Numerical Ability Test - 12

Concept :

In a triangle sum of all interior angles is 180°.

Option: 1 

2 right angles and an acute angle

⇒ 90° + 90° + An acute angle (90° > Acute angle > 0°)

∴ The Sum of all angles is more than 180°. Triangle can't be formed.

Option: 2

2 obtuse angles and 1 acute angle

⇒ An obtuse angle is greater than 90° 

∴ The Sum of all angles is more than 180°. Triangle can't be formed.

Option: 3

1 acute angle, 1 obtuse angle, and 1 right angle is given

⇒ An acute angle (90° > Acute angle > 0°) + An obtuse angle (> 90°) + 90° 

∴ The Sum of all angles is more than 180°. Triangle can't be formed.

Option: 4

If all sides of the triangle are equal then the measurement of each interior angle will be 60°. The Sum of all interior angles is 180°. Triangle can be formed.

The correct option is 4 i.e. 3 equal sides given.

Given

Ratio of the angles of a triangle = 5 : 15 : 16

Concept

Sum of the angles of a triangle is 180°

Calculation

Let the anges are 5x, 15x and 16x

5x+ 15x + 16x = 180

36x = 180

x = 180/36

x = 5

Angles are 25°, 75° and 80°

∴ Largest angles is 80°

Given:

In ΔABC –

AB = BC

∠C = ∠A = x°

∠B = (2x + 12)°

Concept Used:

The sum of the three angles of any triangle is equal to 180°.

Calculation:

The sum of the three angles of any triangle is equal to 180°.

∠A + ∠B + ∠C = 180°

⇒ x° + x° + 2x + 12)° = 180°

⇒ 4x° + 12° = 180°

⇒ 4x° = 168°

⇒ x° = 42°

∴The value of ∠A is 42°

Concept:

Area theorem: It states that '' The ratios of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

i.e., \(\rm \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=(\frac{AB}{PQ})^2 =(\frac{BC}{QR})^2=(\frac{AC}{PR})^2\)

From the figure, Δ ABC ∼ Δ QPR implies AB ∼ QP, BC ∼ QR, and CA ∼ RP where ∼ is read as '' corresponding to ''

Calculation:

Given Δ ABC ∼ Δ QPR and \(\rm \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=\frac{1}{16}\)

Also AB = 12 cm

Consider, \(\rm \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=(\frac{AB}{PQ})^2 \)

⇒ \(\frac{1}{16}=(\frac{12}{PQ})^2 \)

⇒ \(\frac{1}{4}=\frac{12}{PQ}\)

⇒ QP = 48 cm

Hence, the correct answer is option 1).

Given:

\(\frac{KP}{PM} = \frac{4}{13}\) and KN = 20.4 cm

Concept used:

In a triangle ΔKMN, when PQ is parallel to MN then \(\frac{KP}{PM} = \frac{KQ}{QN}\) 

Calculation:

Now, \(\frac{KP}{PM} = \frac{4}{13}\) and \(\frac{KP}{PM} = \frac{KQ}{QN}\) 

⇒ \(\frac{KQ}{QN} = \frac{4}{13}\)

Let KQ = 4x and QN = 13x 

Given KN = 20.4 cm and KN = KQ + QN

⇒ KQ + QN = 4x + 13x = 20.4 cm

⇒ 17x = 20.4

 ⇒ x = 1.2

∴ KQ = 4x = 4 × 1.2 = 4.8 cm

Given:

∠ABC = 70°  and ∠BCA = 60° 

Concept:

Incentre of a triangle = Incentre of the triangle is the point of intersection of all the three interior angle bisector of the triangle

Sum of all interior angle of triangle = 180° 

Calculation:

⇒ In ΔABC, ∠CAB is

⇒ ∠ABC = ∠B = 70° and ∠BCA = ∠C = 60° 

⇒ ∠CAB = 180° - 70° - 60° = 180° - 130° = 50° 

⇒ Line IB and IC is the interior bisector of ∠B and ∠C respectively

⇒ The ∠IBC = ∠B/2 = 70°/2 = 35° 

⇒ The ∠ICB = ∠C/2 = 60°/2 = 30° 

⇒ In ΔIBC ∠BIC,

⇒ ∠BIC = 180° - ∠IBC - ∠ICB = 180° - 35° - 30° = 180° - 65° = 115° 

∴ The required result will be 115°. 

Given:

ΔABC ~ ΔPQR

The Perimeter of ΔABC = 24 cm

The perimeter of ΔPQR = 40 cm

Length of side AB = 5 cm

Concept used:

The ratio of the perimeter of any two similar triangles is equal to the ratio of their corresponding sides.

Perimeter of ΔABC/Perimeter of ΔPQR = AB/PQ = BC/QR = AC/PR

Calculation:

Let the length of side PQ be x.

ΔABC ~ ΔPQR

Perimeter of ΔABC/Perimeter of ΔPQR = AB/PQ

⇒ 24/40 = 5/x

⇒ 3/5 = 5/x

⇒ x = 25/3

∴ The length of the side PQ is 25/3.

Given:-

Angle between bisectors of ∠B and ∠C = 132°   

Concept:

If bisector of angles ∠B and ∠C of a triangle ABC meets at I then ∠BIC = 90° + 1/2 ∠A

Calculation:

∠BIC = 132° 

∠BIC = 90° + 1/2 ∠A

∠A/2 = (132° – 90°)

⇒∠A/2 = 42° 

⇒ ∠A = 42 × 2 

⇒ ∠A = 84° 

∴ The value of ∠A is 84° 

Given:

In Δ ABC, 

∠B = 90° and ∠A = 45°

Concept used:

1. In a triangle, if two angles are equal then their opposite sides are also equal.

2. Sum of the internal angle of a triangle is 180°.

Formula used:

In right angled triangle

(Hypotenuse)² = (Base)² + (Perpendicular)²

Calculation:

In Δ ABC, 

⇒ ∠A + ∠B + ∠C = 180° 

⇒ 45° + 90° + ∠C = 180° 

⇒ ∠C = 180° - 135°

⇒ ∠C = 45° 

and 

Given triangle is right angled triangle because ∠B = 90° then,

⇒ (8)2 = AB2 + BC2

∠A and ∠C are equal so opposite sides of ∠A and ∠B also equal

⇒ AB = BC

⇒ (8)² = AB² + AB²

⇒ 2AB² = 64

⇒ AB² = 32

⇒ AB = 4√2

∴ The length of any other two sides (in units) is 4√2.

Given:

ΔPQR ∼ ΔDEF

The sides of ΔPQR and ΔDEF are in the ratio 5 ∶ 6.

ar(PQR) = 75 cm²

Concepts used:

The ratio of area of similar triangles is equal to the square of the ratio of sides of corresponding triangles.

Calculation:

ΔPQR ∼ ΔDEF

ar(PQR)/ar(DEF) = (Side of ΔPQR/Side of ΔDEF)²

⇒ 75/ar(DEF) = (5/6)²

⇒ ar(DEF) = 108 cm²

∴ Area of ΔDEF is equal to 108 cm².