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Given :

XL is the centroid of a triangle XYZ and G is the centroid of the of the median 

Concept used :

In a triangle ABC if a median AD is drawn from A to side C, then centroid G divides the AD  in the ratio 2 : 1

Calculations :

XL = XG + GL

XG : GL = 2 : 1 (G is the centroid)

⇒ XG = (2/3) of XL 

⇒ XG = (2/3) × 18

⇒ XG = 12 

∴ XG is 12 cm.

Concept:

Following are the properties of similarities of triangle:

1) AAA - When all angle of both the triangle are equal then they are similar.

2) SAS - When two sides and angle between them of two triangle are equal then they are similar.

3) SSS - When all the sides of  both the triangle are in same proportion then they are similar.

Calculation:

Now, We will go through the options.

By option (1),

In all the equilateral triangle all the sides in same proportion.

So they are similar by AAA property.

By option (2) and (3),

If both the triangle has equal angle like 30° – 60° – 90° or 45° – 45° – 90° they are similar by AAA property.

By option (4),

In all isosceles triangles two side are proportional but third side can't be always proportional neither angle between them are always equal.

∴ Two isosceles triangle can not always be similar.

Given

The angles of a triangle are in the ratio 4 : 5 : 6

Concept used

Sum of all angles of triangle is 180° 

Calculation

Let the angles of triangle be 4x, 5x, 6x

⇒ 4x + 5x + 6x = 180° 

⇒ 15x = 180

⇒ x = 12° 

∴ Smallest angle is 4 × 12 = 48° 

(a) In rectangle l² + b² = d² 

b² = d² – l² = (29)² –(20)² 

= 841 – 400 = 441 

b = 21 m 

Perimeter = 2(20 + 21) m = 82 m

Given:

Perimeter = 30 m

Concept used:

Perimeter of an isosceles triangle = 2x + y

Area of an isosceles right angle triangle = 1/2 × x²

x = equal sides of an isosceles triangle 

Calculation:

In an isosceles right angled triangle, the two sides on the right angle are equal

Let the equal sides be a

Hence, hypotenuse of the isosceles right angled triangle = \(√ {{{\rm{a}}^2} + {{\rm{a}}^2}} = √ 2 {\rm{a}}\)

\(\rm a + a\;+ √2a = 30\)

⇒ \(\rm 2a\;+ √2a = 30\)

⇒ \(\rm a(2\;+ √2) = 30\)

⇒ \(\rm a = \frac{30}{(2\;+ √2)}\)

⇒ \(\rm a = \frac{30(2\;-\; √2)}{(2\;+\;√2)(2\;-\; √2)}\)

⇒ \(\rm a = \frac{30(2\;-\; √2)}{2^2\;-\;(√2)^2}\)

⇒ \(\rm a = \frac{30(2\;-\; √2)}{4\;-\;2}\)

⇒ \(\rm a = \frac{30(2\;-\; √2)}{2}\)

⇒ \(\rm a = 15(2\;-\; √2)\)

1/2 × (15(2 – √2))²

⇒ 1/2 × 225(4 + 2 – 4√2)

⇒ 112.5(6 – 5.6568) [√2 = 1.4142]

⇒ 112.5(0.3432)

⇒ 38.626 ≈ 38.63 m

∴ Area of the triangle is 38.63 m (Approximate)

Given:

An isosceles triangle with one right angle is given

Concept Used:

According to Pythagoras theorem In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides

Calculation:

Let suppose PQR is and isosceles triangle right angled at Q and PR is longest

∴ PQ = QR

Now, according to the Pythagoras theorem

∴ PR² = PQ² + QR²

⇒ PR² = 2PQ²

Given :

The ratio of angles of triangle = 2 : 3 : 4

Concept used :

Sum of all angles of a triangle = 180°

Solution :

Let the first angle be 2x

The second angle = 3x

The third angle = 4x

2x + 3x + 4x = 180°

⇒ 9x = 180°

⇒ x = 20°

∵ The greatest angle = 4x

⇒ The greatest angle = 4 × 20°

∴ The greatest angle is 80°

Given :

Ratio of angles of triangle is 3 : 4 : 8  

Concept used :

Sum of all angles of a triangle is 180° 

Calculations :

Let the values of angles be 3x, 4x and 8x

3x + 4x + 8x = 180° 

15x = 180° 

⇒ x = 12° 

3x = 36° , 4x = 48° and 8x = 96° 

One of the angles is more than 90° 

∴ The triangle will be obtuse 

Given:

A point(-13, 4) is given and the equation of the line 20x + 21y + 2 = 0.

