Q: A rectangle of certain dimensions is chopped off from one corner of a larger rectangle as shown. $AB = 8$ cm and $BC = 4$ cm. The perimeter of the figure $ABCPQRA$ (in cm) is <img src="/api/images/rectangle-chopped-off-corner-larger-rectangle-ab…

Concept & Geometric Transformation: When a rectangular corner is removed from a larger rectangle, the perimeter of the resulting shape remains exactly the same as the original un-chopped rectangle. Step-by-Step Solution: • The total perimeter of the figure is the sum of its outer boundary: AB + BC + CP + PQ + QR + RA. • Notice the horizontal segments: The sum of the chopped horizontal segment CP and the inner horizontal segment QR is exactly equal to the full bottom length AB. CP + QR = AB = 8 cm • Notice the vertical segments: The sum of the inner vertical segment PQ and the chopped vertical segment RA is exactly equal to the full right-side length BC. PQ + RA = BC = 4 cm • Now, substitute these sums into the perimeter equation: Perimeter = AB + BC + (CP + QR) + (PQ + RA) Perimeter = 8 + 4 + 8 + 4 Perimeter = 24 cm Exam Strategy & Shortcut: Visualize "pushing" the inner line segments PQ and QR outward to fill the gap. PQ slides right to complete the right edge, and QR slides up to complete the top edge. The shape becomes a perfect 8 × 4 rectangle. Perimeter = 2(8 + 4) = 24. Common Pitfall: Assuming that removing area must also reduce the perimeter, leading test-takers to guess lower numbers without analyzing the lengths of the new edges created by the cut. Final Answer: Therefore, the correct answer is 24.

Q: A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?

Concept & Linear Equations: Translate the word problem into a system of two linear equations. You need to express the total area, the average area, and the difference between the two parts. Step-by-Step Solution: • Let the areas of the two parts be x and y, where x is the larger part. • The total area is 700 hectares: x + y = 700 • The average of the two areas is: Average = (x + y/2) = (700/2) = 350 • The problem states the difference (x - y) is one-fifth of the average: x - y = (1/5) × 350 x - y = 70 • Now we have a simple system of equations: 1) x + y = 700 2) x - y = 70 • We want to find the smaller area (y). Subtract equation (2) from equation (1): (x + y) - (x - y) = 700 - 70 2y = 630 y = 315 Exam Strategy & Shortcut: You know the total is 700, so x + y = 700. The difference x - y = (1/5) × ((700/2)) = 70. To find the smaller number when given a sum and a difference, subtract the difference from the sum and divide by 2: (700 - 70/2) = (630/2) = 315. Common Pitfall: A common error is taking one-fifth of the *total* area (700) instead of the *average* area (350), which would incorrectly make the difference 140, leading to a wrong smaller area of 280 (which is a trap option). Final Answer: Therefore, the correct answer is 315.

Q: An artist has completed one-fourth of a rectangular oil painting. When he will paint another 100 square centimetres of the painting, he would complete three-quarters of the painting. If the height of the oil painting is 10 cm, determine the length (…

Concept & Fraction Logic: This problem is solved by translating word fractions into a simple algebraic equation to find the total area, and then using the standard area formula for a rectangle to find the missing dimension. Area = Length × Height (Breadth) Step-by-Step Solution: 1. Let the total area of the rectangular oil painting be A square centimeters. 2. The artist has already completed (1/4) of the painting, which is (A/4). 3. Painting another 100 cm^2 brings the total completed portion to (3/4) of the painting. We can set up the equation: (A/4) + 100 = (3A/4) 4. Isolate the A terms on one side to solve for the total area: 100 = (3A/4) - (A/4) 100 = (2A/4) 100 = (A/2) 5. Multiply by 2: A = 200 cm^2 6. We are given that the height (which acts as the breadth of the rectangle) is 10 cm. Let the length be L. L × 10 = 200 L = (200/10) = 20 cm Exam Strategy & Shortcut: You can solve this mentally using parts. The difference between completing (3/4) and (1/4) of the painting is (2/4), which simplifies to (1/2) (half of the painting). If half the painting is 100 cm^2, the full painting is 200 cm^2. Divide by the given height of 10 cm to immediately get a length of 20 cm. Common Pitfall: A potential pitfall is solving for the total area (200 cm^2) and mistakenly selecting an option matching that number if it were available, rather than taking the final step to find the specific dimension (length) requested. Final Answer: Therefore, the correct answer is 20.

