All GMAT Math Resources
Example Questions
Example Question #1 : Calculating Whether Quadrilaterals Are Similar
Refer to the above Trapezoid . There exists Trapezoid such that
Trapezoid Trapezoid , and the length of the midsegment of Trapezoid is 91.
Give the length of .
The length of the midsegment of a trapezoid - the segment that has as its endpoints the midpoints of its legs - is half the sum of the lengths of its legs. Therefore, Trapezoid has as the length of its midsegment
.
Sidelengths of similar figures are in proportion. If the similarity ratio is , then the bases of Trapezoid have length and , so their midsegment will have length
,
meaning that the ratio of the lengths of the midsegments will be the same as the similarity ratio. Since the length of the midsegment of Trapezoid is 91, this similarity ratio is
.
The ratio of the length of to that of corresponding side is therefore , so
Example Question #461 : Problem Solving Questions
Refer to the above Trapezoid . There exists Trapezoid such that
Trapezoid Trapezoid , and has length 66.
To the nearest whole, give the area of Trapezoid .
The similarity ratio of the trapezoids is the ratio of the length of one side to that of the corresponding side of the other. For these trapezoids, we take the ratio of the lengths of corresponding sides and :
.
The ratio of the areas is the square of this, or
.
The area of Trapezoid is one half the product of the height and the sum of bases and :
Multiply this by the area ratio:
.
The correct response is 1,271.
Example Question #5 : Calculating Whether Quadrilaterals Are Similar
Rhombus Rhombus . ; Rhombus has area 90; Rhombus has area 360. What is the length of diagonal ?
The area of a rhombus is half the product of the lengths of its diagonals, so we can take advantage of this to find from the known measurements of Rhombus :
The ratio of the area of Rhombus to that of Rhombus is 4, so their similarity ratio is the square root of this, or 2. We can calculate now:
Example Question #6 : Calculating Whether Quadrilaterals Are Similar
In Quadrilateral , , , , is a right angle, and diagonal has length 24.
There exists Quadrilateral such that Quadrilateral Quadrilateral , and .
Give the length of .
The Quadrilateral with its diagonals is shown below. We call the point of intersection :
The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular. , the diagonal that connects the common vertices of the pairs of adjacent sides, bisects the other diagonal, making the midpoint of . Therefore, .
also bisects the and angles of the kite, so the result is that two 30-60-90 triangles, and , and two 45-45-90 triangles , and , are formed; also, , being isosceles, is a 45-45-90 triangle.
Examine . Since its short leg has length 12, by the 30-60-90 Theorem, its hypotenuse, , has twice this length, or 24.
Examine . Since a leg has length 12, by the 45-45-90 Theorem, its hypotenuse, , has length times this, or .
Since by similarity, corresponding sides are in proportion,
Example Question #462 : Problem Solving Questions
Refer to the above Trapezoid . There exists Trapezoid such that
Trapezoid Trapezoid , and has length 60.
Give the length of .
Construct the perpendicular segment from to and let be its point of intersection with . By construction, the trapezoid is divided into Rectangle and right triangle . Since opposite sides of a rectangle are congruent, and ; as a consequence of the latter, . By the Pythagrean Theorem, the length of the hypotenuse of right triangle can be calculated from the length of legs and :
The figure, with the segment and the calculated measurements, is below.
Since Trapezoid Trapezoid , by proportionality of corresponding sides of similar figures:
Example Question #463 : Problem Solving Questions
Rhombus Rhombus . Rhombus has perimeter 80; Rhombus has perimeter 180; .
Find the length of diagonal .
A rhombus has four sides the same length, so each side of Rhombus has length one fourth of 80, or 20; each side of Rhombus has length one fourth of 180, or 45.
The diagonals of a rhombus are each other's perpendicular bisectors, so, if we let be the point of intersection of the diagonals of Rhombus , and , we form four congruent right triangles.
We will examine . ; and,.
By the Pythagorean Theorem,
and
Since corresponding sides of similar figures are in proportion, so are corresponding diagonals. Therefore,
Example Question #9 : Calculating Whether Quadrilaterals Are Similar
Rhombus Rhombus . ; ; .
Give the ratio of the area of Rhombus to that of Rhombus .
The angles of the rhombus measure:
;
, since opposite angles of a rhombus, as in any other parallelogram, are congruent;
and , since consecutive angles of a rhombus are supplementary (the sum of their degree measures is 180).
The diagonals of a rhombus are each other's perpendicular bisectors as well as the bisectors of the angles, so and , whose point of intersection we will call , divide Rhombus into four 30-60-90 triangles. If we examine , we see that its short leg has length half that of ; , so . By the 30-60-90 Triangle Theorem, long leg has length this, or , and the diagonal has measure twice this, or .
The ratio of the lengths of corresponding diagonals of the rhombuses is the same as the similarity ratios of the sides, so the similarity ratio of Rhombus to Rhombus is
The ratio of the areas is the square of the similarity ratio, which is
That is, 12 to 1.
Example Question #4 : Calculating Whether Quadrilaterals Are Similar
In Quadrilateral , , , , and is a right angle; .
There exists Quadrilateral such that Quadrilateral Quadrilateral , and .
Give the length of .
The Quadrilateral with its diagonals is shown below. We call the point of intersection :
The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, bisects the and angles of the kite. Consequently, two congruent 30-60-90 triangles, and , and two congruent 45-45-90 triangles , and , are formed; also, , being an isosceles right triangle, is a 45-45-90 triangle. , the hypotenuse of , has length 16, so by the 30-60-90 Triangle Theorem, its shorter leg has length half this, or 8. Also, is a leg of , so by the 45-45-90 Theorem, the hypotenuse has length times this, or .
Corresponding sides of similar quadrilaterals are in proportion, so
Example Question #11 : Calculating Whether Quadrilaterals Are Similar
In Quadrilateral , , , , is a right angle.
There exists Quadrilateral such that Quadrilateral Quadrilateral , and .
Which of the following is true about the areas of the two quadrilaterals?
Quadrilateral and Quadrilateral have the same area.
Quadrilateral has area twice that of Quadrilateral .
Quadrilateral has area three times that of Quadrilateral .
Quadrilateral has area twice that of Quadrilateral .
Quadrilateral has area three times that of Quadrilateral .
Quadrilateral has area twice that of Quadrilateral .
We will assume that and have common measure 1 for the sake of simplcity; this reasoning is independent of the actual measure of .
The Quadrilateral with its diagonals is shown below. We call the point of intersection :
The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, bisects the and angles of the kite. Consequently, is a 30-60-90 triangle and is a 45-45-90 triangle. By the 30-60-90 Theorem, since and are the short leg and hypotenuse of ,
.
By the 45-45-90 Theorem, since and are a leg and the hypotenuse of ,
The similarity ratio of Quadrilateral to Quadrilateral can be found by finding the ratio of the length of side to corresponding side :
The ratio of the areas is the square of the similarity ratio:
The correct choice is that Quadrilateral has area twice that of Quadrilateral .
Example Question #231 : Geometry
The area of a trapezoid is 1,600 square centimeters. Its height is 24 centimeters, and one base is one decimeter longer than the other. What is the length of the longer base?
Let be the longer base. Then, since this is ten centimeters (= 1 decimeter) longer than the shorter, the shorter base has length . Substitute in the formula for the area of a trapezoid, and solve for :
centimeters