You are here:Home >>Aptitude Questions>>Geometry

Geometry Aptitude Questions

Quantitative Aptitude Questions and Answers section on “Geometry” with solution and explanation for competitive examinations such as CAT, MBA, SSC, Bank PO, Bank Clerical and other examinations.

1.

The in-radius of an equilateral traingle is of length 3 cm. Then the length of each of its medians is :
[A]4 cm
[B]9 cm
[C]9.5 cm
[D]12 cm

9 cm
In the equilateral triangle centroid, incentre, orthocentre, coincide at the same point.
∴ Height \div 3 = in radius
∴ Height = Median = 3\times3 = 9 cm
Hence option [C] is the right answer.

2.

If the orthocentre and the centroid of a triangle are the same, then the triangle is:
[A]Obtuse angled
[B]Right angled
[C]Scalene
[D]Equilateral

Equilateral
In equilateral triangle orthocentre and centroid lie at the same point.
Hence option [D] is the right answer.

3.

If in a triangle, the circumcentre, incentre, centroid and orthocentre coincide, then the triangle is :
[A]Rigth angled
[B]Equilateral
[C]Isosceles
[D]Acute angled

Equilateral
In an equilateral triangle, centroid, incentre etc lie at the same point.
Hence option [C] is the right answer.

4.

If ABC is an equilateral triangle and D is a point on BC such that AD ⊥ BC, then :
[A]AB : BD = 2 : 1
[B]AB : BD = 1 : 2
[C]AB : BD = 1 : 1
[D]AB : BD = 3 : 2

AB : BD = 2 : 1
Let AB = 2x units
=> BD=DC = x units
∴ AB : BD = 2 : 1
Hence option [C] is the right answer.

5.

The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ∠PSR is :
[A]45°
[B]30°
[C]15°
[D]60°

30°
∠PRQ = 60°
∠PRS = 180° – 60° = 120°
=>∠PSR + ∠RPS = 60°
As RS = PR
∴ ∠PSR = ∠RPS
∴ ∠PSR = \frac{60^{\circ}}{2} = 30^{\circ}
Hence option [C] is the right answer.

6.

ABC is an equilateral triangle and CD is the internal bisector of ∠C. If DC is produced to E such that AC = CE, then ∠CAE is equal to :
[A]15°
[B]45°
[C]30°
[D]60°

15°
∠BCD = ∠DCA = 30°
∠DCE = 180°
∴ ∠ACE = 180° – 30° = 150°
AC = CE
∴ ∠CAE = ∠CEA = \frac{30^{\circ}}{2} = 15^{\circ}
Hence option [A] is the right answer.

7.

G is the centroid in the equilateral △ ABC. If AB = 10 cm then length of AG is :
[A]\frac{20\sqrt{3}}{3}cm
[B]10\sqrt{3}cm
[C]\frac{10\sqrt{3}}{3}cm
[D]5\sqrt{3}cm

\mathbf{\frac{10\sqrt{3}}{3}cm}
AB = 10cm
BD = 5 cm
∠ADB = 90°
∴ AD = \sqrt{AB^{2} - BD^{2}}
=> \sqrt{10^{2}-5^{2}} = \sqrt{100-25}
=> \sqrt{75} = 5\sqrt{3}cm
AG = \frac{2}{3} AD= \frac{2}{3}\times 5\sqrt{3}
= \frac{10\sqrt{3}}{3}cm
Hence option [C] is the right answer.

8.

The radius of the incircle of the equilateral triangle having each side 6 cm is
[A]3cm
[B]\sqrt{3}cm
[C]2\sqrt{3}cm
[D]6\sqrt{3}cm

\mathbf{\sqrt{3}cm}
E = In-center, AD ⊥ BC
AB = 6 cm, BD = 3 cm
∠ADB = 90°
∴ AD = \sqrt{AB^{2} - BD^{2}}
=> \sqrt{6^{2}-3^{2}} = \sqrt{36-9}
=> \sqrt{27} = 3\sqrt{3}cm
∴ In-radius = \frac{1}{3}AD
=>\frac{1}{3}\times 3\sqrt{3} = \sqrt{3}cm
Hence option [B] is the right answer.

9.

If ABC is an equilateral triangle and P, Q, R respectively denote the middle points of AB, BC, CA then
[A]PQR must be a right angled triangle
[B]PQR must be an equilateral triangle
[C]PQ + QR + PR = AB
[D]PQ + QR + PR = 2AB

PQR must be an equilateral triangle
The line segments joining the mid points of the sides of a triangle form four triangles, each of which is similar to the original triangle.
So PQR must be an equilateral triangle.

10.

If the incentre of an equilateral triangle lies inside the triangle and its radius is 3 cm, then the side of the equilateral triangle is:
[A]6 cm
[B]9\sqrt{3}cm
[C]6\sqrt{3}cm
[D]3\sqrt{3}cm

\mathbf{6\sqrt{3} cm}
In radius =
= \frac{Side}{2\sqrt{3}}
=> 3 = \frac{Side}{2\sqrt{3}}
=> Side = 3\times 2\sqrt{3} = 6\sqrt{3} cm
Hence option [C] is the right answer.



Advertisement

12