Question 1

An angle is equal to two-thirds of its supplementary angle. What is the measure of this angle in degrees?

Accepted Answer

Introduction / Context: This is a simple algebraic problem about supplementary angles. Two angles are supplementary if their sum is 180 degrees. The question tells you that one angle is two-thirds of its supplementary angle, and you must find the measure of that angle. Such problems help reinforce the relationship between angles that form a straight line and your ability to set up equations from word statements. Given Data / Assumptions: Let the required angle be x degrees. Its supplementary angle is 180° - x. The given relationship is: x is two-thirds of its supplementary angle. We must find the value of x in degrees. Concept / Approach: Use the definition of supplementary angles and translate the phrase "two-thirds of its supplementary angle" into an equation. The basic idea is: x = (2 / 3) * (180° - x). Then solve this linear equation in x. This approach is very standard for problems involving fractions of angles and their complements or supplements. Step-by-Step Solution: Step 1: Let the required angle be x degrees. Step 2: Its supplementary angle is 180° - x. Step 3: The condition given is: x = (2 / 3) * (180° - x). Step 4: Multiply both sides by 3 to clear the fraction: 3x = 2 * (180° - x). Step 5: Expand the right side: 3x = 360° - 2x. Step 6: Add 2x to both sides: 5x = 360°. Step 7: Divide both sides by 5: x = 360° / 5 = 72°. Step 8: Therefore, the angle measures 72 degrees. Verification / Alternative check: Compute the supplementary angle: 180° - 72° = 108°. Now check the relationship: two-thirds of 108° is (2 / 3) * 108° = 72°, which matches x. This quick check confirms that our algebraic equation accurately captured the given condition and that the arithmetic steps were performed correctly. Why Other Options Are Wrong: If x = 108°, then its supplementary angle is 72°, and x would be greater than its supplement, not two-thirds of it. If x = 54°, then its supplement is 126°, and (2 / 3) * 126° = 84°, which is not 54°. For x = 36°, the supplement is 144°, and (2 / 3) * 144° = 96°, not 36°. An angle of 90° would have a supplement of 90°, and two-thirds of 90° is 60°, not 90°. Only 72° satisfies the given relationship exactly. Common Pitfalls: One common error is reversing the relationship and writing 180° - x = (2 / 3)x instead of x = (2 / 3)(180° - x), which leads to a different equation and the wrong answer. Another mistake is adding the angles to get 90° instead of 180°, confusing supplementary with complementary angles. Some students also mishandle the fraction when multiplying across the equation, leading to incorrect coefficients. Writing the equation carefully from the English description is the key step. Final Answer: The angle whose measure is two-thirds of its supplementary angle is 72°.

Question 2

The area of a trapezium is 21 square centimetres, the distance between its two parallel sides is 3 cm, and one of the parallel sides is 8.3 cm. What is the length of the other parallel side?

