IIFT questions on Geometry - Basics

Let the length, breadth and height of the cubhoid be l, b and h respectively.
Therefore of adjacent faces will be lb, bh and hl.
If lb = p, then bh = q and hl = r.

Volume (V) = lbh

V² = (lbh)²

or, V² = pqr

Given,
AD = DC, this implies angle ACD = angle DAC = x
DC = BC, this implies angle CDB = angle CBD = 2x
∠DCB = 180 - x - 81 = 99 - x
In triangle BCD, sum of the angles is 180 degrees
99 - x + 2x + 2x = 180
3x = 81
x = 27
∠DBC = 2x = 54 degrees
The answer is option C

The situation given in the problem can be summarised as:

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By 30 - 60 - 90 theorem, we can say that P'M = 4 meters and MQ' = 3 meters
Also, triangle P'MQ' would be a right angled triangle.
Hence, P'Q' would be the hypotenuse of length = root(16 + 9) = 5 meters

Since ∠ABC = 30° and ∆DBE is an isosceles triangle, ∠BDE = 120°

Applying the cosine rule in △BDE.

BE² = BD² + DE² - 2 × BD × DE cos(120)

1 = 2BD² + BD²                [since BD = DE]

BD = $\frac{1}{\sqrt{3}}$

Thus, BD = DE = EF = FG = GH = HI = IJ = JC = $\frac{1}{\sqrt{3}}$

Area of △BDE = $\frac{1}{2}$ × BD × DE × sin(120) = $\frac{1}{2}$ × $\frac{1}{\sqrt{3}}$ × $\frac{1}{\sqrt{3}}$ × $\frac{\sqrt{3}}{2}$ = 4$\frac{1}{\sqrt{3}}$

The total area of all smaller isosceles triangles = 4 × 4$\frac{1}{\sqrt{3}}$ = $\frac{1}{\sqrt{3}}$ m²

∠DEF = ∠FGH = ∠HIJ = 120°
Thus, the total area of circular arcs = π × ($\frac{1}{\sqrt{3}}$)² = $\frac{\pi }{3}$m²

Now, side BC = BE + EG + GI + IC = 4m

Using the cosine rule in △ABC, we get AB = AC = $\frac{4}{\sqrt{3}}$.

Thus, Area of △ABC = $\frac{1}{2}$ × $\frac{4}{\sqrt{3}}$ × $\frac{4}{\sqrt{3}}$ × sin(120) = $\frac{4}{\sqrt{3}}$m²

Thus, the area of the shaded region = Area of △ABC - Area of smaller isosceles triangles - Area of the circular sections

= $\frac{4}{\sqrt{3}}$ - $\frac{1}{\sqrt{3}}$ - $\frac{\pi }{3}$ = $\sqrt{3}$ - $\frac{\pi }{3}$

Hence, the answer is option C.

As the three portions of the rod correspond to 12, 18 and 32 equal parts, their length will be equal to LCM of 12, 18 and 32.
LCM of 12, 18 and 32 = 288
Since there are three equal parts of the rod, total length = 288 × 3 = 864
Hence, option (b).

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Let the side of isosceles triangle be x.

∴ The side of octagon x$\sqrt{2}$.

∴ x + x + x$\sqrt{2}$ = 2

∴ 2x + x$\sqrt{2}$ = 2

∴ x = $\frac{2}{2+\sqrt{2}}$

∴ Side of octagon = x$\sqrt{2}$ = $\frac{2\sqrt{2}}{2+\sqrt{2}}$ = $\frac{2}{1+\sqrt{2}}$

∴ Perimeter of octagon = 8 × $\frac{2}{1+\sqrt{2}}$ = $\frac{16}{1+\sqrt{2}}$ units

Area of octagon = Area of square – 4 × Area of isosceles triangle

= 2² - 4 × $\frac{1}{2}$ x²

= 4 - 4 × $\frac{1}{2}$ × ${\left(\frac{\sqrt{2}}{1+\sqrt{2}}\right)}^2$

= 4 - $\frac{4}{3+2\sqrt{2}}$ = $\frac{8(1+\sqrt{2})}{3+2\sqrt{2}}$ sq. units

Hence, option (d).

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Area of square 1 = a²

Area of circle 1 = π${\left(\frac{a}{\sqrt{2}}\right)}^2$

∴ D₁ = $\frac{a^2}{2}$(π - 2)

Area of square 2 = (a$\sqrt{2}$)²

Area of circle 2 = πa²

∴ D₂ = a²(π - 2)

Similarly,

D₃ = 2a² (π - 2), and so on.

∴ D₁ + D₂ + D₃ + ... = a²(π - 2) [0.5 + 1 + 2 + 4 + ...]

= Infinite

Hence, option (c).