Very elementary geometry is tested in the IBM IPAT. Most of it just involves the algebraic manipulations of basic geometry formulas. So let’s dig right in with a few examples but before that here is a refresher of the formulas you will encounter in the IPAT.

Square | Perimeter = 4 x a Area = a ^{2}where a is the side length |

Rectangle | Perimeter = 2(l+b) Area = l x b where l is the length and b is the width of the rectanle |

Circle | Perimeter = 2 x pi x r Area = pi x r ^{2}where r is radius of circle |

Triangle | Perimeter = Sum of all sides Area = 1/2 x height x base |

Cylinder | Curved area = 2 x pi x radius x height Total area = top circle area + bottom circle area + curved area Volume = pi x radius ^{2} x height |

Cuboid | Surface area = 2 x ( length x base + base x height + height x length) Volume = length x base x height |

Cone | Curved area = pi x radius x slant height slant height ^{2} = radius^{2}+height^{2}Volume = 1/3 x pi x radius ^{2} x height |

Sphere | Surface area = 4 x pi x radius^{2}Volume = 4/3 x pi x radius ^{3} |

Pythagoras Theorem | In a right triangle, Longest side ^{2} = base^{2} + height^{2} |

Let’s try out two examples which have come in the IPAT.

Q) If the length of a rectangle is increased by 7 cm and its width increased by 5 cm, the resulting rectangle is 100 cm^{2} greater in area than before. If the perimeter of the rectangle is 20 cm, what is the length of the original rectangle?

A] 4.4 cm

B] 3.6 cm

C] 3.2 cm

D] 2.5 cm

E] 2.6 cm

Solution

**Solution – D] 2.5 cm**

Let l be length of the rectangle and its width be w. Converting the information given to us into linear equations:

New rectangle area – old rectangle area = 100 cm, therefore (l+7)(w+5)-(l)(w) = 100 cm^{2}

7w+5l = 65

Perimeter of original rectangle = 20 cm

2(l+w) = 20 therefore l+w = 10

Hence we are left with, l+w = 10 and 5l+7w = 65

Solving the two equations we have length of original rectangle = 2.5 cm

Q) If the dimensions of a rectangle are in the ration 7:5 and the total perimeter reduction achieved by a shrinking length and width in their original ratio has been 30 cm. What is the mathematical difference between the old width and new width?

A] 4.02 cm

B] 6.25 cm

C] 7.50 cm

D] 8.80 cm

E] 9.40 cm

Solution

**Solution – B] 6.25 cm**

Change in perimeter = 30 cm

Change in[ 2(l+b) ] = 30 cm

2 x Change in (l+b) = 30 cm

Change in (l+b) = 15 cm

Since the reduction in dimensions follow the original ratio, 7: 5, change in width = 15 cm x 5/(5+7) = 15 x 5/12 = 6.25 cm which is our answer.

We wanted to give you a quick taste of the type of problems to expect in the IPAT geometry section. As always, for more practice do check out our comprehensive IPAT word problem guide here which supports a full length practice format to train yourself before the test.