Mensuartion 2D — RRB NTPC Quantitative Aptitude

• Given: Isosceles right-angled triangle. Perimeter = $30$. Area = ?

• Solution:
Step 1: Let the equal sides be $x$. The hypotenuse is $x\sqrt{2}$.
Step 2: Perimeter = $x + x + x\sqrt{2} = x(2 + \sqrt{2}) = 30$.
Step 3: $x = \frac{30}{2 + \sqrt{2}}$.
Step 4: Area = $\frac{1}{2} x^2 = \frac{1}{2} \left( \frac{30}{2 + \sqrt{2}} \right)^2 = \frac{1}{2} \left( \frac{900}{4 + 2 + 4\sqrt{2}} \right) = \frac{450}{6 + 4\sqrt{2}}$.
Wait, $(2+\sqrt{2})^2 = 4 + 2 + 4\sqrt{2} = 6 + 4\sqrt{2}$.
Area = $\frac{450}{2(3 + 2\sqrt{2})} = \frac{225}{3 + 2\sqrt{2}}$.
Rationalize: $\frac{225(3 - 2\sqrt{2})}{9 - 8} = 225(3 - 2(1.414)) = 225(3 - 2.828) = 225(0.172)$.
Calculation: $225 \times 0.172 = 38.7$.

• Exam Trick: Perimeter of isosceles right triangle is $P = x(2+\sqrt{2})$. Area is $\frac{1}{2}x^2$. Alternatively, $Area = \frac{P^2}{4(3+2\sqrt{2})} = \frac{P^2(3-2\sqrt{2})}{4}$.
Area = $\frac{900(3-2.828)}{4} = 225(0.172) = 38.7$.

• Final Answer: $38.7 \text{ cm}^2$ (Option 1)

• Given: Sides of triangle = 60, 112, 164. Cost rate = ₹10.50/m². Total cost = ?

• Solution:
Step 1: Check if it's a right-angled triangle. Divide sides by 4: 15, 28, 41.
Step 2: Check if 15, 28, 41 is a Pythagorean triplet. $15^2 + 28^2 = 225 + 784 = 1009$. $41^2 = 1681$. It's not a triplet. Wait, my sides might be wrong. Let's re-read the source carefully.
The source says: "60, 112, 164 ... 4 से काटा 15, 28, 41 ... पाइथागोरियन ट्रिप्लेट नहीं है". Okay, so Hero's formula must be used.
Let's recalculate with Hero's Formula for 60, 112, 164.
$s = \frac{60 + 112 + 164}{2} = \frac{336}{2} = 168$.
$s-a = 168 - 60 = 108$.
$s-b = 168 - 112 = 56$.
$s-c = 168 - 164 = 4$.
Area = $\sqrt{168 \times 108 \times 56 \times 4}$.
$168 = 56 \times 3$. So $\sqrt{56 \times 3 \times 108 \times 56 \times 4} = 56 \times \sqrt{324 \times 4} = 56 \times 18 \times 2 = 56 \times 36$.
Area = $2016$ m².
Step 3: Total Cost = Area $\times$ Rate = $2016 \times 10.50 = 2016 \times \frac{21}{2} = 1008 \times 21 = 21168$.

• Exam Trick: Find area using Hero's formula quickly by splitting factors: $s=168$. Area = $\sqrt{168 \times 108 \times 56 \times 4} = 56 \times 18 \times 2 = 2016$. Cost = $2016 \times 10.5$. Notice 10.5 has a factor of 3 and 7. The Area 2016 is a multiple of 9. So answer must be a multiple of 9. Let's check options: $2+1+1+6+8 = 18$ (Yes). $2+2+1+4+0 = 9$ (Yes). If multiple options match, calculate $1008 \times 21 = 21168$.

• Final Answer: ₹21,168 (Option 2)

• Given: Sides of triangle = 48, 55, 73. Area = ?

• Solution:
Step 1: Check if the sides form a Pythagorean triplet. The numbers 48, 55, 73 form a standard Pythagorean triplet.
Verify: $48^2 + 55^2 = 2304 + 3025 = 5329$. $73^2 = 5329$. Yes, it is a right-angled triangle.
Step 2: The area of a right-angled triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
Step 3: Area = $\frac{1}{2} \times 48 \times 55 = 24 \times 55$.
Step 4: $24 \times 55 = 24 \times \frac{110}{2} = 12 \times 110 = 1320 \text{ cm}^2$.

• Exam Trick: Memorize important triplets like 48, 55, 73. Once identified as a right triangle, area is just half the product of the two smaller sides. Area = $24 \times 55 = 1320$.

• Final Answer: $1320 \text{ cm}^2$ (Option 1)

• Given: Sides of triangle = 96, 110, 146. Cost rate = ₹5.60/m². Total cost = ?

• Solution:
Step 1: Check for Pythagorean triplet by taking ratio. Divide by 2: 48, 55, 73.
Step 2: We know 48, 55, 73 is a standard Pythagorean triplet. Thus, the original triangle is a right-angled triangle with base 96 and height 110 (or vice versa), and hypotenuse 146.
Step 3: Area of the park = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 96 \times 110 = 48 \times 110 = 5280 \text{ m}^2$.
Step 4: Total Cost = Area $\times$ Rate = $5280 \times 5.60$.
$5280 \times \frac{56}{10} = 528 \times 56 = 29568$.

