Geometry And Mensuration Notes Free PDF In English

By Anil kumawat Last updated Mar 24, 2023

Geometry And Mensuration Notes Free PDF In English

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Geometry & Mensuration Bilingual Notes pdf Download:- Today we are sharing a very important and easy PDF of Mensuration and Mensuration Formulas PDF. We have also included some of the most important questions related to Download Geometry and Mensuration notes pdf in English for your better preparation for all the government exams U.P.P, UPSI, UPTGT, PGT, UPTET/CTET, HTET, RTET, UDA/LDA, RO/ARO, Bed, LLB, RRB, UPSC, PATWARI, VDO, ACCOUNTANT, LDC|UDC, HIGH COURT LDC, POLICE, SI, DSSSB, MCD, DDA, NET, ALL STATE PCS EXAM.

If you are preparing for your exams in the last few days then this Download Geometry and Mensuration notes pdf in English 2023 for Free is very important for you.

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Questions on Mensuration with Solutions

Q 1. A rhombus having diagonals of length 10 cm and 16 cm, respectively. Find its area.

Solution: d1 = 10 cm

d2 = 16 cm

Area of rhombus = ½ d1 d2

A = ½ x 10 x 16

A= 80 cm2

Q 2. The area of a trapezium shaped field is 480 m2, the distance between two parallel sides is 15 m and one of the parallel sides is 20 m. Find the other parallel side.

Solution: One of the parallel sides of the trapezium is a = 20 m, let another parallel side be b, height h = 15 m.

The given area of trapezium = 480 m2

We know, by formula;

Area of a trapezium = ½ h (a+b)

480 = ½ (15) (20+b)

20 + b = (480×2)/15

b = 64 – 20 = 44 m

Total Surface Area of Solid Shapes

TSA of Cuboid = 2(lb + bh + hl)

TSA of Cube = 6l2

TSA of Cylinder = 2πr (r + h)

Q 3. The height, length and width of a cuboidal box are 20 cm, 15 cm and 10 cm, respectively. Find its area.

Solution: Total surface area = 2 (20 × 15 + 20 × 10 + 10 × 15)

TSA = 2 ( 300 + 200 + 150) = 1300 cm2

Q 7. If a cube has its side-length equal to 5cm, then its area is?

Solution: Given,

l = 5 cm

Area = 6l2 = 6 x 5 x5 = 150 sq.cm

Q 4. Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm2.

Solution: : Let height of the cylinder = h, radius = r = 7cm

Total surface area = 2πr (h + r)

TSA = 2 x (22/7) x 7 x (7+h) = 968

h = 15 cm

Volume of Cube, Cuboid and Cylinder

V of cube = l3

Volume of cuboid = l × b × h

Volume of cylinder = πr2h

Check: Mensuration Formulas Class 10

Q 5. Find the height of a cuboid whose volume is 275 cm3 and base area is 25 cm2.

Solution: Volume of cuboid = l × b × h

Base area = l × b = 25 cm2

Hence,

275 = 25 × h

h = 275/25 = 11 cm

Q 6. A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.

Solution: Length of the paper will be the perimeter of the base of the cylinder and width will be its height.

Circumference of base of cylinder = 2πr = 11 cm

2 x 22/7 x r = 11 cm

r = 7/4 cm

Volume of cylinder = πr2h = (22/7) x (7/4)2 x 4

= 38.5 cm3

Q 7. If the area of an equilateral triangle is 36 $\sqrt{3}$ sq.m., Find its Perimeter.

ANSWER:

Area of the equilateral triangle = $36 \sqrt{3} sq m$
The area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} (side)^{2}$ .
$i . e ., \frac{\sqrt{3}}{4} (side)^{2} = 36 \sqrt{3} \Rightarrow (side)^{2} = 36 \sqrt{3} \times \frac{4}{\sqrt{3}} \Rightarrow (side)^{2} = 144 \Rightarrow s ide = 12 m$

∴ Perimeter of the equilateral triangle = 3 × side
= 3 × 12 m
= 36 m

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Questions on Mensuration with Solutions

Q 1. A circle has a radius of 21 cm. Find its circumference and area. (Use π = 22/7)

Solution: We know,

Circumference of circle = 2πr = 2 x (22/7) x 21 = 2 x 22 x 3 = 132 cm

Area of circle = πr2 = (22/7) x 212 = 22/7 x 21 x 21 = 22 x 3 x 21

Area of circle with radius, 21cm = 1386 cm2

Q 2. If one side of a square is 4 cm, then what will be its area and perimeter?

