# Geometry and Mensuration

Geometry and Mensuration: Geometry Formulas and Mensuration Formulas. Circle Geometry. Mensuration Meaning. Mensuration Problems. Geometry Problems.

## Geometry and Mensuration: Geometry Formulas and Mensuration Formulas. Circle Geometry. Mensuration Meaning. Mensuration Problems. Geometry Problems.

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## Learning Pundits Content Team

__Cracking Aptitude Questions on Geometry__

__Cracking Aptitude Questions on Geometry__Looking for Questions instead of tips?- You can directly jump to Aptitude Test Questions on Geometry and Mensuration

__Part – I__: Triangles

### Angles and Sides:

1. The sum of any two sides of a triangle is greater than the third side.

2. The line joining the mid-point of a side of a triangle to the opposite vertex is called the median. The median of a triangle divides it into two triangles of the same area.

3. The point of intersection of the 3 medians of a triangle is called its *centroid.* The centroid divides each of the medians in the ratio 2: 1.

4. The perimeter of a triangle of sides of length a, b and c is (a + b + c).

**Question: **In triangle PQR length of the side QR is less than twice the length of the side PQ by 2 cm. Length of the side PR exceeds the length of the side PQ by 10 cm. The perimeter is 40 cm. The length of the smallest side of the triangle PQR is:

**Solution:**

QR = 2PQ – 2 and PR = PQ + 10

PQ + QR + PR = 40

=> PQ + 2PQ - 2 + PQ +10 = 40

=> 4PQ = 32

PQ = 8 cm which is the smallest side of the triangle.

**Question: **In an isosceles triangle ABC, ∠ A = 90^{0}, AL is drawn perpendicular to BC. Find ∠ BAL.

**Solution:**

∠A + ∠B + ∠C = 180^{0}

∠B = ∠C = (180^{0} – 90^{0}) / 2 = 45^{0}

∠BAL + ∠BLA + ∠B = 180^{0}

=> **∠BAL = 180**^{0 }**– 90**^{0}** – 45**^{0}** = 45**^{0}** **

### Area:

1. Area of a triangle = Base x Height / 2.

2. Area of a triangle is **√**s(s – a)(s – b)(s – c) where s = semi perimeter = (a + b + c) / 2.

3. The area of the triangle formed by joining the mid-points of the sides of a given triangle is one-fourth of the area of the given triangle.

**Question: **Triangle ABC is inscribed in a square of side 20 cm. Find the area of the triangle.

**Solution:**

From the figure, we can see that

Base = 20 cm, Height = 20 cm.

**Thus, area of the triangle = 20 x 20 / 2 = 200 cm**^{2 }** **

**Question: **What is the height of the triangle?

I. The area of the triangle is 20 times its base.

II. Perimeter of the triangle = Perimeter of a square of side 10 cm.

*Options:*

A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

D. Both I and II are not sufficient to answer

E. Both I and II are necessary to answer

**Solution:**

I. 1/2 x B x H = 20 x B or, H = 40

I alone gives the answer. **II**. Gives, perimeter of the triangle = 40 cm but this does not give the height of the triangle.

**Correct answer is (A).**

### Types and Properties:

**1. Equilateral Triangle: **

** **

a) The angles of the triangle are equal to 60⁰ each.

b) All the sides of the triangle are equal.

c) Altitude of an equilateral triangle = √3a / 2

d) Area of an equilateral triangle = √3a^{2}/4

**2. Isosceles Triangle: **

** **

a) The two base angles of the triangle are equal.

b) Two sides of the triangle are equal.

**3. Scalene Triangle: **

** **

a) No two angles have the same value.

b) No two sides have the same length.

**4. Acute-angled Triangle: **

a) All angles of the triangle are lesser than 90⁰.

**5. Right-angled Triangle: **

** **

a) One angle of the triangle is 90⁰ while the other 2 are lesser than 90⁰ each.

b) a^{2} = b^{2} + c^{2}

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__Part – II__: Quadrilaterals

### Types:

1. * Trapezium: *A quadrilateral having exactly one pair of parallel sides is called a trapezium.

** **

2. *I sosceles Trapezium: *A trapezium whose non-parallel sides are equal in length is called an isosceles trapezium.

3. * Parallelogram: *A quadrilateral is said to be a parallelogram if both pairs of its opposite sides are parallel.

4. * Rhombus:* A parallelogram having all sides equal is called a rhombus.

5. * Rectangle: *A quadrilateral in which each angle is a right angle is called a rectangle.

6. * Square: *A square is a quadrilateral in which all sides are equal and each angle measures 90⁰.

