Quiz-3

Let the area of backyard be x² this year and y² last year
∴ X² - Y² = 131
=> (X+Y)*(X-Y) = 131
Now, 131 is a prime number (a unique one too. Check out its properties on Google). Also, always identify the prime number given in a question. Might be helpful in cracking the solution.
=> (X+Y)*(X-Y) = 131 x 1
=> X+Y = 131
X-Y = 1
=> 2X = 132 => X = 66
and Y = 65
∴ Number of tomatoes produced this year = 662 = 4356

The question is "How many tomatoes did Anil produce this year?"

Hence, the answer is 4356

Here,
Let the radius of PQRS be 2r
∴ Radius of each of the smaller circles = 2r22r2 = r
∴ Area A can be written as
A = π (2r)² – 4 x π(r)² (Area of the four smaller circles) + B (since, B has been counted twice in the previous subtraction)
=> A = 4πr² - 4πr² + B
=> A = B
=> ABAB = 1

The question is " PQRS is a circle and circles are drawn with PO, QO, RO and SO as diameters areas A and B are shaded A/B is equal to "

Hence, the answer is 1

AP = PB = 2cm
∴ AB = √(2² + 2²) = √8 = 2√2 cm

Similarly, BE = EH = 2√222√22 = √2 cm
∴ EH = √(√2² + √2²)
= √4 = 2 cm
If we write the infinite series of area of squares,
= 42 + (2√2)² + 2² + ……. infinite
Since it is a decreasing series sum of infinite terms can be approximated.
= 16+8+4+………infinite
∴ a (first term) = 16 , r = 1212 (in the infinite G.P)
Sum of an infinite G.P = a1−ra1−r
= 161−1/2161−1/2
= 32 cm²

The question is " Find the sum of all the squares. "

Hence, the answer is 32 cm²

Since neither angles nor sides are given in the question, immediately the sum of angles of pentagon should come in mind. To use it,
We know the area of the sectors of a circle is given as,
Central Angle x π∗r²/360°
In this case,
Sum of shaded region = ∠P∗π∗r²/360° + ∠Q∗π∗r²/360° ………… + ∠T∗π∗r²/360°
= (P + Q +…. T) x π∗r²/360°
= Sum of interior angles of pentagon x π∗r²/360°
= 3 x 180° x π∗r²/360°(Sum of interior angles = (n – 2) X 180°)
= 3 * π∗r²/2
Note => The above concept is applicable for a polygon of n sides.

The question is Find the area of the portion of circles falling inside the pentagon.

Hence, the answer is 1.5πr²

∴ R = 2 + r’
We know from the properties of equilateral triangle,
r’ = Sides√3Sides√3 = 4√34√3
(This can easily be derived using trigonometry. However, please remember this formula. It is useful at places)
∴ R = 2 + 4√34√3
∴ Area = π * R² = π * (2√3+4)2√32(2√3+4)2√32
= π3π3 * (4 + 2√3)²

The question is  Find the area of the circle, circumscribing the given figure.

Hence, the answer is π/3 (4 + 2√3)²

From the figure, we can see that the diagonal of the square inscribed in innermost circle is the diameter of the innermost circle.
Side = √32
Or Diagonal = √32 * √2 = 8 cm
Since the circles are spaced at 1.25 cm, the distance between innermost and outermost circles = 1.25 * 4 = 5 cm
Therefore the diameter of the outermost circle = 5 cm + diagonal of inscribed square + 5cm = 18 cm
Now this 18 cm will be the diagonal of the square that needs to be inscribed in outermost circle.
Or a √2 = 18
a = 9 √2 cm
Area = 9 √2 * 9 √2 = 162 sq. cm

The question is  If a square needs to be inscribed in the outermost circle, what will be its area?

Hence, the answer is 162 sq.cm.

Area of the ball = 4πr²
= 4π * 14 * 14
= 784π
Cumulative area of the 2 pieces = 3/28 more than the ball's surface area.
Therefore, extra area = 3/28 * 784π = 84π
Now this extra area = areas of 2 new circles of those 2 pieces
Let area of each new circle = πr₁²
πr₁² = 84π / 2 = 42π
Or r₁² = 42
Now, r₁, x and r form a right angled triangle.
r² = x² + r₁²
Or x² = 14*14 - 42
= 196 – 42 = 154
Or x = 12.4 cm

The question is "What should be the value of “x” such that the cumulative surface area of the newly formed pieces is 3/28 more than the rubber ball’s original surface area? "

Hence, the answer is 12.4 cm