IIFT questions on Geometry - Triangles

AE is altitude from A on BC. ∠AEC = 90°
It is given,

∠ABC = 60°, ∠ACB = 45°

In triangle ABE,
AE = $\sqrt{3}$x
As ∠EAC = ∠ECA , EC = $\sqrt{3}$x
In triangle AED, ∠EAD = 60°
Therefore, CD = 3x - $\sqrt{3}$x
It is given,

BC ∙ CD = $\frac{3^4.2^3.11}{19}$

x = ($\sqrt{3}$ + 1)$\sqrt{3}$($\sqrt{3}$ - 1)x = $\frac{3^4.2^3.11}{19}$

x = $\frac{3^4.2^3.11}{19}$

AE = 9*2 = 18
AE² = 324
Answer is option B.

If ACB is a right angled isosceles triangle and AB is hypotenuse with length 7*($\sqrt{2}$) length, then sides AC and BC = 7 cm
Now, the circle inscribed is an incircle.
Radius of in-circle = {7 + 7 - [7*($\sqrt{2}$)]}/2 = 7 - $\frac{7\sqrt{2}}{2}$
Now, if center of circle is marked as O, then quadrilateral OZCY is a square.
Hence, CY = CZ = 7 - $\frac{7\sqrt{2}}{2}$
Hence, AZ = 7 - 7 - $\frac{7\sqrt{2}}{2}$ = $\frac{7\sqrt{2}}{2}$
Also, BX = BY and BY = BC - CY = $\frac{7\sqrt{2}}{2}$
Hence, BX = $\frac{7\sqrt{2}}{2}$

Hence, the given ratio becomes 1 : 1 : [($\sqrt{2}$) - 1]

Let the three angles be a − d, a and a + d.
∴ (a − d) + a + (a + d) = 180
∴ a = 60.

Since (a + d) is the largest angle it will be opposite the largest side i.e., 13 and a will be opposite second largest side i.e., 12. 

The triangle along with the angles are shown below.

Applying the cosine rule, we get;
 
Cos 60 = (x² + 13² − 12²)/(2 × x × 13)
 
∴ x = 10.65, 2.34.
 
Inradius = Area/Semiperimeter.
 
When x = 10.65, the semiperimeter = (10.65 + 13 + 12)/2 = 17.825.
 
∴ Inradius = [(1/2) × 10.65 × 13 × Sin60]/17.825 = 3.36.
 
When x = 2.34, the semiperimeter = (2.34 + 13 + 12)/2 = 13.67.
 
∴ Inradius = [(1/2) × 2.34 × 13 × Sin60]/13.67 = 0.963 ≈ 1.
 
Hence option (d).

Applying cosine rule in ∆NPL, we get;
 
Cos60 = 1/2 = [NP² + NL² − PL²]/(2 × NP × NL)
 
∴ 1/2 = [(2x)² + (3x)² − PL²]/(2 × 2x × 3x)
 
∴ PL = x√7.
 
Using angle bisector theorem, MN/NL = MP/PL
 
So, MN/3x = MP/x√7
 
∴ MN = MP × (3/√7) ...(I)
 
Applying cosine rule in ∆MNP, we get;
 
Cos60 = 1/2 = [MN² + NP² − MP²]/(2 × MN × NP)
 
Let MN = a, so using (I) MP = (√7/3) a or MP² = (7/9)a².
 
∴ 1/2 = [a² + 4x²− (7/9)a² ]/(2 × a × 2x)
 
∴ a² − 9ax + 18x² = 0  
 
∴ a = 6x or 3x. 
 
If a = 3x, then MN = NL = 3x, so ∆MNL becomes isosceles and hence ∠NPL = 90°, which means that NL² = NP² + PL².
 
NL² = 9x² and  NP² + PL² = 4x² + (x²/7) = (29/7)x² ≠  9x². Hence a ≠ 3x.
 
∴ a = 6x.
 
NP : NL: MN = 2x : 3x : 6x = 2 : 3 : 6.
 
Hence option (b).
         
Note: We can use the direct formula for the length of the angle bisector L. Consider the triangle drawn below

L = (2abCosQ)/(a + b)

So, 2x = (2 × MN × 3x × Cos60)/(MN + 3x)

Solving, we get; MN = 6x.