Accepted Answer

Introduction / Context: This problem deals with the area of a trapezium (trapezoid), which has two parallel sides called bases and a distance between them called height. You are given the area, the height (the distance between the parallel sides), and one base, and you must find the other base. Questions of this type are very typical in geometry-based aptitude tests and help reinforce the area formula for trapeziums. Given Data / Assumptions: Area of the trapezium A = 21 square centimetres. Distance between the parallel sides (height) h = 3 cm. One of the parallel sides, say a, is 8.3 cm. The other parallel side, say b, is unknown. We use the standard area formula for a trapezium. Concept / Approach: The area A of a trapezium with parallel sides a and b and height h is given by: A = (1 / 2) * (a + b) * h. We are given A, h, and a, and we must solve for b. Rearranging the formula allows us to isolate a + b and then find b by subtracting a. Step-by-Step Solution: Step 1: Write the area formula: A = (1 / 2) * (a + b) * h. Step 2: Substitute A = 21, h = 3, and a = 8.3: 21 = (1 / 2) * (8.3 + b) * 3. Step 3: Simplify the factor (1 / 2) * 3 = 3 / 2. Step 4: So 21 = (3 / 2) * (8.3 + b). Step 5: Multiply both sides by 2 / 3 to clear the fraction: 8.3 + b = (21 * 2 / 3). Step 6: Compute (21 * 2 / 3) = 14. Step 7: So 8.3 + b = 14, hence b = 14 - 8.3 = 5.7 cm. Step 8: Therefore, the other parallel side has length 5.7 cm. Verification / Alternative check: Check by plugging back into the area formula. Compute (a + b) = 8.3 + 5.7 = 14. Then area A = (1 / 2) * 14 * 3 = 7 * 3 = 21 square centimetres, which matches the given area exactly. This confirms that the calculated length b = 5.7 cm is correct. Why Other Options Are Wrong: If the other side were 12.7 cm, then a + b would be 21.0, giving area (1 / 2) * 21 * 3 = 31.5, which is too large. If b were 7.7 cm, then a + b = 16, giving area (1 / 2) * 16 * 3 = 24, again too large. For 10.7 cm, the sum is 19, giving area 28.5. A value of 9.7 cm also gives an incorrect area. Only b = 5.7 cm restores the area to 21 square centimetres. Common Pitfalls: A common mistake is to forget the factor of 1 / 2 in the area formula or to treat 8.3 as an approximate 8, which produces incorrect arithmetic. Another error is solving for a + b correctly but then mis-subtracting the known side, leading to a wrong length. Careful handling of decimal subtraction (14 - 8.3) is important to avoid minor but crucial numerical errors. Final Answer: The length of the other parallel side of the trapezium is 5.7 cm.

Question 3

Triangles ABC and PQR are similar, and the ratio of corresponding sides is AB : PQ = 2 : 3. AD is the median to side BC in triangle ABC and PS is the median to side QR in triangle PQR. What is the value of (BD/QS)^2?

Accepted Answer

Introduction / Context: This problem combines the concepts of similar triangles and medians. It tests whether you understand that when triangles are similar, not only their sides but also certain segments such as medians and altitudes are in the same ratio. You must then use that information to compute a squared ratio involving half segments of corresponding sides. Given Data / Assumptions: Triangle ABC is similar to triangle PQR. The ratio of corresponding sides is AB : PQ = 2 : 3. AD is a median to side BC in triangle ABC, so D is the midpoint of BC. PS is a median to side QR in triangle PQR, so S is the midpoint of QR. We must find the value of (BD / QS)^2. Triangles are assumed to be in a Euclidean plane and similarity is strict. Concept / Approach: When two triangles are similar, the ratio of any pair of corresponding linear measures is constant. This applies to sides, medians, and altitudes. Because AD and PS are medians, their lengths are proportional to the corresponding sides: AD / PS = AB / PQ. Also, a median divides the opposite side into two equal segments, so BD is half of BC and QS is half of QR. Since BC and QR are corresponding sides of similar triangles, BC / QR = AB / PQ. Therefore BD / QS has the same ratio as BC / QR. Step-by-Step Solution: Step 1: From similarity, AB / PQ = 2 / 3. Step 2: For similar triangles, the ratio of corresponding sides is constant, so BC / QR = 2 / 3 as well. Step 3: Because AD is a median, BD = BC / 2. Step 4: Because PS is a median, QS = QR / 2. Step 5: Compute BD / QS: BD / QS = (BC / 2) / (QR / 2) = BC / QR. Step 6: So BD / QS = BC / QR = 2 / 3. Step 7: Now compute (BD / QS)^2 = (2 / 3)^2 = 4 / 9. Step 8: Thus, the required value is 4 / 9. Verification / Alternative check: We can verify by thinking numerically. Suppose AB = 2 units and PQ = 3 units. Then take BC = 2k and QR = 3k for some positive k. Then BD = BC / 2 = k and QS = QR / 2 = 1.5k. Therefore BD / QS = k / (1.5k) = 2 / 3. Squaring gives (BD / QS)^2 = 4 / 9, which matches the analytic reasoning and is independent of the particular value of k. Why Other Options Are Wrong: The value 3 / 5 does not arise from the square of the 2 / 3 ratio. The option 2 / 3 is the ratio BD / QS, not its square. The fraction 4 / 7 is unrelated to the given side ratio 2 : 3. The option 5 / 9 also does not correspond to squaring 2 / 3 or any direct consequence of the given similarity ratio. Only 4 / 9 correctly matches (2 / 3)^2. Common Pitfalls: A frequent error is to confuse BD / QS with AB / PQ itself and forget to square the ratio at the end. Some students also mix up which segments are halves of which sides and mistakenly think BD and QS are in a different ratio. Others attempt to introduce unnecessary variables for the medians, which complicates the problem. Remembering that medians to corresponding sides scale in the same way as the sides themselves and that halving both sides cancels out in the ratio is crucial. Final Answer: The value of (BD / QS)^2 is 4/9.

