Mensuration 2D — BSSC Quantitative Aptitude

Solution:

• The sides $3, 4, 5$ form a Pythagorean triplet, meaning it is a right-angled triangle.

• The perpendicular legs are $3 ext{ cm}$ and $4 ext{ cm}$.

• Area = $rac{1}{2} imes ext{base} imes ext{height} = rac{1}{2} imes 3 imes 4 = 6 ext{ cm}^2$.

Exam Trick:

• Recognizing standard triplets ($3,4,5$) saves time. No need for Heron's formula.

Solution:

• The formula for the area of a square using its diagonal $d$ is $rac{d^2}{2}$.

• Given $d = 10\sqrt{2}$.

• Area = $rac{(10\sqrt{2})^2}{2} = rac{200}{2} = 100 ext{ cm}^2$.

Exam Trick:

• Use the direct diagonal formula $rac{d^2}{2}$ instead of finding the side first.

Solution:

• Let breadth $= b$, then length $= b + 2$.

• Perimeter $= 2(l + b) = 2(b + 2 + b) = 24$.

• $2b + 2 = 12 \Rightarrow 2b = 10 \Rightarrow b = 5 ext{ m}$.

• Length $= 5 + 2 = 7 ext{ m}$.

• Area $= l imes b = 7 imes 5 = 35 ext{ m}^2$.

Exam Trick:

• Half perimeter is $l+b = 12$. Two numbers differing by $2$ and summing to $12$ are $7$ and $5$. Area is $7 imes 5 = 35$.

Solution:

• Diagonals of a rhombus bisect each other at $90^\circ$.

• Half-diagonals are $rac{16}{2} = 8 ext{ cm}$ and $rac{12}{2} = 6 ext{ cm}$.

• Side $a = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10 ext{ cm}$.

• Perimeter $= 4a = 4 imes 10 = 40 ext{ cm}$.

Exam Trick:

• Diagonals in ratio $16:12 = 4:3$. The side forms the hypotenuse of a $3-4-5$ right triangle, giving a side of $5 imes 2 = 10$. Perimeter is $4 imes 10 = 40$.

Solution:

• Area of Trapezium = rac{1}{2} imes ( ext{Sum of parallel sides}) imes ext{height}.

• Area = rac{1}{2} imes (10 + 15) imes 6.

• Area = rac{1}{2} imes 25 imes 6 = 75 ext{ cm}^2.

Exam Trick:

• Average of parallel sides is $rac{25}{2} = 12.5$. Multiply directly by height $6$ to get $75$.

Solution:

• Circumference $= 2\pi r = 44$.

• 2 imes rac{22}{7} imes r = 44 \Rightarrow r = 7 ext{ cm}.

• Area = \pi r^2 = rac{22}{7} imes 7^2 = 154 ext{ cm}^2.

Exam Trick:

• Memorize the standard base values: when radius $r=7$, Circumference $= 44$, Area $= 154$. Direct answer.

Solution:

• A regular hexagon consists of $6$ equilateral triangles of side $a$.

• Area = 6 imes rac{\sqrt{3}}{4} imes a^2.

• Area = 6 imes rac{\sqrt{3}}{4} imes 4^2 = 6 imes 4\sqrt{3} = 24\sqrt{3} ext{ cm}^2.

Exam Trick:

• Mentally calculate the area of one equilateral triangle of side $4$, which is $rac{\sqrt{3}}{4} imes 16 = 4\sqrt{3}$. Multiply by $6$ to get $24\sqrt{3}$.

Solution:

• A rectangle is defined as a quadrilateral with four right angles.

• A square has four equal sides and four right angles.

• Therefore, every square satisfies the conditions of a rectangle, but not every rectangle satisfies the conditions of a square.

Exam Trick:

• No calculation needed; a square is a special, strict subset of rectangles.

Solution:

• Area of a sector = rac{ heta}{360} imes \pi r^2.

• Area = rac{60}{360} imes rac{22}{7} imes 7^2.

• Area = rac{1}{6} imes 154 = rac{77}{3} ext{ cm}^2.

Exam Trick:

• Since $r=7$, the full circle area is $154$. A $60^\circ$ sector is exactly $rac{1}{6}$ of the circle, so the area is $rac{154}{6} = rac{77}{3}$.

Solution:

• Area of circle = \pi r^2 = 616 \Rightarrow rac{22}{7} imes r^2 = 616 \Rightarrow r^2 = 196 \Rightarrow r = 14.

• Length of wire (Circumference) = 2 imes rac{22}{7} imes 14 = 88 ext{ cm}.

• The wire is bent into a square, so perimeter of square $= 88 ext{ cm}$.

• Side of square a = rac{88}{4} = 22 ext{ cm}.

• Area of square $= 22^2 = 484 ext{ cm}^2$.

Exam Trick:

• For the same perimeter, Circle Area > Square Area. Using proportionality, if Area $154 ightarrow$ Square Area $121$, then Area $616$ ($4$ times) $ ightarrow$ Square Area $484$ ($4$ times $121$).