Geometry - Basics - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Geometry - Basics. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

1. IIFT 2013 QA | Geometry - Basics

A rod is cut into 3 equal parts. The resulting portions are then cut into 12, 18 and 32 equal parts, respectively. If each of the resulting portions have integer length, the minimum length of the rod is

A.
6912 units
B.
864 units
C.
288 units
D.
240 units

2. IIFT 2010 QA | Geometry - Basics

In a square of side 2 meters, isosceles triangles of equal area are cut from the corners to form a regular octagon. Find the perimeter and area of the regular octagon.

A.
$\frac{16}{2 + \sqrt{2}}$ ; $\frac{4 (1 + \sqrt{2})}{3 + 2 \sqrt{2}}$
B.
$\frac{8}{2 + \sqrt{2}}$ ; $\frac{2 (1 + \sqrt{2})}{3 + 2 \sqrt{2}}$
C.
$\frac{16}{1 + \sqrt{2}}$ ; $\frac{3 (1 + \sqrt{2})}{3 + 2 \sqrt{2}}$
D.
None of these

Answer: Option D

Explanation :

Let the side of isosceles triangle be x.

∴ The side of octagon x $\sqrt{2}$ .

∴ x + x + x $\sqrt{2}$ = 2

∴ 2x + x $\sqrt{2}$ = 2

∴ x = $\frac{2}{2 + \sqrt{2}}$

∴ Side of octagon = x $\sqrt{2}$ = $\frac{2 \sqrt{2}}{2 + \sqrt{2}}$ = $\frac{2}{1 + \sqrt{2}}$

∴ Perimeter of octagon = 8 × $\frac{2}{1 + \sqrt{2}}$ = $\frac{16}{1 + \sqrt{2}}$ units

Area of octagon = Area of square – 4 × Area of isosceles triangle

= 2² - 4 × $\frac{1}{2}$ x²

= 4 - 4 × $\frac{1}{2}$ × ${(\frac{\sqrt{2}}{1 + \sqrt{2}})}^{2}$

= 4 - $\frac{4}{3 + 2 \sqrt{2}}$ = $\frac{8 (1 + \sqrt{2})}{3 + 2 \sqrt{2}}$ sq. units

Hence, option 4.

3. IIFT 2009 QA | Geometry - Basics

Let A₁ be a square whose side is ‘a’ metres. Circle C₁ circumscribes the square A₁ such that all its vertices are on C₁. Another square A₂ circumscribes C₁. Circle C₂ circumscribes A₂, and A₃ circumscribes C₂, and so on. If D_N is the area between the square A_N and the circle C_N, where N is a natural number, then the ratio of the sum of all D_N to D₁ is:

A.
1
B.
$\frac{π}{2}$ - 1
C.
Infinity
D.
None of the above

Answer: Option C

Explanation :

Area of square 1 = a²

Area of circle 1 = π ${(\frac{a}{\sqrt{2}})}^{2}$

∴ D₁ = $\frac{a^{2}}{2}$ (π - 2)

Area of square 2 = (a $\sqrt{2}$ )²

Area of circle 2 = πa²

∴ D₂ = a²(π - 2)

Similarly,

D₃ = 2a² (π - 2), and so on.

∴ D₁ + D₂ + D₃ + ... = a²(π - 2) [0.5 + 1 + 2 + 4 + ...]

= Infinite

Hence, option 3.

Geometry - Basics - Previous Year CAT/MBA Questions

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