The surface area of the prism below is 264 square units . find the missing dimensions of this prism. ( picture is in the picture).

The missing dimension is 4 units.

Step-by-step explanation:

In the given prism,

Of all the surfaces,

There are three rectangular surfaces and two triangular surfaces

1)The two rectangles with dimensions of 15 units and 5 units:

The area of each rectangle = Length of rectangle Breadth of rectangle

From the diagram,

Length of rectangle = 15 units,

Breadth of rectangle = 5 units.

Area of each rectangle = 15 5 ;

Area of each rectangle = 75 square units.

Sum of areas of the two rectangles = 2 75

Sum of areas of the two rectangles = 150 square units (equation 1)

2) The rectangle with the dimensions 15 units and 6 units:

The area of thus rectangle=  Length of rectangle Breadth of rectangle

From the diagram,

Length of rectangle = 15 units,

Breadth of rectangle = 6 units.

Area of the rectangle = 15 6 ;

Area of the rectangle = 90 square units. (equation 2)

From equation 1 and equation 2,

Therefore sum of areas of all rectangles in the prism = 150 + 90

Sum of areas of rectangles = 240 square units.

We also know,

Total surface area of the prism = Sum of areas of triangles + Sum areas of rectangles

Given, Total surface area of prism = 264 square units.

Therefore from the formula,

Sum of areas of triangles = Total surface area - sum of areas of reactangles

Sum of areas of triangles = 264 - 240

Sum of areas of triangles = 24 square units. (equation 3)

Let the missing dimension be 'h units'

From diagram,both triangles are congruent,hence have the same area

Area of a triangle =

From diagram,

Base = 6 units,

Height = h units.

Area of a triangle = = 3h square units.

Sum of the areas of both triangles = 3h+3h = 6h square units.

Using equation 3, we get

6h = 24;

h =

Therefore,

h = 4 units.

Therefore,

The missing dimension is 4 units.

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