• Exam Trick: Recognize the triplet ratio: $96:110:146 = 48:55:73$. So, Area = $\frac{1}{2} \times 96 \times 110 = 5280$. Cost = $5280 \times 5.6$. Check unit digit: $8 \times 6 = 48$, so it must end in 8. Only Option 1 ends in 8.

• Final Answer: ₹29,568 (Option 1)

• Given: Triangle ABC. D is on BC. BD:DC = 4:3. Area(ABD) = 40. Area(ABC) = ?

• Solution:
Step 1: Triangles ABD and ADC share the same height from vertex A.
Step 2: Therefore, the ratio of their areas is equal to the ratio of their bases.
Ratio of Area(ABD) : Area(ADC) = BD : DC = 4:3.
Step 3: Area(ABD) = 4 units = 40 cm². So, 1 unit = 10 cm².
Step 4: Total base BC = BD + DC = 4 + 3 = 7 units.
Step 5: Total Area(ABC) = 7 units = $7 \times 10 = 70 \text{ cm}^2$.

• Exam Trick: Area is directly proportional to the base when height is constant. Base ratio is 4:3. 4 parts = 40, so total 7 parts = 70.

• Final Answer: $70 \text{ cm}^2$ (Option 1)

• Given: Area $L \times W = 600$. New dimensions $(L+10)$ and $(W-2)$ give same area 600. Original length $L = ?$

• Solution:
Step 1: Equation 1: $L \times W = 600 \Rightarrow W = \frac{600}{L}$.
Step 2: Equation 2: $(L + 10)(W - 2) = 600$.
Step 3: Expand Equation 2: $LW - 2L + 10W - 20 = 600$.
Step 4: Substitute $LW = 600$: $600 - 2L + 10W - 20 = 600 \Rightarrow -2L + 10W = 20 \Rightarrow -L + 5W = 10$.
Step 5: Substitute $W = \frac{600}{L}$: $-L + 5\left(\frac{600}{L}\right) = 10 \Rightarrow -L^2 + 3000 = 10L \Rightarrow L^2 + 10L - 3000 = 0$.
Step 6: Factorize: $(L + 60)(L - 50) = 0$. Since length must be positive, $L = 50$.

• Exam Trick: Check options. Area must be 600.
If $L=50$, $W=12$. New dimensions: $L=60, W=10$. New Area = $60 \times 10 = 600$. Matches perfectly!
If $L=40$, $W=15$. New dimensions: $L=50, W=13$. New Area = $50 \times 13 = 650 \neq 600$.

• Final Answer: $50 \text{ m}$ (Option 1)

• Given: Inner rectangle $L = 30$, $W = 20$. Path width $x = 2$ on the outside. Area of path = ?

• Solution:
Step 1: The outer dimensions will be $(L + 2x)$ and $(W + 2x)$.
Outer Length = $30 + 2(2) = 34 \text{ m}$.
Outer Width = $20 + 2(2) = 24 \text{ m}$.
Step 2: Area of outer rectangle = $34 \times 24 = 816 \text{ m}^2$.
Step 3: Area of inner rectangle = $30 \times 20 = 600 \text{ m}^2$.
Step 4: Area of path = Outer Area - Inner Area = $816 - 600 = 216 \text{ m}^2$.

• Exam Trick: Use the direct formula for outer path area: $2x(L + W + 2x)$.
Area = $2(2)(30 + 20 + 2(2)) = 4(50 + 4) = 4 \times 54 = 216 \text{ m}^2$.

• Final Answer: $216 \text{ m}^2$ (Option 1)

• Given: Rhombus diagonals $d_1 = 24$, $d_2 = 10$. Perimeter = ?

• Solution:
Step 1: The diagonals of a rhombus bisect each other at $90^\circ$.
Step 2: The lengths of the bisected diagonals form the legs of four right-angled triangles.
The legs are $\frac{24}{2} = 12$ and $\frac{10}{2} = 5$.
Step 3: The side of the rhombus is the hypotenuse. Side $a = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.
Step 4: Perimeter = $4 \times a = 4 \times 13 = 52 \text{ cm}$.

• Exam Trick: Recognize the 5-12-13 Pythagorean triplet immediately. Diagonals halves are 12 and 5, so side is 13. Perimeter = $4 \times 13 = 52$.

• Final Answer: $52 \text{ cm}$ (Option 1)

• Given: Area of rhombus = 216. Diagonal $d_1 = 18$. $d_2 = ?$

• Solution:
Step 1: Area of rhombus = $\frac{1}{2} \times d_1 \times d_2$.
Step 2: $216 = \frac{1}{2} \times 18 \times d_2$.
Step 3: $216 = 9 \times d_2$.
Step 4: $d_2 = \frac{216}{9} = 24 \text{ cm}$.

• Exam Trick: $d_2 = \frac{2 \times \text{Area}}{d_1} = \frac{2 \times 216}{18} = \frac{432}{18} = 24$.

• Final Answer: $24 \text{ cm}$ (Option 1)

• Given: Trapezium parallel sides $a = 14$, $b = 26$. Height $h = 10$. Area = ?

• Solution:
Step 1: Area of trapezium = $\frac{1}{2} \times (a + b) \times h$.
Step 2: Area = $\frac{1}{2} \times (14 + 26) \times 10$.
Step 3: Area = $\frac{1}{2} \times 40 \times 10 = 20 \times 10 = 200 \text{ cm}^2$.

• Exam Trick: Area = Average of parallel sides $\times$ height = $\left(\frac{14+26}{2}\right) \times 10 = 20 \times 10 = 200$.

• Final Answer: $200 \text{ cm}^2$ (Option 1)