Since S is the midpoint of QR, A(∆PSR)

= A(∆PQR)/2

= 5.8/2 = 2.9 sq.cm

Now A(∆PSR) = A(∆PSX) + A(∆SXR)

Also, A(∆RTX) = A(∆TSX) + A(∆SXR)

Now, triangles PSX and TSX lie within the same parallel lines – SX and PT – and hence, have the same area.

∴ A(∆PSX) = A(∆TSX)

∴ A(∆PSR) = A(∆RTX) = 2.9 sq.cm

Hence, option (a).

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AC = ED = 7.6 and AB = 6.4
From the given diagram ; BC² = 7.6² – 6.4² = 16.8
∴ BC ≈ 4.1 m
When the ladder slips 1.2 m, its top edge would be at D at a height of 4.1 – 1.2 ≈ 2.9 m
∴ BE² = 7.6² – 2.9² ≈ 49
∴ BE ≈ 7
∴ Ladder shifts approximately 7 – 6.4 = 0.6 m.
Hence, option (b).

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When the boat was at C, the angle of depression = 60° ⇒ m∠ BAC = 30°

When the boat was at D, the angle of depression = 45° ⇒ m∠DAB = 45° ⇒ BD

= AB = 210 m

In ∆ABC, by theorem of 30° – 60° – 90°,

BC = 70√3 ≈ 121.1

DC ≈ 210 – 121.2 = 88.8 m

Speed of the boat = 3 km/ hr = 50 m/min

Time taken = 88.8/ 50 = 1.776 minutes

Hence, option (d).

Let AB = a³ and AC = 3ⁿ

Also

AC = 3 × AB

∴ 3ⁿ = 3 × a³

∴ 3^{(n – 1)} = a³

Now, let the perimeter be equal to p

p = BC + AC + AB

= 397 + 3ⁿ + 3^{(n – 1)}

(p – 397) = 3^{(n – 1)} × (3 + 1)

= 3^{(n – 1)} × 4

Thus the LHS of the above equation should be a multiple of 3 and 4. Substitute the value of perimeter given in the options and verify this. Among the options, only (3313 – 397) is divisible by 3 and 4..

Hence, option (d).

All the three options given satisfy the perimeter criteria. The hypotenuse is the greatest side.

∴ Option 1 is eliminated.

Now, 97² = 9409

Option 2:

97² – 72² = 4225 = 65²

Option 3:

97² – 80² = 3009 ≠ 57²

Hence, option (b).

If the coordinates of the vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃) then its area is given by

A = $\frac{1}{2}$$\left|\begin{array}{ccc}{x}_1& x_2& x_3\\ y_1& y_2& y_3\\ 1& 1& 1\end{array}\right|$

∴ If the third vertex of the given triangle is (x, y)

6 = $\frac{1}{2}$$\left|\begin{array}{ccc}1& 4& x\\ 1& -1& y\\ 1& 1& 1\end{array}\right|$

∴ 12 = 1(−1 − y) − 4(1 − y) + x(1 + 1)

∴ 12 = −1 – y − 4 + 4y + 2x

∴ 2x + 3y = 17                        …(i)

But (x, y) also lies on y = x + 5 …(ii)

Solving (i) and (ii)

x = $\frac{2}{5}$, y = $\frac{27}{5}$

Hence, option (a).

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2x² is a power of 2.

Hence, x² is a power of 2.

Hence, x is a power of 2.

Also by triangle inequality 3x² > 295

∴ x² > 98.33

∴ x ≥ 10

Hence, x is a power of 2 which is greater than 10.

We try with x = 16. Then we have,

AB = 256, AC = 512

∴ Perimeter = 256 + 512 +295= 1063, which is there in the options.

Hence, option (c).

Refer to the following figure:

​​​​​​​

The vertical distance between the warship and the submarine = 25 km

∴ When the angle of depression of the submarine is 60°, the distance between the warship and the submarine is

$\frac{25}{\sin 60°}$ = $\frac{50}{\sqrt{3}}$ km

Hence, option (a).

Let the speed of the submarine be x kmph.

Then the speed of the warship is 2x kmph.

Distance travelled by the warship in 30 minutes = x km

Distance travelled by the submarine in 30 minutes = 0.5x km

​​​​​​​

From the figure,

x = 25$\sqrt{3}$ + 0.5x - $\frac{25}{\sqrt{3}}$

∴ 0.5$\sqrt{3}$x = 50

∴ x =$\frac{100}{\sqrt{3}}$ kmph

∴ The speed of the warship = $\frac{200}{\sqrt{3}}$ kmph.

Hence, option (d).