Question 4

AB and AC are two tangents drawn from an external point A to a circle of radius 6 cm. If ∠BAC = 60°, what is the value, in centimetres, of √(AB² + AC²)?

Accepted Answer

Introduction / Context: This is a geometry problem involving tangents drawn from an external point to a circle. The tangents have equal length and form a given angle at the external point. You are asked to find the square root of the sum of the squares of their lengths, which essentially uses properties of isosceles triangles and the Pythagoras theorem, together with a key fact about tangent segments from a common external point. Given Data / Assumptions: A circle has centre O and radius r = 6 cm. AB and AC are tangents drawn from an external point A to the circle. ∠BAC = 60°. We must find √(AB² + AC²). Since AB and AC are tangents from the same point, AB = AC. Concept / Approach: First, note that AB = AC because tangents from a common external point are equal in length. Thus we need √(AB² + AC²) = √(2AB²) = AB * √2. To find AB, we use the geometry of tangents and chords. The angle between two tangents from an external point is supplementary to the angle subtended by the chord joining the points of contact at the centre. We then use this central angle to find the chord length BC, and finally relate BC to AB in triangle ABC, which is isosceles. Step-by-Step Solution: Step 1: Let B and C be the points of contact on the circle, and O be the centre. Step 2: Tangent segments AB and AC are equal, so AB = AC. Step 3: The angle between tangents at A is ∠BAC = 60°. Step 4: The angle at the centre subtended by chord BC is ∠BOC = 180° - ∠BAC = 180° - 60° = 120°. Step 5: In triangle BOC, OB = OC = radius = 6 cm. It is an isosceles triangle. Step 6: Use the cosine rule to find BC: BC² = OB² + OC² - 2 * OB * OC * cos(∠BOC). Step 7: Substitute values: BC² = 6² + 6² - 2 * 6 * 6 * cos(120°). Step 8: cos(120°) = -1/2, so BC² = 36 + 36 - 72 * (-1/2) = 72 + 36 = 108, hence BC = 6√3. Step 9: Now consider triangle ABC. It is isosceles with AB = AC and included angle ∠BAC = 60° and side BC opposite this angle. Step 10: Apply cosine rule in triangle ABC: BC² = AB² + AC² - 2 * AB * AC * cos(60°). Step 11: Let AB = AC = x, then BC² = x² + x² - 2x² * (1 / 2) = 2x² - x² = x². Step 12: Thus BC = x, so x = BC = 6√3 and therefore AB = AC = 6√3. Step 13: Compute √(AB² + AC²) = √((6√3)² + (6√3)²) = √(108 + 108) = √216 = 6√6. Verification / Alternative check: As a check, we can compute AB² + AC² directly. With AB = AC = 6√3, we have AB² = AC² = 108, so the sum is 216. The square root of 216 can be simplified by factoring: 216 = 36 * 6, so √216 = √36 * √6 = 6√6. This matches our final result. Also, the relationship BC = AB in the isosceles triangle with vertex angle 60° is consistent with the law of cosines, further confirming the internal geometry. Why Other Options Are Wrong: The value 4√6 would correspond to a smaller tangent length and would not match the chord length we found. The values 9√3 and 8√3 do not simplify to √(AB² + AC²) when AB = AC = 6√3. The numeric option 12 is not compatible with the computed sum of squares. Only 6√6 is consistent with a careful application of geometry and the Pythagoras theorem. Common Pitfalls: Students often forget that angle between tangents at an external point is supplementary to the angle at the centre, leading to an incorrect central angle. Others confuse the lengths of the tangent segments and the radius, or misapply the cosine rule. Some also try to compute AB directly using right triangles with the centre without correctly identifying the angles. A clear diagram and consistent stepwise application of known formulas avoid these errors. Final Answer: Therefore, the value of √(AB² + AC²) is 6√6.

