Test Level 3: Mensuration - CAT MCQ

We label the points A, B and C as shown in the diagram. From symmetry, we note that ΔBAC is an isosceles right-angled triangle.
From A, we draw a line perpendicular to BC meeting the line at point D.

This construction allows us to conclude that ΔABD is also a right-angled isosceles triangle and specifically that BD = DA.
Since BD = DA; BD = DC = 20; we find DA = 20 cm.
This makes the depth of the water equal to (50 - 20) cm or 30 cm.

As AF and BF are both radii of equal arcs, AF = BF = AD = 1 unit.
∴ AB = 2 units
Area of the rectangle = 2 sq. units

Note that BE is the diagonal of square FBCE.
Therefore, BE = √2 units
∴ EQ = BE - BQ = √2 - 1 (∵ BQ = BC = 1)
∴ Area of semicircle with centre E

Also, total area of the two segments with centres A and B 

∴ Area of the shaded portion

Volume of the cylindrical container = 10 × volume of each cone


Diameter of each ice-cream cone = 6 cm

 Volume of the cone

Volume of all chocolate bullets = Volume of water that spilt out 

Volume of each chocolate bullet 

Number of chocolate bullets

Length of cylinder = 21 cm
Internal radius = 4 cm
External radius = 4.4 cm

= 66 (19.36 – 16) = 221.76 cm3
Hence, Mass of cylinder = Volume of metal × 8 = 221.76 × 8 = 1774 gm

Let the radius of the bigger circle be R.
OC = R √2 cm
Let r be the radius of the smaller circle.
Then, R√2 = R + r + r√2 (From the figure)


Area of the shaded region 

Refer to the figure. △AB'E and △CDE form the area which has not doubled up.

 

Both triangles are similar since ∠AB'E = ∠CDE = 90°.
∠AEB' = ∠DEC  (vertically opposite angles)
Also, since CD and AB', the corresponding sides of the triangles are equal (= 12 cm), the triangles are congruent.
Let  and DE = B'E = x and CE = AE = 16 - x
Then, x² + 12² = (16 - x)²

If a slice is cut parallel to the surface of the cube at the stage where the upper corners of the cube just touch the sphere.
The 3-D picture of the cube will look as shown below.

The length of a side of the cube is 2a and the radius of the sphere is R.

By Pythagoras theorem, R2 = (√2a)² + a² = 3a²
So, surface area of the sphere = 4 × πR² = 4 × π × 3a² = 12πa²
Now, the radius of the circular face of the cylinder = a.
Height of the cylinder = 2a
So, outer surface area of the cylinder = 

Thus, the desired ratio 

Let 'R' be the radius of the semicircular piece of paper.
Let the radius of the base of the cone be r.
Let the height of the cone be h.
Circumference of base of cone = 2πr = πR
So, r = R/2
Now, the slant height of the cone so formed = R
Applying Pythagoras theorem, we get

Volume of the cone,

Putting the values of r and h, we get

Putting the value of R, we get 

Height of the frustum (h) = 8 cm

Radii of the base of the frustum are:
r₁ = 6 cm and r₂ = 2 cm
Height of the original cone = h₁ = 8 + h₂ … (i)
∵ ΔOPB and ΔOQD are similar
[∵QOD = POB (same angle) and OQD = OPB = 90°]
So, the ratio of their corresponding sides is equal.
Thus,  h₁/h₂ = 6/2 = 3/1  or h₁ = 3h₂
Putting the value h₁ = 3h₂ into equation (i), we get
 ⇒ 3h₂ = 8 + h₂ or 2h₂ = 8 ⇒ h₂ = 4 cm
Again putting h₂ = 4 into (i), we get
⇒ h₁ = 8 + 4 = 12 cm
So, height of the original cone = 12 cm