Formula Used:

Distance (d) = |(Ax₁ + By₁ + C)/(√(A² + B²)|

Where Ax + By + C is the given equation of the line, and (x₁,y₁) is the given point from where the distance is calculated.

Calculation:

By using the above formula we get

⇒ Distance (d) = |(20 × -13 + 21 × 4 + 2)/√(20² + 21²)|

⇒ d = |(-260 + 86)/ 29| = 174/29

∴ Distance is 6 unit.   

Given:

Two lines are,

y – √3x = 2 and \(\frac{1}{{\surd 3}}\)x – y = –4 

Formula Used:

y = mx + c,

where,

m is the slope of a line.

tanθ = |\(\frac{{m1 - m2}}{{1 + \left( {m1 \times m2} \right)}}\)|

where,

m₁ is the slope of the first line. 

m₂ is the slope of the second line.

Calculation:

For 1^st equation,

⇒ y – √3x = 2 

⇒ y = 2 + √3x

⇒ m₁ = √3

For 2^nd equation,

⇒ \(\frac{1}{{\surd 3}}\)x – y = –4 

⇒ –y = –4 – \(\frac{1}{{\surd 3}}\)x

⇒ y = 4 + \(\frac{1}{{\surd 3}}\)x

⇒ m₂ = 1/√3

tanθ = |\(\frac{{m1 - m2}}{{1 + \left( {m1 \times m2} \right)}}\)| 

⇒ tanθ = |\(\frac{{√ 3 \; - \;1/\surd 3}}{{1 + \left( {√ 3 \times 1/\surd 3} \right)}}\)|

⇒ tanθ = |(2/√3)/2 |

⇒ tanθ = |1/√3 |

⇒ tanθ = 1/√3

We know that, tan 30° = 1/√3

⇒ θ = 30°    

∴ Angle between two lines is 30°.

Given:

A line is perpendicular to line 4x - 3y + 8 = 0.

Formula used:

y = mx + c

Where m is the slope of the line.

The product of slope of two perpendicular lines is –1

m₁ × m₂ = –1

Calculation:

Let us find the slope of line 4x - 3y + 8 = 0

⇒ y = (4/3)x + (8/3)

The slope of the equation is (4/3)

The product of slope of two perpendicular lines is –1

Let slop of the perpendicular line is m

⇒ m × (4/3) = –1

⇒ m = –3/4

∴ The slope of the line perpendicular is –3/4.

Given:

Point (–1, 6)

Slope = 3

Formula Used:

Equation of a line in slope form

y = mx + C

m = slope , C = constant

Calculation:

Equation of line y = 3x + C 

For the value of C, point (–1, 6) will satisfy the equation

⇒ 6 = 3(–1) + C 

⇒ C = 6 + 3 

⇒ C = 9

Now, the equation of line is y = 3x + 9

∴ The equation of the line passing through the point (–1, 6) and slope 3 is y = 3x + 9

The number of capital letters of the English alphabets having no line of symmetry is 10 ( F, G, J, L, N, P, Q, R, S, Z)

Concept used :

Number of diagonals in polygon = [n(n - 3)]/2   (where n is the number of sides)

Calculations :

Here n will be 8 as octagon has 8 sides. 

Number of polygons in octagon = [8 × (8 - 3)]/2 

⇒ (8 × 5)/2 

⇒ 40/2 = 20 

∴ Option 1 will be the right choice. 

Given:

The area of the incircle of an equilateral triangle is 49π cm².

Formula used:

Inradius of equilateral triangle = a/(2√3)

Circumradius of equilateral triangle = a/(√3)

Where,

a = side of an equilateral triangle

Calculation:

Inradius of equilateral triangle = a/(2√3)

Area of incircle = π(a/2√3)²

⇒ π(a/2√3)²  = 49π

⇒ (a²)/12 = 49

⇒ a² = 49 × 12

⇒ a = 14√3

Circumradius of equilateral triangle = a/√3

⇒ (14√3)/√3

⇒ 14

Area of Circumcircle

⇒ π(14)²

⇒ 196π cm²

GIVEN:

Three statements.

