Test: Mensuration - CAT MCQ

Let the edges of the brick be a, b, and c such that
a < b < c
a² + b² = 3² = 9 ...(i)
a² + c² = ( 2√3 )² = 12 ...(ii)
b2 + c2 = ( √15 )² = 15 ...(iii)
Adding all three equations. We get
2(a² + b² + c²) = 9 + 12 + 15 = 36
a² + b² + c² = 18 ...(iv)
From (i) and (iv), c² = 18 – 9 ⇒ c = 3
From (iii) and (iv) a² = 18 – 15 ⇒ a = 3
Therefore, required ratio =

An equilateral triangle formed by joining the midpoints of the sides of a given equilateral triangle will have its side equal to half the side of the given equilateral triangle.
Now, side of T₁ = 24 cm
Side of T₂ = 12 cm
Side of T₃ = 6 cm
and so on.
Sum of the areas of all the triangles =

Let the length and the breadth of the rectangle be l and b respectively.
Diameter of the circle = Diagonal of the rectangle

Hence, possible values of l and b are 24 and 10 respectively.

The radius of the cone is 4 feet.
The tip of the cone is a cone of height 3 feet.
By similarity, its radius is 1 foot.
The volume of the remaining part of the cone
= Volume of the cone – Volume of the tip of the cone

 

Volume ratio of 5 smaller and original cube = 1:1 : 8 : 27 : 27 : 64
Side ratio of 5 smaller and original cube = 1 : 1 : 2 : 3 : 3 : 4
Surface area ratio of 5 smaller and original cube = 1 : 1 : 4 : 9 : 9 : 16
Sum of smaller cube (surface area) = 24 parts.
Big cube = 16 parts.
The sum is 50% greater.

Let the number of sides be 2n. Let the length of the side of S and the length of the perpendicular from the centre to each side be P. Since the number of sides is even, the opposite sides will be parallel and the distance between any two opposite sides is equal to 2P.
Also, area of the polygon (A) =  ...(i)
According to question,
S(2P) = A/4 or SP = A/8
∴ (1) ⇒ A = n (A/8)
⇒ n = 8 or 2n = 16
∴ number of sides = 16.

Here, HCF of 24 and 56 = 8 then, the side of the identical square plots = factors of 8.
The factors of 8 are 1, 2, 4 and 8
If side of the square plot is 1 m. the length of fencing material required is (25 × 56 + 57 × 24) = 2768 m
But 2768 m > 2700 m. (So, It is not possible)
If side of the square plot is 2 m. the length of fencing material required is
(29 × 24+13 × 56) =1464m < 2700 m
∴ minimum fencing material to be left the side  of identical square plot = 2m.

PQRS is a rhombus, so the centre of the inscribed circle will be the center of the rhombus PQRS

PR = SQ = 8 units
⇒ PQRS is a square

Area of trapezium =   8 = 64 sq. unit.
Required ratio = 

Let length of smaller side of rectangle = a
∴ Radius = a/2 of circle
According to question,

So, length of smaller side of rectangle = 

Here, ABC represents cone; DEBC represents frustum which was made by cone and BRC represents conical cavity.
So, Volume of material = cone ABC – (cone ADE + conical cavity of BRC).
Now, ΔAPE and ΔAQC are similary,

⇒ So, AP = 8 cm.

Required volume of solid =

= 132π cm³.

Let the length of the other two sides of the triangle be 'x' and 'y' respectively. then,
x² + y² = (240)² (hypotenuse of triangle = 240)
Also, x + y + 240 must be perfect square (because perimeter of triangle is perfect square).
When the two sides other than hypotenuse are equal length than perimeter of right angled triangle is maximized.
Therefore,
The maximum perimeter of right angled triangle

Also,
The perimeter of triangle should be greater than twice the length of hypotenuse of triangle.
Then, the perimeter of triangle should be greater than 480 and less than 579. according to question.
Perimeter of triangle should be 484, 529 and 576
(because perimeter is perfect square)
Again, according to question, Perimeter of triangle is greater than 550, So perimeter = 576.
Here, perimeter of triangle = 576.
According to pythogorean triples.
The value of x and y will be 192 and 144 units.

Let the length and breadths of rectangle be a units and b units.
According to question,
ab = 2a + 2b
a = 2b / (b – 2)
Here, Two cases are possible
(i) a = 4, b = 4
(ii) a = 3 and b = 6
Thus, two different rectangles are possible.

Let the semi vertical angle of the cone be θ.

The vertical angle of the cone = 60°
Therefore, the triangle ABC is an equilateral triangle.
Therefore, the center (O) of the circle will coincide with the centroid of the triangle ABC.
In triangle ACM, OC is angle bisector of ∠ACM.
∴ ∠OCM = 30°
In ΔOCM,

Hence, the ratio of the volume of the sphere to that of

∴ Required ratio = 4 : 9

Required number of traingles = 
Here, n = number of items and
k = number you are picking at time
∴ n = 5, k = 3

Let, the radius of circle be 'x' cm.

∴ CM = CN = CR =  x
Given that GN = 4 cm
∴ GP = 4 cm
Also, DP = DM = QN = 1 cm
∴ GD = 5 cm
GQ = GN – QN = 4 – 1 = 3 cm
In ΔDGQ : DQ = 
⇒ 2x = 4 cm (∵ DQ = MN)
∴  x = 2 cm
∴ radius or circle = 2 cm

Let the lengths and breaths of each,
Small rectangles be 'x' meter and 'y' meter respectively.

∵ Perimeter = 88 m
∴ 5y + 4x = 88 and 3y = 2x
​∴ 5y + 3y × 2 = 88
5y + 6y = 88  ⇒ 11y = 88

​∴ 5 × 8 + 4x = 88
4x = 88 – 40 = 48

∴ Perimeter of each small rectangle = 2 (x + y)
= 2 (12 + 8) = 40 meter.

Let  EF = s unit
In Δ DEF : DF = 2EF cos30° = √3s
Semi perimeter of D DEF (in units) = S

Area of Δ DEF (in unit² = Δ

Inradius of Δ DEF (in units) = r₂

In radius of triangle DEF (in units) = r₂

Centre of the circle will be mid-point of the hypotenuse of the triangle.
So, hypotenuse = 2 × 6.5 = 13 cm.
According to Pythagorean triples.
other sides are 5 cm and 12 cm because all sides are natural number.
∴ So perimeter = 5 + 12 + 13 = 30 cm

Since cubic shape will be the greatest volume (as volume = side³)

Hence, 3b = 2a

In right ΔAPB,
x² + (x + 7)² = 13²
⇒ x = 5 ∴ x + 7 = 12.