### Angles and Sides:

1. The diagonals of a parallelogram bisect each other.

2. Perimeter of a rectangle = 2(Length + Breadth)

3. Perimeter of a square = 4 x length of a side

4. Each diagonal of a parallelogram divides it into triangles of the same area.

5. The diagonals of a rectangle are equal and bisect each other.

6. The diagonals of a square are equal and bisect each other at right angles.

7. The diagonal of a square of side a = a√2.

8. The diagonals of a rhombus are unequal and bisect each other at right angles.

9. The sum of angles of a quadrilateral is 360 °

**Question: **What is the smaller angle of a parallelogram?

I. Ratio of the angles of a triangle is 3:4:5 and the larger angle of the parallelogram is 34^{0} greater than the largest angle of the triangle.

II. Larger angle of the parallelogram is 38° more than its smaller angle.

A. I alone sufficient while II alone not sufficient to answer

B. II alone sufficient while I alone not sufficient to answer

C. Either I or II alone sufficient to answer

D. Both I and II are not sufficient to answer

E. Both I and II are necessary to answer

**Solution: **

I. The sum of the angles of a triangle is 180^{0}. Thus, the angles can be determined and hence, larger angle of parallelogram. Larger angle + Smaller angle = 180^{0}. So I alone is sufficient.

II. Sum and difference of angles is known. Both angles can be calculated. II alone is sufficient.

**Thus, the correct answer is C.**

### Area:

1. Area of a rectangle = (Length x Breadth).

2. Area of a square = (side)^{2} = ½ x (diagonal)^{2}.

3. Area of parallelogram = (Base x Height).

4. Area of a rhombus = ½ x (Product of diagonals).

5. Area of a trapezium = ½ x (sum of parallel sides) x distance between them.

**Question: **An error of 2% in excess is made while measuring the side of a square. Calculate the percentage of error in the calculated area of the square.

**Solution:**

Let the actual length of side be x. Then measured length = 1.02x.

Actual area = x^{2}. Measured area = (1.02x)^{2}.

Error in area calculation = (1.02^{2} – 1^{2})x^{2} / x^{2} = (1.02 + 1)(1.02 – 1) = 2.02 x 0.02 = 0.0404

**Percentage error = 0.0404 x 100% = 4.04%**

**Question: **A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

**Solution:**

Let the width of the road be x meters.

Then, (60 – 2x)(40 – 2x) = 2109

=> 2400 – 200x + 4x^{2} = 2109

=> 4x^{2} – 200x + 291 = 0

=> x = (200 ± 188) / 8 = 1.5 or 48.5

But width of road cannot be greater than breadth of the park. Thus, **width of road = 1.5m.**

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__Part – III__: Circles – Angles, Radii and Area

1. Area of a circle = πR^{2}, where R is the radius

2. Circumference of a circle = 2πR

3. Length of an arc = 2πRθ / 360, where θ is the central angle.

4. Area of a sector = ½ x (arc length x R) = πR^{2}θ / 360

5. Area of semi-circle = πR^{2} / 2

**Question: **A lawn is in the shape as shown. Find the area of the lawn.

**Solution: **

The lawn is in the form of a rectangle with two semicircles on opposite ends of the width.

Area of the rectangle = 200m x 14m = 2800 sq. m

Radius of the semicircles = 14/2 m = 7 m

Area of the two semicircles = 2 x (πR^{2} / 2) = 22 / 7 x 7 x 7 = 154 sq. m.

Total Area of the lawn = 2954 sq. m.

**Question: **ABCD is a square with one vertex at the center of the circle and two vertices on the circle. What is the length of the diagonal of the square if the area of the circle is 100 square cm?

**Solution:**

Area of circle = πR^{2} = 100 sq. cm

R^{2} = 100/π Or, R = 10/√π cm

From the figure, side of square, a = Radius of circle, R = 10/√π cm

Diagonal of the square = a√2 = 10√(2/π) cm

__Cracking Aptitude Questions on Volume and Surface Area__

__Cracking Aptitude Questions on Volume and Surface Area____Part – I: __Cube

Let the length of each edge of a cube be ‘a’. Then,

1. Volume = a^{3} cubic units.

2. Surface area = 6a^{2} sq. units.

3. Diagonal = a x √3 units.

**Question: **A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?

**Solution:**

Volume of the larger cube = Total volume of the smaller cubes

= (3^{3} + 4^{3} + 5^{3}) cm^{3} = 216 cm^{3}

Side of the cube = (216)^{1/3} = 6 cm

Surface area of the larger cube = 6 x 6 x 6 = 6 x 36 cm^{2}

Total surface area of the smaller cubes = 6 x (3^{2} + 4^{2} + 5^{2}) cm^{2} = 6 x 50 cm^{2}

**Required ratio = 50 / 36 = 25:18.**

**Question: **A 4cm cube is cut into 1 cm cubes. What is the total surface area of the new cubes?

**Solution:**

Number of cubes that can be formed = Total volume / Volume of each cube

= (4 x 4 x 4) / (1 x 1 x 1) = 64.