CONCEPT:

Geometry

FORMULA USED:

Number of diagonals in a polygon of side ‘n’ is given by n (n - 3) / 2.

CALCULATION:

A:

Number of diagonals in a polygon of side ‘n’ is given by n (n - 3) / 2.

B:

Number of diagonals in a polygon of side is 12 = 12 (12 - 3) / 2

= (12 × 9) / 2

= 54

C:

Any quadrilateral in which opposite angles are supplementary is called a cyclic quadrilateral.

GIVEN:

Four statements.

CONCEPT:

Geometry

FORMULA USED:

∠BIC = 90° + (∠A / 2)

∠BOC = 2 × ∠A

∠BHC = 180° - ∠A

∠BXC = 90° - (∠A / 2)

CALCULATION:

A:

∠BIC = 90° + (∠A / 2)

= 90° + (70° / 2)

= 90° + 35°

= 125°

B:

∠BOC = 2 × ∠A

= 2 × 70°

= 140°

C:

∠BHC = 180° - ∠A

= 180 - 70°

= 110°

D:

∠BXC = 90° - (∠A / 2)

= 90° - (70° / 2)

= 90° - 35°

= 55°

GIVEN:

∠A - ∠B = 33°, ∠B - ∠C = 18° 

CALCULATION:

∠A - ∠B = 33°, ∠B - ∠C = 18°

⇒ ∠A = 33° + ∠B and ∠C = ∠B - 18° 

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠B + 33° + ∠B + ∠B - 18° = 180° 

⇒ 3∠B = 180° - 15° 

⇒ 3∠B = 165° 

⇒ ∠B = 55° 

⇒ ∠A = 33° + 55° = 88° 

⇒ ∠C = 55° - 18° = 37° 

⇒ The largest angle A = 88° and the smallest angle C = 37°  

⇒ Sum of the smallest and the largest angle = (88° + 37° ) = 125° 

∴  Sum of the smallest and the largest angle is 125° 

Congruent Figures: Two plane- figures (figures drawn in a plane) are said to be congruent if the traced copy of one coincides with the other exactly. To make the work logical, we will give some definitions and some conditions under which figures (triangles) become congruent. The symbol that represents is congruent to is ≅.

• Congruence of line-segments: If the lengths of two line segments are equal, then the line segments are said to be congruent. Thus two congruent line segments are of equal in length.
• Congruence of angles: Two angles are congruent if their measures are equal. Thus two congruent angles have equal measures.
• Congruence of triangles: Two triangles are said to be congruent if their all sides and angles are equal corresponding to the other triangle’s all sides and angles. There are four ways in which a triangle can be proved or said congruent.
  • SSS (Side-Side-Side) Rule: It is when we are given that all the three sides of one triangle are equal to three sides of another triangle correspondingly.
  • SAS (Side-Angle-Side) Rule: It states that two sides and the angle included between those sides of one triangle are equal to two sides and the angle included between them of the other triangle.
  • ASA (Angle-Side-Angle) Rule: If the two angles and one side between them of one triangle is equal to the corresponding angles and side, the triangle is said to be congruent.
  • AAS (Angle-Angle-Side) Rule: It states that if two consecutive angles and one adjacent side of a triangle is equal to two consecutive angles and one adjacent side of another triangle.
  • Right- Hand Side (RHS) Rule: It states that if in two right-angled triangles, the length of the hypotenuse and one of the sides of one triangle is equal to the length of the hypotenuse and one of the sides of another triangle, the triangle is said to be congruent.

Hence, we conclude that both statements are true.

Given:

In ΔABC, PQ || BC.

AP = 5 cm, AB = 15 cm and PQ = 10 cm

Concepts used:

Rule of similarity - AA, that is Angle-angle.

If two triangles are similar, the ratio of corresponding sides is equal.

Calculation:

In ΔAPQ and ΔABC,

∠APQ = ∠ABC (Correspnding angles are equal when PQ || BC)

∠PAQ = ∠BAC (Common angle)

∴ ΔAPQ ∼ ΔABC (By AA).

