All GMAT Math Resources
Example Questions
Example Question #4 : Calculating The Perimeter Of A Rectangle
The diagonal of the rectangle is cm long and is cm long. What is the perimeter of the rectangle?
We can see that the hypotenuse AD of triangle ABD is 15 cm. We should check whether ABD is a Pythagorean Triple, whose sides are in the ratio , where is a constant. Since AD is 15 and BD is 9, the triangle must be a Pythagorean Triple with , therefore AB must be 12 cm long. Now we know all the lengths of the sides of our triangle, we can find the perimeter which will be given by , which gives us 42, our final answer.
Example Question #121 : Geometry
The area of rectangle is , and the diagonal has length . What is the perimeter of rectangle ?
Given the information we are provided, we could set up an quadratic equation, however, this equation would be really complicated to solve, instead we should try to find whether the diagonal of this rectangle can be the hypotenuse of Pythagorean triangle. The hypotenuse AD of triangle ADC, has length 10, therefore if it is a Pythagorean Triple, its other lengths must be 6 and 8, since the sides are in ratio where is a constant. By testing, we see that if and we get an area of 48 for this rectangle, by multiplying DC by AC. This is the original area of our rectangle, therefore, these must be the right lenghts and the final answer is given by , which is 28.
Alternatively, we could have found the missing lengths with trial and error, by doing so, we would have picked length for which, the area is 48 and tried to see whether the diagonal would remain 10, until we get the working set 6 and 8.
Example Question #2 : Calculating The Perimeter Of A Rectangle
The width of a rectangle is twice the length. Find an equation representing the perimeter.
The perimeter of a rectangle can be found by adding up all the sides:
We are told the width of the rectangle is twice its length so , inserting this into our equation for the perimeter leaves us with:
, which is II
Example Question #2 : Calculating The Perimeter Of A Rectangle
Find the perimeter of a rectangle whose width is and length is .
To solve, simply use the formula for the perimeter of a rectangle:
Example Question #8 : Calculating The Perimeter Of A Rectangle
Note: figure NOT drawn to scale
Give the perimeter of the above rectangle.
The perimeter of a rectangle is the sum of the lengths of its sides. Since a rectangle has two pairs of sides of the same lengths, this will be
Example Question #1 : Calculating The Length Of The Side Of A Rectangle
The area of a rectangle is 85; its length is . What is its width?
The product of the length and the width of a rectangle is its area, so divide area by length to get width. This is simply .
Example Question #2 : Calculating The Length Of The Side Of A Rectangle
The perimeter of a rectangle is ; its width is . Which of the following expressions is equal to the length of the rectangle?
Substitute and solve for :
Example Question #124 : Geometry
Rectangle A and Rectangle B have the same area. Rectangle A has length 80% that of Rectangle B, and its width is 120 centimeters. What is the width of Rectangle B?
Not enough information is given to answer the question.
Let and be the length and width of Rectangle B.
Then the width of Rectangle A is 80% of , or . Its length is 120.
The area shared by the two can be expressed as both and . We can set the two equal to each other and calculate :
, or .
Example Question #125 : Geometry
Craig is building a fence around his rectangular back yard. He knows that the yard is feet longer than twice the width. What is the width of the yard if Craig needs feet of fencing to completely enclose the yard?
The length is 5 more twice the width. Let y be the length of the yard and w the width.
The perimeter of the yard is 160, so we can write:
The yard is 25 feet wide and 55 feet long.
Example Question #126 : Geometry
A rectangle has an area of . If the width of the rectangle is , what is its length?
Using the formula for the area of a rectangle, we can solve for the unknown length. We are given the area and the width, so we can solve for the length as follows: