The length of the longest rod that can be placed in a room which is 12 m long, 9 m broad and 8 m high is
To find the length of the longest rod that can be placed in a room, we need to calculate the length of the space diagonal of the cuboid formed by the room's dimensions. This is a standard application of the 3D Pythagorean theorem.
Thus, the length of the longest rod that can be placed in the room is 17 m.
Correct Option: C) 17 m
A metal cube of edge 12 cm is melted and formed into three smaller cubes. If the edges of two smaller cubes are 6 cm and 8 cm, then find the edge of the third smaller cube.
When a metal cube is melted and recast into smaller cubes, the total volume of the metal remains constant. This means the volume of the original large cube is equal to the sum of the volumes of the three smaller cubes. We can use the formula for the volume of a cube, \(V = a^3\), where \(a\) is the length of the edge.
Correct Option: A) 10 cm is the correct edge length for the third smaller cube, as calculated by the conservation of volume principle.
A wire is bent down to form a square of side 22 cm. If the wire is rebent to form a circle, then its radius will be
The problem involves a wire that is first bent into a square and then rebent into a circle. The key concept here is that the length of the wire remains constant. Therefore, the perimeter of the square will be equal to the circumference of the circle. We will calculate the perimeter of the square and then use this value to find the radius of the circle.
Given that the side of the square is \(22 \text{ cm}\).
The perimeter of a square is given by the formula:
\[ \text{Perimeter of square} = 4 \times \text{side} \]\[ \text{Perimeter of square} = 4 \times 22 \text{ cm} \]\[ \text{Perimeter of square} = 88 \text{ cm} \]Since the same wire is rebent to form a circle, the length of the wire (perimeter of the square) will be equal to the circumference of the circle.
\[ \text{Circumference of circle} = \text{Perimeter of square} \]\[ \text{Circumference of circle} = 88 \text{ cm} \]The formula for the circumference of a circle is:
\[ \text{Circumference} = 2\pi r \]Where \(r\) is the radius of the circle and \(\pi\) is approximately \(\frac{22}{7}\).
Substitute the circumference value into the formula:
\[ 2\pi r = 88 \]\[ 2 \times \frac{22}{7} \times r = 88 \]\[ \frac{44}{7} \times r = 88 \]To find \(r\), multiply both sides by \(\frac{7}{44}\):
\[ r = 88 \times \frac{7}{44} \]\[ r = 2 \times 7 \]\[ r = 14 \text{ cm} \]Correct Option: C) 14 cm
A cistern 6 m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is
To find the total area of the wet surface of the cistern, we need to calculate the area of the base and the area of the four side walls that are in contact with the water. The cistern is a cuboid, and the water fills it up to a certain depth. We will use the given dimensions to find the area of the bottom and the four wet sides.
Convert depth to meters:
The depth of the water is given as 1 m 25 cm. We convert this to meters:
\[ 1 \text{ m } 25 \text{ cm} = 1 + \frac{25}{100} \text{ m} = 1 + 0.25 \text{ m} = 1.25 \text{ m} \]Identify dimensions:
Calculate the area of the base:
The base of the cistern is a rectangle with length \(L\) and width \(W\).
\[ \text{Area of base} = L \times W = 6 \text{ m} \times 4 \text{ m} = 24 \text{ m}^2 \]Calculate the area of the two longer side walls:
These walls have length \(L\) and height equal to the water depth \(H\).
\[ \text{Area of two longer sides} = 2 \times (L \times H) = 2 \times (6 \text{ m} \times 1.25 \text{ m}) \]\[ = 2 \times 7.5 \text{ m}^2 = 15 \text{ m}^2 \]Calculate the area of the two shorter side walls:
These walls have width \(W\) and height equal to the water depth \(H\).
\[ \text{Area of two shorter sides} = 2 \times (W \times H) = 2 \times (4 \text{ m} \times 1.25 \text{ m}) \]\[ = 2 \times 5 \text{ m}^2 = 10 \text{ m}^2 \]Calculate the total wet surface area:
The total wet surface area is the sum of the area of the base and the areas of the four wet side walls.
\[ \text{Total wet surface area} = \text{Area of base} + \text{Area of two longer sides} + \text{Area of two shorter sides} \]\[ = 24 \text{ m}^2 + 15 \text{ m}^2 + 10 \text{ m}^2 = 49 \text{ m}^2 \]Correct Option: A) 49 m2
The area of a right angled isosceles triangle whose hypotenuse is equal to 270 m is
To find the area of a right-angled isosceles triangle, we first need to determine the lengths of its equal sides. Since it's a right-angled isosceles triangle, the two perpendicular sides (legs) are equal in length. We can use the Pythagorean theorem to relate the legs to the hypotenuse, and then use the formula for the area of a right-angled triangle.
Define the sides:
Let the two equal sides (legs) of the right-angled isosceles triangle be \(x\) meters each. The hypotenuse is given as \(270\) meters.
Apply the Pythagorean theorem:
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Solve for \(x^2\):
We need \(x^2\) to calculate the area. Divide both sides by 2.
Calculate the area of the triangle:
The area of a right-angled triangle is given by the formula:
Substitute the value of \(x^2\):
\[ \text{Area} = \frac{1}{2} \times 36450 \]\[ \text{Area} = 18225 \text{ m}^2 \]B) 18225 m2 is the correct area calculated using the Pythagorean theorem and the area formula for a right-angled triangle.
A 5100 sq.cm trapezium has the perpendicular distance between the two parallel sides 60 m. If one of the parallel sides be 40m then find the length of the other parallel side.
The problem asks us to find the length of one of the parallel sides of a trapezium, given its area, the perpendicular distance between the parallel sides (height), and the length of the other parallel side. We will use the formula for the area of a trapezium to solve this problem.
The area of a trapezium is given by the formula:
\[ \text{Area} = \frac{1}{2} \times (a + b) \times h \]where \(a\) and \(b\) are the lengths of the parallel sides, and \(h\) is the perpendicular distance between the parallel sides (height).
B) 130 m is the length of the other parallel side. This matches our calculated value.
