² + 15x  =  1000

 x² + 15 x - 1000  =  0

(x - 25) (x + 40)  =  0

x - 25  =  0

x  =  25

x + 40  =  0

x  =  -40

Here x represents length of altitude. So, negative value is not possible.

If x  =  25, then

y  =  x + 15

y  =  25 + 15

y  =  40 cm

Therefore the altitude is 25 cm and base is 40 cm.

After having gone through the stuff given above, we hope that the students would have understood how to solve word problems involving geometric shapes.

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