Test: Geometry & Solid Geometry - GMAT MCQ

The lengths of the four sides, AB, BC, CD and DA are all equal to√41.
Hence, the given quadrilateral is either a Rhombus or a Square.
The diagonals of a square are equal. The diagonals of a rhombus are unequal. Compute the lengths of the two diagonals AC and BD.
The length of AC is √162 and the length of BD is √2.
As the diagonals are not equal and the sides are equal, the given quadrilateral is a Rhombus.

If two sides of a triangle and the included angle 'y' is known, then the area of the triangle = 1/2 * (Product of sides) * sin y Substituting the values in the formula, we get 1/2*8*12*sin150

Note: Sin 150^o = 1/2
Area = 24 sq units.

9, 40 and 41 is Pythagorean triplet. So, the triangle is right angled.
n a right angled triangle, the radius of the circle that circumscribes the triangle is half the hypotenuse.
In the given triangle, the hypotenuse = 41.
Therefore, the radius of the circle that circumscribes the triangle = 41/2 = 20.5 units

The sum of an exterior angle and an interior angle of a polygon = 180^o.
The sum of all the exterior angles of a polygon = 360^o.
From the question stem: sum of all interior angles of the given polygon = 1440^o.
Therefore, sum of all the interior and all exterior angles of the polygon = 1440 + 360 = 1800^o.

The number of diagonals of an n-sided convex polygon = 
This polygon has 63 sides. Hence, n = 63.
Therefore, number of diagonals = 

Finding the answer to this question requires that you know this key property about sides of an acute triangle.
If 'a', 'b', and 'l' are the 3 sides of an acute triangle where 'l' is the longest side then l² < a² + b²
The sides are 10, 12, and 'x'.
Scenario 1: Among the 3 sides 10, 12, and x, for values of x ≤ 12, 12 is the longest side.
Scenario 2: For values of x > 12, x is the longest side
Possibilities in scenario 1:
When x ≤ 12, let us find the number of values for x that will satisfy the inequality 12² < 10² + x²
i.e., 144 < 100 + x²
The least integer value of x that satisfies this inequality is 7.
The inequality will hold true for x = 7, 8, 9, 10, 11, and 12. i.e., 6 values.

Possibilities in scenario 2:
When x > 12, x is the longest side.
Let us count the number of values of x that will satisfy the inequality x² < 10² + 12²
i.e., x² < 244
x = 13, 14, and 15 satisfy the inequality. That is 3 more values.

Hence, the values of x for which 10, 12, and x will form sides of an acute triangle are x = 7, 8, 9, 10, 11, 12, 13, 14, 15. A total of 9 values.

Given ABC is a triangle

Statement 1: a²+b²>c²

Condition for a triangle to be obtuse is
c²>a²+b² (with "c" largest side)
Since nothing is given about which is the largest side
Not sufficient

Statement 2: The center of the circle circumscribing the triangle does not lie inside the triangle.

Circum-center lies on the triangle = Right-angled triangle
Circum center lies outside the triangle = Obtuse Triangle
Hence two possible cases, not sufficient

Combined:
Still not sufficient

The statement provides us with one valuable information: we can form a triangle with sides a², b², c².

For any triangle we know that sum of two sides is greater than the third side.

So, we can infer that a² < b² + c².

The inequality above is the condition to be met if the triangle with sides a, b and c were to be an acute triangle.
Eliminate choices B, C and E. Choices narrow down to A or D.

When a 5 cc cube is sliced into 1 cc cubes, we will get 5 * 5 * 5 = 125 cubes of 1 cubic centimeter.

In each side of the larger cube, the smaller cubes on the edges will have more than one of their sides painted.
Therefore, the cubes which are not on the edge of the larger cube and that lie on the facing sides of the larger cube will have exactly one side painted.

In each face of the larger cube, there will be 5 * 5 = 25 cubes.
Of these, the cubes on the outer rows will be on the edge. 16 such cubes exist on each face.
If we count out the two outer rows on either side of a face of the cube, we will be left with 3 * 3 = 9 cubes which are not on the edge in each face of the cube.
Therefore, there will be 9 cubes of 1-cc volume per face that will have exactly one of their sides painted.
In total, there will be 9 * 6 = 54 such cubes.

The radius of the wheel measures 21 cm.
In one rotation, the wheel will cover a distance which is equal to the circumference of the wheel.
∴ in one rotation this wheel will cover 2 * π * 21 = 132 cm.

In a minute, the distance covered by the wheel = circumference of the wheel * rpm
∴ this wheel will cover a distance of 132 * 600 = 79200 cm in a minute.

In an hour, the wheel will cover a distance of 79200 * 60 = 4752000 cm.

Therefore, the speed of the car = 4752000 cm/hr = 47.52 km/hr

Let 'a' meters be the length of a side of the square field.

Therefore, its area = a² square meters. --- (1)

The length of the diagonal 'd' of a square whose side is 'a' meters =√2a ---(2)
From (1) and (2), we can deduce that the square of the diagonal = d² = 2a² = 2(area of the square)

The shape of the area used for growing cabbages has remained a square in both the years.
Let the side of the square area used for growing cabbages this year be X ft.
Therefore, the area of the ground used for cultivation this year = X² sq.ft.

Let the side of the square area used for growing cabbages last year be Y ft.
Therefore, the area of the ground used for cultivation last year = Y² sq.ft.

As the number of cabbages grown has increased by 211, the area would have increased by 211 sq ft because each cabbage takes 1 sq ft space.

Hence, X² - Y² = 211
(X + Y)(X - Y) = 211.
211 is a prime number and hence it will have only two factors. i.e., 211 and 1.
Therefore, 211 can be expressed as product of 2 numbers in only way = 211 * 1
i.e., (X + Y)(X - Y) = 211 * 1
So, (X + Y) should be 211 and (X - Y) should be 1.
Solving the two equations we get X = 106 and Y = 105.

Therefore, number of cabbages produced this year = X² = 106² = 11236.
The area in both the years are squares of two numbers.
That rules out choice D. 12696 is not the square of any number.
Check Choice A: If this year's produce is 11236, last year's produce would have been 11236 - 211 = 11025
11025 is the square of 105. 
So, 11236 is the answer.

The cow can graze the area covered by the circle of radius 19m initially, as the length of the rope is 19m.

Area of a circle = π * (radius)²
Therefore, the initial area that the cow can graze = 
When the length of the rope is increased to 30m, grazing area becomes = 
The additional area it could graze when length is increased from 19m to 30m