Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.1 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.1 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.1: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 7 Exercise 7.1 – Mensuration on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 7 Exercise 7.1 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 7.1

TN Samacheer Kalvi 10th Maths Solutions Chapter 7 – Mensuration

Board	TNSCERT Class 10th Maths
Class	10
Subject	Maths
Chapter	7 (Exercise 7.1)
Chapter Name	Mensuration

TNSCERT Class 10th Maths Pdf | all Exercise Solution

 

UNIT – 7.1

MENSURATION

 

(1) The radius and height of a cylinder are in the ratio 5:7 and its curved surface area is 5500 sq.cm. Find its radius and height.

Solution:

Given that, the radius and height ratio 5:7

Let, the radius of the cylinder is 5x and the height 18 7x.

We know that a cylinder curved surface area is 2πrh, where r = radius and h = height.

Now, 2πrh = 5500

2×22/7 × 5x × 7x = 5500 [∵π = 22/7]

x² = 5500/2×22×5

x² = 25

x = 5

∴The cylinder radius = 3x = 5×5 = 25cm

The cylinder height = 7x = 7×5 = 35cm.

Thus, the cylinder radius and height are 25cm and 35cm.

 

(2) A solid iron cylinder has total surface area of 1848 sq. m. Its curved surface area is five – sixth of its total surface area. Find the radius and height of the iron cylinder.

Solution:

Let, the cylinder radius and height are r m and h m.

We know that the cylinder total surface area is 2πr (h + r) sq. m.

Now, 2πr (h + r) = 1848

r (h + r) = 1848×7/22×2

r (h + r) = 294 —- (i)

Now, curved surface area = 5/6 total surface area

2πrh = 5/6 2πr (h + r)

2πrh/2πr (h + r) = 5/6

h/h+r = 5/6

5 (h + r) = 6h

5r = 6h – 5h

h = 5r —- (ii)

5 (h + r) = 6h

From (i), putting h = 5r, we get

r (r + h) = 294

r (r + 5r) = 294

r× 6r = 294

6r² = 294

r² = 49

r = 7m

From (ii), r = 7 putting, we get

h = 5r = 5×7 = 35 m

Thus, the cylinder radius and height are 7m and 35m

 

(3) The external radius and the length of a hollow wooden log are 16 cm and 13 cm respectively. If its thickness is 4 cm then find its T.S.A.

Solution:

Given that the hollow wooden height (h) = 13 cm

External radius (R) = 16cm

Thickness = 4cm

Now, Internal radius (r) = R – thickness

= (16 – 4) cm

= 12 cm

We know, that the hollow cylinder T.S.A = 2π (R + r) (R – r + h) Sq. cm

= 2 × 22/7 (16 + 12) (16 – 12 + 13) sq. cm

= 2 × 22/7 × 28 × 17 sq. cm

= 2992 sq. cm.

Thus, the hollow wooden T.S.A is 2992 sq. cm.

 

(4) A right angled triangle PQR where ∠Q = 90° is rotated about QR and PQ. If QR=16 cm and PR=20 cm, compare the curved surface areas of the right circular cones so formed by the triangle.

Solution:

Given that,

R = 16cm

PR = 20cm

Now, ∆PQR, ∠Q = 90°

∴ (PR)² = (QR)² + (PQ)²

(PQ)² = (PR)² – (QR)²

(PQ)² = (20)² – (16)²

(PQ)² = 400 – 256

(PQ)² = 144

PQ = 12 cm

Now, when QR is rotated

∴r = 16cm and l = 20 cm

C.S.A of the cone

= πrl sq. cm

= 22/7× 16 × 20

= 320π sq. cm

When PQ is rotated

∴ r = 12 cm and l = 20cm

C.S.A of the cone

= πrl sq. cm

= π × 12 × 20 sq. cm

= 240π sq cm

When QR is rotated C.S.A is 80π sq.cm larger.

Thus, C.S.A of the cone when rotated about QR is larger.