Solution: Given,

Length of side of square = 4 cm

Area = side2 = 42 = 4 x 4 = 16 cm2

Perimeter of square = sum of all its sides

Since, all the sides of the square are equal, therefore;

Perimeter = 4+4+4+4 = 16 cm

Area of quadrilateral = ½ d (h1+h2)

Where d is the diagonal of quadrilateral dividing it into two triangles

h1 and h2 are the heights of two triangles falling on the same base (diagonal of quad.)

Area of Rhombus = ½ d1 d2

Where d1 and d2 are diagonals of rhombus

Area of trapezium = ½ h (a+b)

Where a and b are the two parallel sides of trapezium

h is the distance between a and b.

Q 3. Suppose a quadrilateral having a diagonal of length 10 cm, which divides the quadrilateral into two triangles and the heights of triangles with diagonals as the base, are 4 cm and 6 cm. Find the area of the quadrilateral.

Solution: Given,

Diagonal, d = 10 cm

Height of one triangle, h1 = 4cm

Height of another triangle, h2 = 6cm

Area of quadrilateral = ½ d(h1+h2) = ½ x 10 x (4+6) = 5 x 10 = 50 sq.cm.

Q 4. If the area of an equilateral triangle is 363–√ sq.m., Find its Perimeter.

ANSWER:
Area of the equilateral triangle = 363–√ sq m
The area of an equilateral triangle is given by 3√4(side)2.
i.e., 3√4(side)2 = 363–√⇒(side)2 = 363–√×43√⇒(side)2 = 144⇒ side = 12 m

∴ Perimeter of the equilateral triangle = 3 × side
= 3 × 12 m
= 36 m

Q 2. The area of an equilateral triangle is 813–√ sq.m., Find its height.

ANS. Area of the equilateral triangle = 813 −−√ sq m
The area of an equilateral triangle is given by 3√4(side)2.
i.e., 3√4(side)2 = 813–√⇒(Side)2 = 813–√×43√ = 324⇒Side = 18 m

Side of the equilateral triangle = 18 m
Area of the equilateral triangle = 12×Base×Height

⇒813–√ = 12×base×height⇒813–√ = 12×18×height⇒Height = 813√9 m = 93–√m

∴ The height of the equilateral triangle is 93–√ m.

Q 3. The base of an isosceles triangle measures 80 cm and its area is 360 sq.m., Find the perimeter of the triangle.

ANS. Base of an isosceles triangle = c = 80 cm
Its area = 360 m2

Let the congruent sides of an isosceles triangle be x m.
i.e., a = b = x m

Using Heron’s formula, we have:

s = a+b+c2 = x+x+802 =2x+802 = x+40

Area of an isosceles triangle = s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√

⇒ 360 = (x+40)(x+40−x)(x+40−x)(x+40−80)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√⇒ 360 = (x+40)(40)(40)(x−40)−−−−−−−−−−−−−−−−−−−−√⇒ 360 = 40(x2−402)−−−−−−−−−√⇒36040 = x2−1600−−−−−−−−√⇒92= x2−1600⇒ x2 = 1681or, x = 41 m

Length of the congruent sides of an isosceles triangle = 41 m
∴ Perimeter of an isosceles triangle =Sum of all sides
= (41 + 41 + 80) cm
= 162 cm

Q 4. The perimeter of an isosceles triangle is 42 cm and its base 112 times its congruent sides. Find
(i) the length of congruent sides of a triangle

ANS.(i) Let the length of the congruent sides be x cm.
Then length of the base = 112x = 32x cm
Perimeter of the isosceles triangle = sum of all sides
⇒ 42 = x + x + 32x
⇒ 42 = 2x+32x⇒42 = 7×2⇒x = 12 cm

∴ Length of the congruent sides = 12 cm

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