Volume of concrete required to build the first step = 1/4 × 1/2 × 50 = 1/4 × 25 m³ Volume of concrete required to build the second step = 2/4 × 1/2 × 50 = 2/4 × 25 m³ Volume of concrete required to build the third step = 3/4 × 1/2 × 50 = 3/4 × 25 m³ Volume of steps are increasing in AP, whose firstRead more
Volume of concrete required to build the first step = 1/4 × 1/2 × 50 = 1/4 × 25 m³
Volume of concrete required to build the second step = 2/4 × 1/2 × 50 = 2/4 × 25 m³
Volume of concrete required to build the third step = 3/4 × 1/2 × 50 = 3/4 × 25 m³
Volume of steps are increasing in AP, whose first term a = 1/4 × 25 and the last term
a₁₅ = 15 /4 × 25. The common difference d = 2/4 × 25 – 1/4 × 25 = 1/4 × 25.
Therefore, the total volume of concrete required to build the terrace = S₁₅
= 15/2[(1/4 × 25) + (15 – 1)(1/4 × 25)]
= 15/2[25/2 + 175] = 15/2 ×200/2 = 750m³
Hence, the total 750m³ valume of concrete required to build the terrace.
The distance between top and bottom rungs is 2(1/2)m and the distance between two successive rungs is 25 cm, therefore The number of rungs = 250/25 + 1 = 11 The lenght of rungs is increasing from 25 cm to 45 cm in the from of AP, whose first term a = 25 and the last term a₁₁ = 45 Let the common diffRead more
The distance between top and bottom rungs is 2(1/2)m and the distance between two successive rungs is 25 cm, therefore
The number of rungs = 250/25 + 1 = 11
The lenght of rungs is increasing from 25 cm to 45 cm in the from of AP, whose first term a = 25 and the last term a₁₁ = 45
Let the common difference of this AP be d.
Therefore, a₁₁ = 45
⇒ a + (11 – 1)d = 45
⇒ 25 + 10d = 45
⇒ d = 20/10 = 2
The lenght of wood required
S₁₁ = 11/2[2a + (11 -1)d]
= 11/2[2(25) + 10(2)]
= 11 × 35 = 385 cm
Hence, 385 cm lenght of the wood is required for the rungs.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. Calculate the total volume of concrete required to build the terrace.
Volume of concrete required to build the first step = 1/4 × 1/2 × 50 = 1/4 × 25 m³ Volume of concrete required to build the second step = 2/4 × 1/2 × 50 = 2/4 × 25 m³ Volume of concrete required to build the third step = 3/4 × 1/2 × 50 = 3/4 × 25 m³ Volume of steps are increasing in AP, whose firstRead more
Volume of concrete required to build the first step = 1/4 × 1/2 × 50 = 1/4 × 25 m³
See lessVolume of concrete required to build the second step = 2/4 × 1/2 × 50 = 2/4 × 25 m³
Volume of concrete required to build the third step = 3/4 × 1/2 × 50 = 3/4 × 25 m³
Volume of steps are increasing in AP, whose first term a = 1/4 × 25 and the last term
a₁₅ = 15 /4 × 25. The common difference d = 2/4 × 25 – 1/4 × 25 = 1/4 × 25.
Therefore, the total volume of concrete required to build the terrace = S₁₅
= 15/2[(1/4 × 25) + (15 – 1)(1/4 × 25)]
= 15/2[25/2 + 175] = 15/2 ×200/2 = 750m³
Hence, the total 750m³ valume of concrete required to build the terrace.
Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO=CO/DO.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO=CO/DO. Show that ABCD is a trapezium.
S and T are points on sides PR and QR of Triangles PQR such that angles P = angles RTS. Show that Triangles RPQ ~ Triangles RTS.
A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2(1/2)m apart, what is the length of the wood required for the rungs?
The distance between top and bottom rungs is 2(1/2)m and the distance between two successive rungs is 25 cm, therefore The number of rungs = 250/25 + 1 = 11 The lenght of rungs is increasing from 25 cm to 45 cm in the from of AP, whose first term a = 25 and the last term a₁₁ = 45 Let the common diffRead more
The distance between top and bottom rungs is 2(1/2)m and the distance between two successive rungs is 25 cm, therefore
See lessThe number of rungs = 250/25 + 1 = 11
The lenght of rungs is increasing from 25 cm to 45 cm in the from of AP, whose first term a = 25 and the last term a₁₁ = 45
Let the common difference of this AP be d.
Therefore, a₁₁ = 45
⇒ a + (11 – 1)d = 45
⇒ 25 + 10d = 45
⇒ d = 20/10 = 2
The lenght of wood required
S₁₁ = 11/2[2a + (11 -1)d]
= 11/2[2(25) + 10(2)]
= 11 × 35 = 385 cm
Hence, 385 cm lenght of the wood is required for the rungs.