Total surface area of all cubes = Number of cubes x Surface area of 1 cube = 64 x (6 x 1 x 1) = **384 cm**^{2}

__Part – II:__ Cuboid

Let the length, breadth and height be ‘l’, ‘b’ and ‘h’ respectively. Then,

1. Volume = (l x b x h) cubic units.

2. Surface area = 2(lb + bh + lh) sq. units.

3. Diagonal = √ (l^{2} + b^{2} + h^{2)} units.

**Question: **50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the average displacement of water by a man is 4 m^{3}, then calculate the rise in the water level.

**Solution:**

Total volume of water displaced = 50 x 4 m^{3} = 200 m^{3}

Change in volume of the tank occupied = length x breadth x change in height.

=> ** Change in height = 200 m**^{3}** / (40 m x 20 m ) = 0.25 m = 25 cm.**

**Question: **A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. What is the surface area of the wet surface?

**Solution:**

Surface area of wet surface = Area occupied – Area of top surface (since it is an open surface)

= Area of cuboid – Area of base

= 2(lb + bh + lh) – lb

= 2(6x4 + 4x1.25 + 6x1.25) – 6x4

= 2x1.25x10 + 24 = **49 m**^{2}

__Part – III__: Cylinder

Let radius of the base of a cylinder be ‘r’ and its height be h. Then,

1. Volume = π (r^{2}h) cubic units.

2. Curved surface area = (2 π rh) sq. units.

3. Total surface area = 2 π r(h + r) sq. units.

**Question: **A hollow iron pipe is 21 cm long and its external diameter is 8 cm. If the thickness of the pipe is 1 cm and iron weighs 8 g/cm^{3}, what is the weight of the pipe?

**Solution:**

External radius R = 4 cm, Internal radius r = (4 – 1) cm = 3 cm

Volume of the pipe = Total solid volume – Volume of hollow part

= π(R^{2} – r^{2})h

= 22 / 7 x (4x4 – 3x3) x 21 = 462 cm^{3}.

Mass of the pipe = density x volume

**= 8 x 462 = 3696 gm = 3.696 kg.**

**Question: **66 cm^{3} of silver is drawn into a wire 1 mm in diameter. Find the length of the wire in meters.

**Solution: **

Radius of the wire r = 0.5 mm = 0.05 cm and let length = l.

Volume of the wire = 66 cm^{3} = π r^{2} l = 22 / 7 x 0.05 x 0.05 x l

=> ** l = 66 x 7 / (22 x 0.05 x 0.05) = 8400 cm = 84 m.**

__Part – IV__: Sphere and Hemi-Sphere

Let the radius of the sphere be r. Then,

1. Volume = (4 / 3) π r^{3} cubic units.

2. Surface area = (4 π r^{2}) sq. units.

For a hemisphere of radius r, we have

1. Volume = (2 / 3) π r^{3} cubic units.

2. Curved surface area = (2 π r^{2}) sq. units.

3. Total surface area = (3 π r^{2}) sq. units.

**Question: **The volume of a spherical ball is 121000 cm^{3}. What is the radius of the ball?

**Solution:**

(4 / 3) π r^{3} = 121000

=> r^{3} = 121000 x 3 / 4 x 7 / 22 = 28875

=> **r = (28875)**^{1/3}** = 30.68 cm.**

**Question: **Calvin is making a model Earth in his science class. The radius of the inner core of the model Earth is 4 cm and the radius of the entire model is 10 cm. How many times larger is the volume of the entire model than the volume of the inner core?

**Solution: **

Volume of core = (4 / 3) π x 4 x 4 x 4 cm^{3}

Volume of earth = (4 / 3) π x 10 x 10 x 10 cm^{3}

Volume of entire model = (1000 / 64) times the volume of the core

= **15.625 times the volume of the core.**

__Part – V:__ Cone

Let radius of base = r and Height = h. Then,

1. Slant height, l = √(h^{2} + r^{2}) units.

2. Volume = (π r^{2}h / 3) cubic units.

3. Curved surface area = (π rl) sq. units.

4. Total surface area = πrl + π r^{2} sq. units.

**Question:** A right triangle with sides 3 cm, 4 cm and 5 cm is rotated along the side of 3 cm to form a cone. Find the volume of the cone so formed.

**Solution:**

From the question, we can say that r = 3 cm, h = 4 cm and l = 5 cm.

Thus, **volume of the cone = π x 3 x 3 x 4 / 3 **

** = 12 π cm**^{2}**.**

**Question:** The slant height of a right circular cone is 10 m and its height is 8 m. Find the area of its curved surface.

**Solution: **

Given, l = 10 m and h = 8 m.

r = √(l^{2} – r^{2}) = √(100 – 64) = 6 m

**Curved surface area = π x r x l **

** = π x 6 x 10 = 60 π m**^{2}**.**

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# Geometry and Mensuration

Geometry and Mensuration: Geometry Formulas and Mensuration Formulas. Circle Geometry. Mensuration Meaning. Mensuration Problems. Geometry Problems.

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