5 GMAT Geometry Practice Questions & Explanations

Here are 5 GMAT Geometry questions with answers and explanations:

Question 1 : 

In the figure above, ABCDEF is a regular hexagon. If area of ?ACE 100√3 cm² what is the area of the hexagon?

(A) 150

(B) 150Ö3

(C) 200

(D) 200Ö3

(E) None of these

Ans:

It is given that ABCDEF is a regular hexagon in which all the sides are equal and all the angles are also equal. Hence, ABCDEF is a symmetrical figure due to which we can break it down into identical figures.

Let’s say O is the centre of the regular hexagon. Hence, if we connect the vertices with the centre of the regular hexagon the figure will look like as under:

If we see, AOEF is a parallelogram in which AE is the diagonal. And, we know that diagonal of a parallelogram divides it into two triangles of equal areas.

So, Area of ?AOE = Area of ?AEF ............................(1)

Similarly,

Area of ?COE = Area of ?CED .....................................(2)

And, Area of ?AOC = Area of ?ABC .................................(3)

Now area of triangle ACE = Area of ?AOE + Area of ?COE + Area of ?AOC = 100√3 cm²

Area of hexagon = Area of ?AOE + Area of ?COE + Area of ?AOC + Area of ?AEF + Area of ?ABC + Area of ?CED

Using equation (1), (2) & (3);

Area of Hexagon = 2 (Area of ?AOE + Area of ?COE + Area of ?AOC)

= 2 (Area of Triangle ACE)

= 2 (100√3) = 200√3 cm²

The correct answer is D.

Question 2: What is the perimeter of an isosceles triangle PQR with integer sides if PQ = 6cm?

I. QR = 3cm

II. PR < 4 cm

Ans:

PQR is an isosceles triangle with integer sides. PQ = 6cm. We need to find the Perimeter.

1) QR = 3cm. Since its an isosceles triangle, the third side PR has to be equal to one of the other two sides. So, PR can be either 3 cm or 6 cm.

But in a triangle, sum of two sides has to be greater than the third side.

If we consider PR = QR = 3cm, then PR + QR (=6cm) is not greater than PQ(=6cm). So, PR can’t be 3cm.

Hence, PR will be 6 cm. And the perimeter = 6+6+3 = 15 cm;SUFFICIENT.

2) PR < 4cm. Considering the other two sides PQ and QR to be equal (=6cm), we still do not know the exact length of PR. It can be 1cm, 2cm or 3cm.

Hence, we can’t find out the perimeter of triangle PQR; NOT sufficient.

The correct answer is A; statement 1 alone is sufficient.

Question 3: 

squares are placed one inside another, leaving a strip of uniform width around each square. If

PQ = 2Ö2 units, QR = 2 units and RS = 10 units, what is the area of the shaded region?

(A) 16

(B) 28

(C) 32

(D) 48

(E) 64

Ans:

GIVEN:

Width of strip is uniform around each square. PQ = 2√2 units, QR = 2 units and RS = 10 units.

First of all we will draw a perpendicular PU from vertex P on side QT.

So in Triangle PUQ:-

Let PU = QU = x units. (width of the shaded region)

By applying Pythagoras theorem,

(PQ)² = (PU)² + (QU)²

(2√2)² = x² + x²

2x² = 8

x² = 4

x = 2 units.

As, RS =10 units.

So, QT = RS – QR = 10 – 2 = 8 units.

And, PV = QT – 2(PU) = 8-2(2) =4 units.

Area of Shaded region = (Area of square with side QT) – (Area of square with side PV) = (8 × 8) – (4 × 4)= 48 sq units.

The correct answer is D.

Question 4:

The figure above shows the setup of a park DEF where a concert has to be arranged. ABC represents the stage (and the shaded region represents the alley for attendees). If the area of the park is 72Ö3 square metres, and AB = BF = DA, what is the area of the stage? (in square metres)

(A) 4

(B) 8

(C) 10

(D) 12

(E) 16

Ans:

The sides in a 90-45-45 degree triangle are in particular ratio of √2:1:1

Similarly, the sides in a 90-30-60 degree triangle are in particular ratio of 2:1:√3

Using this concept in the question:

For the park which is representing a right angle triangle with angles as 90-30-60 degrees, the sides will be in the ratio of 2:1:√3.

Let’s say the length of side DE = a,

That makes the lengths of FE=√3a and DF=2a, thus maintaining the same ratio of 2:1:√3.

As, the area of the park is 72√3, thus using the formula for area of triangle,

1/2×DE×FE=72√3,

or,1/2×a×a√3=72√3,

or,a² =144

or,a=12

Thus, DF=2a=24

Now, as AB=BF=DA,

And AB+BF+DA=DF

Or, 3AB=24

Or, AB=8

As, the stage is representing a right angle triangle with angles as 90-45-45 degrees, the sides will be in the ratio of √2:1:1

Now, as AB=8, therefore,

AC=BC=8/(√2),thus maintaining the same ratio of √2:1:1

Hence, the area of the stage will be,

=1/2×AC×BC

or,= 1/2×8/(√2)×8/(√2)

or,= 64/4

or,area of stage=16 square metres

The correct answer is E.

Question 5 : 

In the figure above, AB and CD are diameters of the circle with centre as O. AEB is arc of circle with centre as D and AFB is an arc of the circle with centre as C. If AB = 20cm, what is the area of the shaded region?

(A) 50

(B) 100

(C) 150

(D) 200

(E) 250

Ans:

Considering the figure as under:

If we connect the diameter AB with the point D, it will create right angled triangle ADB with right angled at D {using the property, if we connect the point on the circumference with the diameter it will always create right angled triangle}

Now, first we will find the area of the sector AEBD and then from the sector AEBD we will subtract the area of triangle ABD which will give us the area of the segment AEBO.

Since, OB = OD {radius of the circle}

BD²=OB²+OD²

Or,BD²=10²+10²

Or,BD=10√2

Since, AEB is the arc of the circle with centre at D, thus BD will be the radius of this arc.

Hence, Area of sector AEBD = (π r² θ)/360

=(π (10√2)² 90)/360=50π

Now, Area of the triangle ABD = 1/2×AB×OD

=1/2×20×10=100

Thus, area of the segment AEBO = Area of sector AEBD – Area of triangle ABD

=50π-100

Similarly, we can find the area of segment AFBO which will also be,

=50π-100

Thus, total area of the segment AEBF = Area of segment AEBO + Area of segment AFBO

=50π-100+50π-100

=100π-200

Hence, the Area of the shaded region = Area of the complete circle – Area of the segment AEBF

Thus, Area of shaded region will be,

=π10²-(100π-200)

=π100-π100+200

=200

The correct answer is D.

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