As ΔAPQ ∼ ΔABC,

⇒ AP/AB = PQ/BC (Ratio of corresponding sides of similar triangles is equal)

⇒ 5/15 = 10/BC

⇒ 1/3 = 10/BC

⇒ BC = 3 × 10 

⇒ BC = 30 cm.

∴ The length of side BC is 30 cm.

Given:

ΔDEF ∼ ΔPQR

Height of ΔDEF/Height of ΔPQR = 48 cm/24 cm

Concepts used:

If two triangles are similar, the ratio of corresponding sides is equal to ratio of corresponding heights.

Calculation:

ΔDEF ∼ ΔPQR

Height of ΔDEF/Height of ΔPQR = DE/PQ

⇒ 48/24 = DE/6 

⇒ 2 = DE/6

⇒ DE = 2 × 6

⇒ DE = 12 cm.

∴ The value of side DE is 12 cm.

Given:

The ratio of angles of triangle = 4 ∶ 3 ∶ 2

Formula used:

Sum of all angles of triangle = 180° 

The supplementary angle of x = 180° – x

Calculation:

Let the angles be 4x, 3x and 2x

Sum of all angles of triangle = 180° 

4x + 3x + 2x = 180° 

⇒ 9x = 180° 

⇒ x = 20° 

Angles of the triangle are 4x = 80°, 3x = 60°, and 2x = 40° 

Largest angle of triangle = 80° 

Supplementary angle of 80° = 180° – 80° 

⇒ Supplementary angle of 80° = 100° 

∴ Supplementary angle of largest angle of the triangle is 100° 

Given:

In ΔABC

∠ABC = 90º 

AC = 10 cm 

BC = 8 cm 

Formula Used:

Pythagoras therom 

AC2 = AB2 + BC2

Calculation:

Using pythagoras therom

10² = AB² + 8²

100 - 64 = AB²

AB² = 36

AB = 6 cm  

 ∴ The length of AB is 6 cm 

Concept used:

The right-angled triangle follow the Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

The hypotenuse is the greatest side of the right-angled triangle

Calculation:

Checking the options one by one 

Option 1- 15² + 32² = 1249 ≠ 37²

It doesn't satisfy the Pythagoras theorem so it is not the side of the right-angled triangle

Option 2- 65² + 72² = 9409 = 97²

It satisfies the Pythagoras theorem so it is the side of the right-angled triangle

Option 3- 20² + 21² = 841 ≠ 31²

It doesn't satisfy the Pythagoras theorem so it is not the side of the right-angled triangle

Option 4- 35² + 77² = 7154 ≠ 88²

It doesn't satisfy the Pythagoras theorem so it is not the side of the right-angled triangle

∴ The correct answer is option 2

Given:

ΔABC is scalene triangle.

AB = 13 cm

Concept:

Since we know that the sum of any two sides of a triangle is greater than the third side, i.e AB + BC > AC

⇒ AB > AC - BC

Therefore, the difference of the other two side must be smaller than the third side.

∴ 13 cm cannot be the difference of the two side.

Given:

In ΔPQR, O is the incentre 

∠R = 42° 

Concept used:

In ΔPQR,

∠QOP = 90° + ∠R/2 

∠QOR = 90° + ∠P/2

∠POR = 90° + ∠Q/2

Calculations:

∠QOP = 90° + 42°/2

⇒ 90° + 21° 

⇒ 111° 

∴ ∠QOP is 111°.

Given:

The three angles of triangle = a°, (a +24)° and (a + 36)° 

Calculation:

 a°, (a +24)° and (a + 36)° = 180°  (angle sum property of a triangle)

⇒ 3a° = 180° – 60° 

⇒ 3a = 120° 

⇒ a = 40° 

⇒ (a + 24)° 

 ⇒ (40 + 24)° 

⇒ 64° 

⇒ (a + 36)° 

⇒ (40 + 36)° 

⇒ 76° 

Given:

(x, y) divides the line segment joining (4, 3) and (2, 5) in the ratio 1 : 1 internally.

Formula Used:

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Calculation:

x = (mx₂ + nx₁)/(m + n)

⇒ x = (1 × 2 + 1 × 4)/(1 + 1)

⇒ x = (2 + 4)/2

⇒ x = 6/2

⇒ x = 3

y = (my₂ + ny₁)/(m + n)

⇒ y = (1 × 5 + 1 × 3)/(1 + 1)

⇒ y = (5 + 3)/2

⇒ y = 8/2

⇒ y = 4

∴ The value of (x, y) is (3, 4).

