Chapter 11: Mensuration

Solution: 10a²

Explanation:

Let “a” be the side of two cubes.

When the two cubes are joined face to face, the figure obtained should be a cuboid having the same breadth and height as the combined cube has a length twice of the length of a cube.

It means that l = 2a, b = a and h = a

Hence, the total surface area of cuboid = 2(lb + bh + hl)

= 2(2a × a + a × a + a × 2a)

Simplify the above expression, we get

= 2[2a² + a² + 2a²]

= 10a²

Solution: 4 times

Explanation:

Let p and q be the two diagonals of the rhombus

We know that the area of a rhombus = pq/2

If the diagonals are doubled, we will get

A= (4p)(4q)/2

Take 4 outside, we will get

A = 4(pq/2)

Solution: volume is h³ and surface area is 6h²

Explanation:

Each side of a cube = h

Thus, the volume of cube = h³

Surface area of a cube = 6 (h²)

Solution: ¼ times

Explanation:

The volume of cylinder = πr²h (when the radius is r and height is “h”)

When the radius is halved, then it becomes

V = π (r/2)²h

V = ¼ (πr²h)

Solution: 50%

Explanation:

The CSA of the cylinder with radius “r” and height “h” is 2πrh

When the height is halved, then the new CSA is 2πr (h/2) = πrh

Hence, the percentage reduction in CSA = [(2πrh – πrh) (100)]/ 2πrh = 50%

Solution: πa³ /4

Explanation:

When the cylinder exactly fits in the cube of side “a”, the height equals to the edges of the cube, and the radius equals to half the edges of a cube.

It means that,

h = a,and r = a/2

Then the volume of a cylinder be = πr²h

= π(a/2)²(a)

= πa³/4

Solution: πb²

Explanation:

When the cylinder exactly fits in the cube of side “b”, the height equals to the edges of the cube and the radius equal to half the edges of a cube.

It means that,

h = b, and r = b/2

Then the CSA of a cylinder be = 2πrh

= 2π (b/2)(b)

= πb²

Solution: ½ (h₁ +h₂) d

Explanation:

Assume that ABCD be a quadrilateral, h₁ and h₂ are the heights on the diagonal BD = d, then the area of a quadrilateral be

= (1/2)(h₁ +h₂) BD

Since the diagonal is doubled and the heights are halved, we will get

= (1/2) [ (h₁/2) +(h₂/2) ] 2d

= ½ (h₁ +h₂) d

Solution: Two times

Explanation:

We know that the perimeter of a rectangle is 2(l+b)

When the length and breadth of the perimeter are doubled, we will get

P = 2(2l +2b)

Now take 2 outside,

P = 2 [2(l+b)]

Solution: 3, equal areas

Explanation:

By using the SSS congruency rule of the triangle, we can show that a trapezium can be divided into three equilateral triangles with equal areas.