AREA OF A SHAPE ON A GRID

Find the area of each triangle.

Problem 1 :

Solution :

Base = 7 units

Height = 4 units

Area of triangle = 1/2 × b × h

= 1/2 × 7 × 4

= 14 units²

So, the area of triangle is 14 units².

Problem 2 :

Solution :

Base = 6 units

Height = 5 units

Area of triangle = 1/2 × b × h

= 1/2 × 6 × 5

= 15 units²

So, the area of triangle is 15 units².

Problem 3 :

Solution :

Base = 11 units

Height = 3 units

Area of triangle = 1/2 × b × h

= 1/2 × 11 × 3

= 16.5 units²

So, the area of triangle is 16.5 units².

Problem 4 : 

Solution :

By drawing the vertical line, we divide the shape into two rectangles.

Solution :

Measures of rectangle in the left :

length = 3, width = 4

Area of rectangle = 3 x 4 ==> 12 cm²  ---(1)

Measures of rectangle in the right :

length = 2, width = 6

Area of rectangle = 2 x 6 ==> 12 cm² ---(2)

(1) + (2)

= 12 + 12

= 24 cm²

Find the area of each parallelogram.

Problem 5 :

Solution :

Base = 4 units

Height = 3 units

Area of parallelogram = base × height

= 4 × 3

= 12 units²

So, the area of parallelogram is 12 units².

Problem 6 :

Solution :

Base = 6 units

Height = 3 units

Area of parallelogram = base × height

= 6 × 3

= 18 units²

So, the area of parallelogram is 18 units².

Problem 7 :

Solution :

Base = 7 units

Height = 5 units

Area of parallelogram = base × height

= 7 × 5

= 35 units²

So, the area of parallelogram is 35 units².

Problem 8 :

Solution :

Area of a circle = πr²

Radius (r) = 2 units

= 22/7 × 2²

= 22/7 × 4

= 12.57 units²

Problem 9 :

Solution :

Area of a circle = πr²

Radius (r) = 4 units

= 22/7 × 4²

= 22/7 × 16

= 50.28 units²

Problem 10 :

You design a tree house using a coordinate plane. You plot the vertices of thefl oor at

J (2, 1), K (2, 8), L(9, 8), and M (9, 1).

The coordinates are measured in feet.

a. What is the shape of the floor?

b. What are the perimeter and the area of the floor?

Solution :

a) Finding the shape of the floor :

Distance between JM = 9 - 2 ==> 7 units

Distance between ML = 8 - 1 ==> 7 units

Distance between KL = 9 - 2 ==> 7 units

Distance between KJ = 8 - 1 ==> 7 units

Since all sides are equal, it must be a square.

b) Perimeter of square = 4(side length)

= 4(7)

= 28 units.

Problem 11 :

In a grid of the exhibits at a zoo, the vertices of the giraffe exhibit are E (0, 90), F (60, 90), G (100, 30), and H (0, 30). The coordinates are measured in feet. What is the area of the giraffe exhibit?

Solution :

Plotting these points in the coordinate plane, we get

Height of the trapezium = 90 - 30 ==> 60 units

Base length (EF) = 60 - 0 ==> 60 units

Base length (GH) = 100 - 0 ==> 100 units

Area of trapezium = (1/2) x h x (a + b)

= (1/2) x 60(60 + 100)

= 30 x 160

= 1800 square units

Area of giraffe exhibit = 1800 square units

Problem 12 :

Jasmine says the area of this shape is 10cm. Explain her mistake.

Solution :

Length of rectangle = 3 units

Width of rectangle = 2 units

Area of rectangle = length x width 

= 3 x 2

= 6 square units

Problem 13 :

Find the area of the fairway between two streams on a golf course.

Solution :

There are several ways to separate the fairway into fi gures whose areas you can fi nd using formulas. It appears that one way is to separate it into a right triangle and a rectangle. Identify each shape and fi nd any missing dimensions.

Area of rectangle + area of triangle 

= length x width + (1/2) x base x height

= 40 x 70 + (1/2) x 40 x 30

= 2800 + 20 x 30

= 2800 + 600

= 3400 square yards

So, the area of the fairway is 2800 + 600 = 3400 square yards.

The triangle and the trapezoid share a 15-inch base and a height of 10 inches.

a. The area of the trapezoid is less than twice the area of the triangle. Find the values of x. Explain your reasoning.

Solution :

a) Area of trapezoid = 2(area of triangle)

height of trapezium = 10 = height of triangle

base of trapezium = base of triangle = 15 inches

1/2 ⋅ 10  ⋅ (15 + x) = 2(1/2)  ⋅ 15  ⋅ 10

5(15 + x) = 15  ⋅ 10

15 + x = 150/5

15 + x = 30

x = 30 - 15

x = 15

So, the value of x is 15 inches.