Given:

(x, y) divides the line segment joining (6, 5) and (4, 7) in the ratio 1 : 1 internally.

Formula Used:

x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Calculation:

x = (mx₂ + nx₁)/(m + n)

⇒ x = (1 × 4 + 1 × 6)/(1 + 1)

⇒ x = (4 + 6)/2

⇒ x = 10/2

⇒ x = 5

y = (my₂ + ny₁)/(m + n)

⇒ y = (1 × 7 + 1 × 5)/(1 + 1)

⇒ y = (7 + 5)/2

⇒ y = 12/2

⇒ y = 6

Concept :

The Sum of any 2 sides should be bigger than 3rd side of the triangle.

Option A:

5.5 cm + 8.5 cm > 7.5 cm

⇒ 14 cm > 7.5 cm  (yes)

 8.5 cm + 7.5 cm > 5.5 cm

⇒ 16 cm > 5.5 cm  (yes)

7.5 cm + 5.5 cm > 8.5 cm

⇒ 13 cm > 8.5 cm  (yes)

The Sum of any 2 sides should be bigger than 3rd side of the triangle. So, the triangle can be formed.

This option is correct

Option B:

2.5 cm + 6.5 cm > 3.5 cm

⇒ 9 cm > 3.5 cm  (yes)

6.5 cm + 3.5 cm > 2.5 cm

⇒ 10 cm > 2.5 cm  (yes)

3.5 cm + 2.5 cm < 6.5 cm

⇒ 6 cm < 6.5 cm  (No)

The Sum of any 2 sides is smaller than 3rd side of the triangle. So, the triangle can't be formed.

Option C:

4.2 cm +  6.2 cm > 1.9 cm

⇒10.4 cm > 1.9 cm   (yes)

6.2 cm + 1.9 cm > 4.2 cm

⇒ 8.1 cm > 4.2 cm   (yes)

4.2 cm + 1.9 cm < 6.2 cm

⇒ 6.1 cm < 6.2 cm  (No)

The Sum of any 2 sides is smaller than 3rd side of the triangle. So, the triangle can't be formed.

Option D: 

3.1 cm + 4.1 cm > 1.0cm

⇒ 7.2 cm > 1.0 cm   (yes)

4.1 cm + 1.0cm > 3.1 cm

⇒ 5.1 cm > 3.1 cm   (yes)

3.1 cm + 1.0 cm = 4.1 cm

⇒ 4.1 cm = 4.1 cm  (No)

The Sum of any 2 sides is smaller than 3rd side of the triangle. So, the triangle can't be formed.

The correct option is 1 i.e. 5.5 cm, 8.5 cm, 7.5 cm                             

Concept

For an equilateral triangle, all the four points (circumcenter, incenter, orthocenter and centroid) coincide.

Orthocenter : The orthocenter is the point of intersection of the three heights of a triangle.

Centroid : The centroid is the point of intersection of the three medians.

Circumcenter : The circumcenter is the point of intersection of the three perpendicular bisectors.

Incenter : The incenter is the point of intersection of the three angle bisectors.

Given:

(3 × supplementary angle) - (7 × complementary angle) =10° 

Formula Used/Concept used:

Sum of complementary angle is 90° 

Sum of supplementary angle is 180° 

Calculation:

Let the angle be x

⇒ {3 × (180° - x)} - {7 × (90° - x)} = 10° .

⇒ 540° - 3x - (630° - 7x) = 10° 

⇒ 540° - 3x - 630° + 7x = 10°

⇒ 4x - 90° = 10°

⇒ 4x = 10° + 90°

⇒ 4x = 100°

⇒ x = 100°/4

⇒ x = 25° 

∴ The required answer is 25°

Concept 

Complementary angles are those angles whose sum is 90º 

Supplementary angles are those whose sum is 180º

Calculation

x is equal to its complement 

⇒ x + x = 90º 

2x = 90º

⇒ x = 45º

y is equal to its supplement 

⇒ y + y = 180º

2y = 180º

⇒ y = 90º

so, 2x + 3y = 2 × 45º + 3 × 90º = 90º + 270º = 360º

Concept 

Complementary angles are those angles whose sum is 90°

Supplementary angles are those whose sum is 180° 

Calculation

x + x + 32° = 180°

⇒ 2x = 180° - 32° = 148°

So, x = 74°

y + y - 24° = 90°

⇒ 2y = 90° + 24° = 114°

So, y = 57°

Then, x + y = 74° + 57° = 131°

Given:

