Important Questions for Class 8 Maths Chapter 9 PDF Download

Radius of the base of the cone=21 cm and slant height (l) = 60 cm $\therefore Curvedsurfacearea=\pi rl=\frac{22}{7}\times 21\times 60=3960c{m}^2$

7. The lateral surface of a cylinder is equal to the curved surface of a cone. If the radius be the same find the ratio of the height of the cylinder and slant height of the cone.

Solution: Let r be the radius of both the cylinder and the cone Again let h and l respectively be the height and slant height of the cylinder and the cone Then lateral surface of the cylinder =curved surface of the cone $\Rightarrow 2\pi rh=\pi rl\Rightarrow 2h=l$ $\Rightarrow \frac{h}{l}=\frac{1}{2}\Rightarrow h:1=1:2$

8. In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.

Solution:

From the question statement draw the diagram. Consider a trapezium of ABCD. Let AB and DC be the parallel sides Now, CM will be the distance between the two parallel sides or the height of the trapezium. We know, Area of trapezium = ½ × sum of parallel sides × height. So, height has to be found. In the diagram, draw CL || AD Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm As AD = CB, CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height. Here, BL = AB – AL = (40 – 20) = 20 cm. So, LM = MB = ½ BL = ½ × 20 = 10 cm Now, in ΔCLM, CL ² = CM ² + LM ² (Pythagoras Theorem) 26 ² = CM ² + 10 ² CM ² = 26 ² – 10 ² Using algebraic identities, we get; 26 ² – 10 ² = (26 – 10) (26 + 10) hence, CM ² = (26 – 10) (26 + 10) = 16 × 36 = 576 CM = √576 = 24 cm Now, the area of trapezium can be calculated. Area of trapezium, ABCD = ½ × (AB + CD) × CM = ½ × (20 + 40) × 24 Or, Area of trapezium ABCD = 720 cm ²

9. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed.

Solution:

The area of the remaining sheet after the smaller circle is removed will be = Area of the entire circle with radius 4 cm – Area of the circle with radius 3 cm We know, Area of circle = πr² So, Area of the entire circle = π(4)² = 16π cm ² And, Area of the circle with radius 3 cm which is cut out = π(3)² = 9π cm ² Thus, the remaining area = 16π – 9π = 7π cm ²

10. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted.

Solution:

Given, Length of the box, l = 2 m, Breadth of box, b = 1 m Height of box, h = 1.5 m We know that the surface area of a cuboid = 2(lb + lh + bh) But here the bottom part is not to be painted. So, Surface area of box to be painted = lb + 2(bh + hl) = 2 × 1 + 2 (1 × 1.5 + 1.5 × 2) = 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 Hence, the required surface area of the cuboidal box = 11 m ²

11. A square and a rectangle have the same perimeter. Calculate the area of the rectangle if the side of the square is 60 cm and the length of the rectangle is 80 cm.

Solution:

Perimeter of square formula = 4 × side of the square Hence, P (square) = 4 × 60 = 240 cm Perimeter of rectangle formula = 2 × (Length + Breadth) Hence, P (rectangle) = 2 (80 + Breadth) = 160 + 2 × Breadth According to the given question, 160 + 2 × Breadth = 240 cm 2 × Breadth = 240 – 160 Breadth = 80/2 The breadth of the rectangle = 40 cm Now, the area of rectangle = Length × Breadth = 80 × 40 = 3200 cm ²

12. A lawnmower takes 750 complete revolutions to cut grass on a field. Calculate the area of the field if the diameter of the lawnmower is 84 cm and the length is 1 m.

Solution:

Given, length of lawnmower = 1m = 100cm Its circumference = π × D = 22/7 × 84 = 264 cm Length of field will be = 264 × 750 = 198000 cm Here, the width of field = length of the lawnmower i.e. 100 cm So, area of field = 198000 × 100 = 19,800,000 cm² Or, 1980 m²

Important Questions for Class 8 Maths Chapter 1	Important Questions for Class 8 Maths Chapter 2
Important Questions for Class 8 Maths Chapter 3	Important Questions for Class 8 Maths Chapter 4
Important Questions for Class 8 Maths Chapter 5	Important Questions for Class 8 Maths Chapter 6
Important Questions for Class 8 Maths Chapter 7	Important Questions for Class 8 Maths Chapter 8
Important Questions for Class 8 Maths Chapter 9	Important Questions for Class 8 Maths Chapter 10
Important Questions for Class 8 Maths Chapter 11	Important Questions for Class 8 Maths Chapter 12

Benefits of Solving Important Questions for Class 8 Maths Chapter 9