 

(5) 4 persons live in a conical tent whose slant height is 19 m. If each person require 22 m² of the floor area, then find the height of the tent.

Solution:-

Given that slant height (l) = 19m

Each person floor area = 22 m²

4 person floor area = 4×22 m²

Now, πr² = 4×22

22/7 × r² = 4 × 22

r² = 28

Now, the tent height (h) = √l² – r²

= √(14)² – (√28)² m

= √361 – 28

= √333 m

= 18.24 m

Thus, the height of the tent is 18.24 m.

 

(6) A girl wishes to prepare birthday caps in the form of right circular cones for her birthday party, using a sheet of paper whose area is 5720 cm², how many caps can be made with radius 5 cm and height 12 cm.

Solution:

Given that radius (r) = 5cm

Height (h) = 12

Let slant height = l cm

We know that

l = √r² + h²

= √5² + 12²

= √25 + 144

= √169

= 13cm

Now, the C.S.A of the cone = πrl sq. cm

= π × 5 × 13 sq. cm

∴ Numbers of the caps is = 5720/π×5×13

= 5720×7/22×5×13

= 28

Therefore, 28 caps can be made.

 

(7) The ratio of the radii of two right circular cones of same height is 1:3. Find the ratio of their curved surface area when the height of each cone is 3 times the radius of the smaller cone.

Solution:-

Given that two right circular cone radius ratio is 1:3

Let smaller cone radius = x

Second cone radius = 3x

Two cone height = 3 × smaller cone radius

= 3x

Now, the ratio of the curved surface area is πr₁l₁:πr₂l₂

r₁√r₁² + h₁² : r₂ √r₂² + h₂²

x √x² + 9x² : 3x √9x² + 9x²

√10x² : 3√18x²

√10x : 3√18 x

√5×2 : 3√9×2

√5 : 3×3

√5 : 9

Therefore, the required ratio is √5 : 9

 

(8) The radius of a sphere increases by 25%. Find the percentage increase in its surface area.

Solution:

Let, the radius of the sphere is r cm then the radius increases by 25%

∴ New radius = r + r × 25/100

= r + r/4

= 5r/4 cm

Now, difference in surfaces Area,

= 4π (5r/4)² – 4π r²

= 4π (25r²/16 – r²)

= 4π × 25r² – 16r²/16

= 4π × 9r²/16

Now, percentage increases in surface area

= 4π×(9r²/16)/4πr²× 100%

= 9×100/16 %

= 56.25%

Therefore, 56.25% increases in surface area.

 

(9) The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost to paint the vessel all over at ₹0.14 per cm².

Solution:

Given, internal radius (r) = 26/2 = 10cm

External radius (R) = 38/2 = 14cm

We know that, the hollow hemisphere

T.S.A = π (3R² + r²) sq. cm

= 22/7 × {3 × (14)² + (10)²} sq. cm

= 22/7 × {588 + 100} sq. cm

= 22/7 × 688 sq. cm

∴The cost of paint the vessal

₹ 22/7 × 688 × 0.14

₹ = 22/7 × 14/100 × 688

₹ = 30272/100 = 302.72

Therefore, cost of painting ₹ 302.72.

 

(10) The frustum shaped outer portion of the table lamp has to be painted including the top part. Find the total cost of painting the lamp if the cost of painting 1 sq.cm is ₹2.

Solution:

Let slant height = l cm

l = √h² + (R – r)²

l = √(8)² + (12 – 6)² [∵ h = 8cm, R = 12cm, r = 6cm]

l = √64 + 36

l = √100

l = 10cm

C.S.A of the Frustum

= π (R + r) l sq. cm

= π (12 + 6) × 10

= 180π sq. cm

Top area = πr² = π×(6)²

= 36π sq. cm

Cost of painting the lamp.

₹ (180π + 36π) × 2

= 216π × 2

= 216 × 22/7 × 2

= 1357.72

Therefore the cost of painting the lamp is ₹1357.72

 

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 7 Mensuration Exercise 7.1

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