In ΔABC, 2∠A = 3∠B = 6∠C

Concept Used:

The sum of all the angles of a triangle is 180°

Calculation:

Let, 2∠A = 3∠B = 6∠C = 6k 

so,  ∠A = 3k,  ∠B = 2k and  ∠C = k

Now, in ΔABC

∠A + ∠B + ∠C = 180°

3k + 2k + k = 180°

6k = 180°

k = 30° 

∠A = 3k = 90° 

∴ ∠A is 90°

Given:

The difference between the two complementary angles = 50° 

Concept used:

Complementary angle:- The sum of two angles is 90° 

Calculation:

Let the two angles be x and y respectively

A/Q

x + y = 90      -----(i)

x - y = 50      ------(ii)

By adding both equations (i) and (ii)

x + y + x - y = 90 + 50

⇒ 2x = 140

⇒ x = 70

By substituting value of x in equation in eq. (i)

x + y = 90

⇒ 70 + y = 90

⇒ y = 90 - 70

⇒ y = 20° 

Hence, x > y

∴ The greater angle is 70° 

Given:

Perpendicular = 2 cm and base = 1 cm

Formula used:

Radius of circumcircle of right angled triangle = Hypotenuse/2

Area of circle = πr²

Calculation:

(Hypotenuse)² = (perpendicular)² + (base)²

⇒ (1)² + (2)²

⇒ √5 cm

Radius of circumcircle = √5/2

Area of circle = π(√5/2)²

⇒ 5π/4 cm²

Given

The perimeter of a circle is 264 cm

Two sides of triangle = 21 cm and 12 cm 

Area of triangle = 21 cm²

Concept

Circumradius (R) = (S1 × S2 × S3)/(4 × Δ )

Where,

S1 , S2  and S3  are sides of triangle

Δ = Area of triangle

Perimeter of circle = 2πR 

Calculation

Perimeter = 2 × (22/7) × R

⇒ 264 = 2 × (22/7) × R

⇒ 6 × 7 = R 

⇒ 42 = R

Now, 

⇒ 42 = (21 × 12 × S3)/(4 × 21)

⇒ S3 = 14 cm.

∴ The third side of the triangle is 14 cm

Given

In a pentagon,

Two angle are 50° and 50° 

The rest are in the ratio = 5 : 2 : 4

Concept

Sum of all interior angles of a regular polygon = (n - 2) × 180° 

Calculation

Number of sides = 5 

⇒ Sum of all interior angles of a pentagon = (5 - 2) × 180° 

⇒ Sum of all interior angles of a pentagon = 3 × 180° 

⇒ Sum of all interior angles of a pentagon = 540° 

Now,

⇒ Two angles = 50° + 50° 

⇒ Two angles = 100° 

⇒ Sum of rest of the angles = 540° - 100° 

⇒ Sum of rest of the angles = 440° 

Let the ratio of sides in x 

⇒ 5x : 2x : 4x

⇒ 5x + 2x + 4x = 440° 

⇒ 11x = 440° 

⇒ x = (440°/11)

