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Selina solutions

CHAPTERS

1. Chapter 1- Rational and Irrational Numbers

2. Chapter 2- Compound Interest [Without Using Formula

3. Chapter 3- Compound Interest [Using Formula

6. Chapter 6- Simultaneous Equations

7. Chapter 7- Indices [exponents]

9. Chapter 9- Triangles [Congruency in Triangles]

10. Chapter 10- Isosceles Triangle

12. Chapter 12- Mid-Point and Its Converse

13. Chapter 13- Pythagoras Theorem

14. Chapter 14- Rectilinear Figures

15. Chapter 15- Construction of Polygons

19. Chapter 19- Mean and Median

20. Chapter 20- Area and Perimeter of Plane Figures

22. Chapter 22- Trigonometrical Ratios

23. Chapter 23- Trigonometrical Ratios of Standard Angles

24. Chapter 24- Solution Of Right Triangles

25. Chapter 25- Complementary angles

26. Chapter 26- Co-ordinate Geometry

A hollow square-shaped tube open at both ends is made of iron. The internal square is 5 cm side and the length of the tube is 8 cm. There are 192 cm3 of iron in this tube. Find its thickness.

Given that the volume of the iron in the tube 192 cm3

Let the thickness of the tube = x cm

Side of the external square = (5 + 2x) cm

The external volume of the tube - its internal volume= volume of iron in the tube

According to the question,

Then, the thickness is 1 cm.

Chapter 1- Rational and Irrational Numbers

Chapter 2- Compound Interest [Without Using Formula

Chapter 3- Compound Interest [Using Formula

Chapter 6- Simultaneous Equations

Chapter 7- Indices [exponents]

Chapter 9- Triangles [Congruency in Triangles]

Chapter 10- Isosceles Triangle

Chapter 12- Mid-Point and Its Converse

Chapter 13- Pythagoras Theorem

Chapter 14- Rectilinear Figures

Chapter 15- Construction of Polygons

Chapter 20- Area and Perimeter of Plane Figures

Chapter 22- Trigonometrical Ratios

Chapter 23- Trigonometrical Ratios of Standard Angles

Chapter 24- Solution Of Right Triangles

Chapter 25- Complementary angles

Chapter 26- Co-ordinate Geometry

Given that the volume of the iron in the tube 192 cm3

Let the thickness of the tube = x cm

Side of the external square = (5 + 2x) cm

The external volume of the tube - its internal volume= volume of iron in the tube

According to the question,

Then, the thickness is 1 cm.

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