Question 5

Two circles have radii 3 cm and 4 cm respectively, and the distance between their centres is 10 cm. What is the ratio of the length of a direct common tangent to the length of a transverse (internal) common tangent?

Accepted Answer

Introduction / Context: This question involves geometric properties of tangents drawn to two circles. You are asked to compare the length of a direct common tangent with the length of a transverse (internal) common tangent. The problem checks whether you know the standard formulas relating the distance between centres and the radii of the circles to the lengths of these tangents. Given Data / Assumptions: Radius of the first circle, r1 = 3 cm. Radius of the second circle, r2 = 4 cm. Distance between the centres of the circles, d = 10 cm. We must find the ratio (length of direct common tangent) : (length of transverse common tangent). Both circles are separate and the tangents considered are the usual external and internal common tangents. Concept / Approach: For two circles with radii r1 and r2 and centre distance d, there are formulas for the lengths of the common tangents: Length of a direct (external) common tangent, L_d = √(d^2 - (r1 - r2)^2). Length of a transverse (internal) common tangent, L_t = √(d^2 - (r1 + r2)^2). We simply substitute the given values into these formulas and then form the ratio L_d : L_t, simplifying if possible. Step-by-Step Solution: Step 1: Identify given values: d = 10, r1 = 3, r2 = 4. Step 2: Compute r1 - r2 = 3 - 4 = -1; the square is (r1 - r2)^2 = (-1)^2 = 1. Step 3: Compute r1 + r2 = 3 + 4 = 7; the square is (r1 + r2)^2 = 7^2 = 49. Step 4: Find L_d: L_d = √(d^2 - (r1 - r2)^2) = √(10^2 - 1) = √(100 - 1) = √99. Step 5: Find L_t: L_t = √(d^2 - (r1 + r2)^2) = √(10^2 - 49) = √(100 - 49) = √51. Step 6: Ratio L_d : L_t = √99 : √51. Step 7: Simplify the ratio by factoring: √99 = √(9 * 11) = 3√11 and √51 = √(3 * 17) = √3 * √17. Step 8: Write both in terms of √3: √99 = √3 * √33, and √51 = √3 * √17. Step 9: Cancel the common factor √3, giving ratio = √33 : √17. Verification / Alternative check: You can also keep the ratio as √99 : √51 and compare it with the options. Noticing that 99 and 51 share a factor 3, divide both under the radical by 3 to get √(99 / 3) : √(51 / 3) = √33 : √17. This directly matches option √33 : √17, confirming that the algebraic simplification is correct without needing to evaluate any decimal approximations. Why Other Options Are Wrong: The option √51 : √68 would require the internal tangent to have length √68, which does not match the given formula and data. The ratio √66 : √51 has the two numbers reversed and does not follow from the calculations. The ratio √28 : √17 does not correspond to any combination of d^2, (r1 - r2)^2 or (r1 + r2)^2 for the given values. The option 1 : 1 would mean the direct and transverse tangents have equal length, which only occurs in special symmetric cases not satisfied here. Only √33 : √17 matches the correct ratio. Common Pitfalls: A common mistake is to swap the formulas and use r1 + r2 where r1 - r2 is needed, or vice versa. Another issue is forgetting to square the differences or sums before subtracting from d^2. Some students also simplify the radicals incorrectly, losing common factors and getting a ratio that does not match any option. Careful substitution and stepwise simplification of the radicals is essential for accuracy. Final Answer: The ratio of the length of the direct common tangent to the length of the transverse common tangent is √33 : √17.

Question 6

Triangle ABC has side lengths AB = 5 cm, AC = √41 cm and BC = 8 cm. AD is drawn perpendicular to BC from vertex A. What is the area, in square centimetres, of triangle ABD?

