# Geometry and Mensuration notes pdf in English

## Geometry and Mensuration notes pdf in English

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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 363–√3 sq.m., Find its Perimeter.

Area of the equilateral triangle =  363–√sqm363 sq m
The area of an equilateral triangle is given by 34(side)234(side)2.
i.e.,34(side)2=363–√(side)2=363–√×43(side)2=144side=12mi.e., 34(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

### 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.

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