हिंदी

A Roller of Diameter 0.9 M and the Length 1.8 M is Used to Press the Ground. Find the Area of the Ground Pressed by It in 500 Revolutions.(Pi=3.14) - Geometry Mathematics 2

Advertisements
Advertisements

प्रश्न

A roller of diameter 0.9 m and the length 1.8 m is used to press the ground. Find the area of the ground pressed by it in 500 revolutions.
`(pi=3.14)`

योग

उत्तर

Diameter = 0.9 m  r = 0.45 m

Height =1.8 m

Number of rotations = 500

Area of the ground pressed by the roller in one rotation = curved surface area of the roller.

= `2pi "rh"`

= 2 × 3.14 × 0.45 × 1.8

= 5.0868 m2

 Area of the ground pressed by the roller in 500 rotations.

= 500 × 5.0868 m2

= 2543.4 m2

 Area of the ground pressed by the roller is 2543.4 m2

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
2015-2016 (July)

APPEARS IN

संबंधित प्रश्न

The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.


In the following figure RP: PK= 3:2, then find the value of A(ΔTRP):A(ΔTPK).


In the given figure, AD is the bisector of the exterior ∠A of ∆ABC. Seg AD intersects the side BC produced in D. Prove that :

\[\frac{BD}{CD} = \frac{AB}{AC}\]

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.


In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`


In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.

  1. `("A"(∆"PQB"))/("A"(∆"PBC"))`
  2. `("A"(∆"PBC"))/("A"(∆"ABC"))`
  3. `("A"(∆"ABC"))/("A"(∆"ADC"))`
  4. `("A"(∆"ADC"))/("A"(∆"PQC"))`

 In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ 

 

 


In trapezium ABCD, side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15 then Find OD. 


In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.


The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.


In fig., TP = 10 cm, PS = 6 cm. `"A(ΔRTP)"/"A(ΔRPS)"` = ?


In fig. BD = 8, BC = 12, B-D-C, then `"A(ΔABC)"/"A(ΔABD)"` = ?


In fig., AB ⊥ BC and DC ⊥ BC, AB = 6, DC = 4 then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?


From adjoining figure, ∠ABC = 90°, ∠DCB = 90°, AB = 6, DC = 8, then `("A"(Δ"ABC"))/("A"(Δ"BCD"))` = ?


In ΔABC, B − D − C and BD = 7, BC = 20, then find the following ratio.

(i) `"A(ΔABD)"/"A(ΔADC)"`

(ii) `"A(ΔABD)"/"A(ΔABC)"`

(iii) `"A(ΔADC)"/"A(ΔABC)"`


If ΔABC ∼ ΔDEF, length of side AB is 9 cm and length of side DE is 12 cm, then find the ratio of their corresponding areas.


In the figure, PQ ⊥ BC, AD ⊥ BC. To find the ratio of A(ΔPQB) and A(ΔPBC), complete the following activity.


Given: PQ ⊥ BC, AD ⊥ BC

Now, A(ΔPQB)  = `1/2 xx square xx square`

A(ΔPBC)  = `1/2 xx square xx square`

Therefore, 

`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×