Accepted Answer

Introduction / Context: This question combines coordinate geometry thinking with classical triangle area formulae. You are given all three side lengths of a triangle and an altitude from one vertex to the opposite side. You must find the area of just one of the smaller triangles formed by this altitude. The problem tests understanding of how altitudes split the base and how area is computed using base and height. Given Data / Assumptions: Triangle ABC has sides AB = 5 cm, AC = √41 cm, and BC = 8 cm. AD is the altitude from A to side BC, so AD is perpendicular to BC. D lies on BC, and we consider triangle ABD. We must find the area of triangle ABD. The triangle is non-degenerate and lies in a Euclidean plane. Concept / Approach: We can place the triangle on a coordinate system to find the height AD easily. Let BC be on the x-axis with B at (0, 0) and C at (8, 0). Then the coordinates of A = (x, y) must satisfy the distance conditions AB = 5 and AC = √41. Solving these gives the height y, which equals the length of AD. Once we know the height, the area of triangle ABD is given by one-half the product of base BD and the height AD. Step-by-Step Solution: Step 1: Place B at (0, 0) and C at (8, 0). Step 2: Let A have coordinates (x, y). Step 3: From AB = 5, we get: x² + y² = 5² = 25. Step 4: From AC = √41, we get: (x - 8)² + y² = 41. Step 5: Subtract the first equation from the second: (x - 8)² - x² = 41 - 25 = 16. Step 6: Expand (x - 8)² = x² - 16x + 64, so x² - 16x + 64 - x² = 16. Step 7: Simplify to get -16x + 64 = 16 ⇒ -16x = -48 ⇒ x = 3. Step 8: Substitute x = 3 into x² + y² = 25: 9 + y² = 25 ⇒ y² = 16 ⇒ y = ±4. Step 9: The altitude length AD is the absolute value of the y-coordinate, so AD = 4 cm. Step 10: D lies on BC at the vertical projection of A, so D has coordinates (3, 0). This means BD = 3 cm and DC = 5 cm. Step 11: Area of triangle ABD is (1 / 2) * base * height = (1 / 2) * BD * AD = (1 / 2) * 3 * 4 = 6 square centimetres. Verification / Alternative check: We can verify by computing the total area of triangle ABC using base BC = 8 cm and height AD = 4 cm. The total area is (1 / 2) * 8 * 4 = 16 square centimetres. The segment BD is 3 cm, so area of ABD is (3 / 8) times the total area, because it uses base 3 instead of 8 with the same height. (3 / 8) * 16 = 6 square centimetres, which matches the value we found earlier. This consistency confirms our answer. Why Other Options Are Wrong: An area of 12 or 20 square centimetres is too large considering that the entire triangle only has area 16 square centimetres. An area of 10 square centimetres would again exceed the total area proportionally when compared with the base segment BD = 3 cm on a base of length 8 cm. The option 8 square centimetres would represent half the total area, which would require BD = 4 cm, not the 3 cm derived from the coordinate solution. Only 6 square centimetres fits all the geometric relationships exactly. Common Pitfalls: Students sometimes incorrectly assume that BD is half of BC or make guesses about its length without using the distance conditions. Another error is to misapply the distance formula or to forget that both equations for AB and AC must be satisfied simultaneously. Some also compute the total area correctly but then divide it into two smaller triangle areas in the wrong ratio. A systematic coordinate approach avoids guesswork and leads to the correct values. Final Answer: The area of triangle ABD is 6 square centimetres.

Question 7

A rectangle has length 8 cm and breadth 6 cm. From each of its four vertices, straight cuts are made so that the remaining figure is a regular octagon (all sides equal). What is the length, in centimetres, of each side of this regular octagon?

