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Shortcut Trick

Height of trapezium = (2 × Area) ÷ Sum of parallel sides = (2 × 180) ÷ (20 + 10) = 12 cm.

Base of the right-angled triangle formed at the ends = (20 − 10) ÷ 2 = 5 cm.

Length of each non-parallel side = √(12² + 5²) = √(144 + 25) = √169 = 13 cm.

∴ The correct answer is 13 cm.

Alternate Method

Given:

Parallel sides of the trapezium are 20 cm and 10 cm.

Area of the isosceles trapezium = 180 cm².

Formula Used:

Area of Trapezium = 1/2 × (Sum of parallel sides) × Height

Pythagoras Theorem: Hypotenuse² = Base² + Perpendicular²

Calculations:

⇒ Area = 1/2 × (AB + CD) × Height

⇒ 180 = 1/2 × (10 + 20) × h

⇒ 180 = 1/2 × 30 × h

⇒ 180 = 15 × h

⇒ h = 12 cm

So, the height of the trapezium (AM = BN) is 12 cm.

In isosceles trapezium ABCD, dropping perpendiculars AM and BN to CD constructs a rectangle ABMN.

⇒ MN = AB = 10 cm

Since it is an isosceles trapezium, the remaining base segments are equal:

⇒ DM = NC = (CD − MN) ÷ 2

⇒ DM = (20 − 10) ÷ 2 = 5 cm

Now, applying Pythagoras theorem in right-angled ΔAMD:

⇒ AD² = AM² + DM²

⇒ AD² = 12² + 5²

⇒ AD = √(144 + 25)

⇒ AD = √169 = 13 cm

∴ The correct answer is 13 cm.

Additional Information

Isosceles Trapezium Properties

In an isosceles trapezium, the non-parallel sides are equal in length, the base angles are equal, and the lengths of the diagonals are identical.

Area of a Trapezium

The area is calculated as Area = 1/2 × (a + b) × h, where 'a' and 'b' are the lengths of the parallel sides and 'h' is the perpendicular distance (height) between them.

Median of a Trapezium

The median is the line segment joining the midpoints of the non-parallel sides. Its length is the average of the parallel sides: m = (a + b) ÷ 2, and Area = m × h.

Concept-

• Diagonals of a rhombus bisect each other at 90°
• Area of rhombus = \( {1} \over {2}\) × Product of diagonals

 

Calculation-

BD = 12 cm 

⇒ BO = 6 cm (half of diagonal)

Using Pythagoras theorem in ΔAOB right angle at O.

AB² = BO² + OA²

⇒ OA = \(\sqrt{AB^{2}-BO^{2}}\)

⇒ OA = \(\sqrt{10^{2}-6^{2}}\) = 8 cm.

AC = 2 OA = 16 cm

Area of rhombus = 1/2 × 12 × 16 

⇒ 96 cm².

∴ The area of a rhombus is 96 cm².

Concept- 

\(Inradius = \frac {Area \ of \ triangle \ } {Semiperimeter}\)

Calculation

\(r = \frac{\Delta }{s}\)

⇒ \(3 = \frac{15 }{s}\) 

⇒ s = 5 cm

Now perimeter will be 2 × semiperimeter

Perimeter = 2 × (5 )

Perimeter = 10 cm.

∴ the perimeter of a triangle is 10 cm.

Calculation-

As -3 is the zeros of the quadratic polynomial  (k - 1)x2 + kx + 1 

On substitute -3 in the equation, we get 0 

⇒ (k - 1)(-3)2 + k(-3) + 1 = 0

⇒ 9k - 9 - 3k + 1 = 0

⇒ 6k = 8

⇒ k = 4/3 

∴ the value of k is 4/3.

Concept-

The angle subtended by the chord at the centre is double the angle subtended by it at any point on the circle.

Calculation-

∠ AGB = 2 × ∠ ACB

⇒ 150° = 2 × ∠ ACB

⇒ ∠ ACB = 75° 

∴ the value of ∠ ACB is 75°

Calculation-

The divisibility rule of 3 - the sum of the digits is divisible by 3.