⇒ x = 40° 

Now, 

⇒ 5 × 40°  = 200° 

⇒ 2 × 40° = 80° 

⇒ 4 × 40°  = 160° 

∴ Biggest angle is 200° 

Given

Two angle are 50° and 50° 

The rest are in the ratio = 15 : 2 : 5

Concept

Sum of all interior angles of a regular polygon = (n - 2) × 180° 

Calculation

Number of side = 5 

⇒ Sum of all interior angles of a pentagon = (5 - 2) × 180° = 3 × 180° = 540° 

Now,

Two angles = 50° + 50° = 100° 

Sum of rest of the angles = 540° - 100° = 440° 

Let the ratio of sides in x 

⇒ 15x + 2x + 5x = 440° 

⇒ 22x = 440° 

⇒ x = 20° 

Now, 

15 × 20°  = 300° 

2 × 20° = 40° 

4 × 20°  = 80° 

∴ Smallest angle is 40° 

Given 

In a pentagon,

Two angle are 40° and 60°

The rest are in the ratio = 5 : 2 : 4

Concept

Sum of all interior angles of a regular polygon = (n - 2) × 180° 

Calculation

Number of side = 5

⇒ Sum of all interior angles of a pentagon = (5 - 2) × 180° 

⇒ Sum of all interior angles of a pentagon = 3 × 180° 

⇒ Sum of all interior angles of a pentagon = 540° 

Now,

⇒ Two angles = 40°  + 60° 

⇒ Two angles = 100° 

⇒ Sum of rest of the angles = 540° - 100° 

⇒ Sum of rest of the angles = 440° 

Let the ratio of sides in x 

⇒ 5x : 2x : 4x

⇒ 5x + 2x + 4x = 440° 

⇒ 11x = 440°

⇒ x = (440°/11)

⇒ x = 40° 

Now, 

⇒ 5 × 40° = 200° 

⇒ 2 × 40°  = 80° 

⇒ 4 × 40°  = 160° 

Now,

⇒ Biggest angle = 200° 

⇒ Smallest angle = 40° 

⇒ Difference between biggest angle and smallest angle = 200° - 40° 

⇒ Difference between biggest angle and smallest angle = 160° 

∴  Difference between biggest angle and smallest angle is 160° 

 Given :

There are two linear equations intersect at a point (a, 4)

4x - 2y = 10 

4x + ky = 2 

Calculations :

4x - 2y = 10      ....(1) 

4x + ky = 2      ....(2) 

Solving equations (1) and (2)  we will get,

-2y - ky = 8 

-y (2 + k) = 8 

put y = 4 as intersecting points were (a,4)

-4 (2 + k) = 8 

⇒ k = -4 

∴ The value of the k will be -4 

Concept :

Square has all 4 sides equal. So it is having rotational symmetry at every ¼ th turn.

Rectangle has opposite sides equal. So it is having rotational symmetry at every ½ turn.

A regular hexagon has all 6 sides equal. So it is having rotational symmetry at every 1/6 th turn.

An equilateral triangle has all 3 sides equal. So it is having rotational symmetry at every 1/3 turn. 

The correct option is 1 i.e. Square.

Given:

Circumference of the circle = 22 cm

Formula used:

Circumference = 2πr

Area of a circle = πr²

Area of the quadrant = (πr2 × 1/4)

r = Radius of the circle

Calculation:

2πr = 22

⇒ πr = 11

⇒ 22/7 × r = 11

⇒ r = 3.5

Area of the circle = πr²

⇒ 22/7 × 7/2 × 7/2

⇒ 154/4

⇒ Area of the quadrant = 154/4 × 1/4

⇒ 154/16

⇒ 9.625

⇒ Area of the quadrant = 9.625 cm²

∴ Area of the quadrant is 9.625 cm²

Given:

Point fall in second quadrant.

Concept used:

Values of x and y in different quadrant.

1^st quadrant = (x, y)

2^nd quadrant = (–x, y)

3^rd quadrant = (–x, –y)

4^th quadrant = (x, –y)

Calculation:

Option 1). (3, –3)

Will fall in 4th quadrant

Option 2). (3, 3)

Will fall in 1st quadrant

Option 3). (–3, 3)

Will fall in 2nd quadrant.

Option 4). (–3, –3)

Will fall in 3rd quadrant.

∴ Point (–3, 3) will fall in 2^nd quadrant.

Given:

Ratio of sides =  \(\frac{1}{2}\)∶ \(\frac{1}{3}\)∶ \(\frac{1}{5}\)

Perimeter = 186 cm

Concept used:

Perimeter of a triangle = a + b + c

Here, a, b and c are the sides of the triangle

Calculation:

L.C.M of 2, 3, 5 = 30

Ratio of sides = 30 × 1/2 : 30 × 1/3 : 30 × 1/5

⇒ 15 : 10 : 6

According to the question

31 → 186 cm

⇒ 1 → 6 cm

⇒ 15 → 90 cm

⇒ largest side = 90 cm

∴ Length of the largest side is 90 cm