In 2143251 sum of digits will be 18 which is divisible by 3

∴ The number 2143251 is divisible by 3.

Shortcut Trick

The average of the first n natural numbers is given by the formula: (n + 1) ÷ 2.

Here, n = 43.

Average = (43 + 1) ÷ 2 = 44 ÷ 2 = 22.

∴ The correct answer is 22.

Alternate Method

Given: The first 43 natural numbers are 1, 2, 3, ..., 43.

Formula Used: Sum of first n natural numbers = [n(n + 1)] ÷ 2 | Average = Sum ÷ n

⇒ Sum = [43 × (43 + 1)] ÷ 2

⇒ Sum = [43 × 44] ÷ 2 = 43 × 22 = 946

⇒ Average = Total Sum ÷ Total count

⇒ Average = 946 ÷ 43 = 22

∴ The correct answer is 22.

Additional Information

Average of Even Numbers

The average of the first n even natural numbers is (n + 1).

Average of Odd Numbers

The average of the first n odd natural numbers is n.

Average of AP

For any Arithmetic Progression, the average is exactly the middle term, or (First term + Last term) ÷ 2.

Calculation- 

sin θ = \(\frac{\sqrt3}{2}\)

⇒ θ = 60° 

Now, cos3θ will be 

⇒ cos 180° = -1 ----(cos nπ = (-1)ⁿ)

∴ The value of cos 3θ is -1.

Concept-

• tan (A + B) = \( {tan A + tan B} \over 1 - {tan A tan B}\)
• tan nπ = 0

 

Calculation-

tan (π + θ) = \( {tan \pi + tan θ } \over 1 - {tan \pi tan θ}\)

tan (π + θ) = \( \frac{0+tan\theta } {1-0}\)

tan (π + θ) = tan θ 

∴ the value of tan (π + θ) is tan θ .

Shortcut Trick

The length of the wire remains constant, meaning the perimeter of the square equals the circumference of the circle.

Perimeter = 4 × √(Area of square) = 4 × √169 = 52 cm.

The area of a circle in terms of its circumference (C) is given by Area = C² ÷ 4π.

Area = (52 × 52) ÷ (4 × 22/7) = (2704 × 7) ÷ 88 = 215.09 cm².

Rounding to the nearest whole number, we get 215.

∴ The correct answer is 215 cm².

Alternate Method

Given:

Area of the square = 169 cm²

Area = 169 sq. cm Square: P = 52

Bent into L = 52 cm

r Area = ? Circle: C = 52

Formula Used:

Area of square = a²

Perimeter of square = 4a

Circumference of circle = 2πr

Area of circle = πr²

Calculations:

⇒ Area of square = a² = 169

⇒ a = √169 = 13 cm

⇒ Perimeter of square = 4 × 13 = 52 cm

⇒ Since the same wire is bent to form a circle, Circumference of circle = 52 cm

⇒ 2 × (22/7) × r = 52

⇒ r = (52 × 7) ÷ 44

⇒ r = 91 ÷ 11 cm

⇒ Area of circle = πr²

⇒ Area = (22/7) × (91/11) × (91/11)

⇒ Area = (2 × 13 × 91) ÷ 11

⇒ Area = 2366 ÷ 11 = 215.09 cm²

⇒ Rounding off to the nearest whole number, we get 215 cm².

∴ The correct answer is 215 cm².

Additional Information

Isoperimetric Property

Among all planar shapes having the same perimeter, the circle always encloses the maximum area. This is why a wire forms a larger area when bent into a circle than a square.

Conservation of Perimeter

When a wire is reshaped from one figure into another, its total length strictly remains constant. The perimeter of the initial shape equals the perimeter (or circumference) of the newly formed shape.

Area and Perimeter Formula Ratio

For a circle, Area = Circumference² ÷ 4π. For a square, Area = Perimeter² ÷ 16. Using these relationships directly can significantly speed up calculations in competitive exams.