Accepted Answer

Introduction / Context: In this geometry question, we start with a rectangle whose length is 8 cm and breadth is 6 cm. Each vertex is cut off in such a way that the remaining figure becomes a regular octagon, meaning all eight sides are equal in length. The main challenge is to express the side length of this octagon in terms of the dimensions of the original rectangle and the amounts cut off at the corners. Given Data / Assumptions: Original rectangle: length = 8 cm, breadth = 6 cm. Four corners are cut to form a regular octagon. Let the amount cut off along the longer side at each corner be x cm. Let the amount cut off along the shorter side at each corner be y cm. The resulting octagon has all sides equal in length. Concept / Approach: The new figure has 8 equal sides. These sides come from: The remaining horizontal portions of the length: 8 - 2x. The remaining vertical portions of the breadth: 6 - 2y. The slant (diagonal) edges formed by cutting off the corners: each diagonal has length √(x^2 + y^2). For a regular octagon, all these three kinds of edges must have the same length. This gives us equations relating x, y and the octagon side length a, and solving these equations yields the required side length. Step-by-Step Solution: Let the side length of the octagon be a. Horizontal side length: a = 8 - 2x. Vertical side length: a = 6 - 2y. Diagonal side length: a = √(x^2 + y^2). From a = 8 - 2x and a = 6 - 2y, we get 8 - 2x = 6 - 2y. So 2 = 2x - 2y → x - y = 1 → x = y + 1. Now a = 6 - 2y. Also a^2 = x^2 + y^2. Substitute x = y + 1: a^2 = (y + 1)^2 + y^2 = 2y^2 + 2y + 1. But a = 6 - 2y, so (6 - 2y)^2 = 2y^2 + 2y + 1. Compute left side: (6 - 2y)^2 = 36 - 24y + 4y^2. Set equal: 36 - 24y + 4y^2 = 2y^2 + 2y + 1. Rearrange: 4y^2 - 2y^2 - 24y - 2y + 36 - 1 = 0 → 2y^2 - 26y + 35 = 0. Divide by 1: 2y^2 - 26y + 35 = 0. Solve quadratic: y = [26 ± √(26^2 - 4 * 2 * 35)] / (2 * 2). Compute discriminant: 26^2 = 676, 4 * 2 * 35 = 280, so √(676 - 280) = √396 = √(36 * 11) = 6√11. Thus y = (26 ± 6√11) / 4 = (13 ± 3√11) / 2. We need y to be less than 3 (so that 6 - 2y > 0), so we choose y = (13 - 3√11) / 2. Now a = 6 - 2y = 6 - (13 - 3√11) = 3√11 - 7. Verification / Alternative check: Approximate √11 ≈ 3.316, so 3√11 ≈ 9.948. Then a ≈ 9.948 - 7 = 2.948 cm, which is a reasonable side length (less than both 6 and 8). Also y ≈ (13 - 9.948) / 2 ≈ 1.526 cm, so the trims are small but positive, which is geometrically valid. Why Other Options Are Wrong: Option B (5√13 – 8) gives a much larger side length that would not fit inside the original 8 cm by 6 cm rectangle once corners are cut. Option C (4√7 – 11) and Option D (6√11 – 9) either give negative or unrealistic dimensions when checked numerically. Option E (2√10 – 5) does not satisfy the algebraic relationship derived from equating horizontal, vertical and diagonal sides. Common Pitfalls: A common mistake is to assume the cuts at each corner are squares (x = y), which only works if the original figure is a square, not a rectangle. Another error is to assume that 8 - 2x automatically equals 6 - 2x, which is impossible and leads to contradiction. Students may also forget to check which root of the quadratic is geometrically valid (positive but small enough so that the remaining sides stay positive). Final Answer: Thus, the length of each side of the regular octagon formed from the 8 cm by 6 cm rectangle is 3√11 – 7 centimetres.

Question 8

The area of a rectangular field is 2100 sq. metres. If the field is 60 metres long, what is its perimeter?  (Bank Recruitment, 2007)

Accepted Answer

Concept & Formula: The perimeter requires both length and breadth. First, deduce the missing dimension (breadth) using the area formula, then apply the perimeter formula. Area = L × B Perimeter = 2(L + B) Step-by-Step Solution: • Given: Area = 2100 sq. m, Length (L) = 60 m. • Step 1: Find the breadth (B). 2100 = 60 × B B = (2100/60) = 35 m • Step 2: Calculate the perimeter. Perimeter = 2 × (60 + 35) Perimeter = 2 × (95) Perimeter = 190 m Exam Strategy & Shortcut: When you calculate the perimeter as 190 m and scan the options, you'll notice 190 m is missing. Do not second-guess your calculation if you are confident; simply select "None of these". Common Pitfall: A frequent error is calculating the breadth (35 m) and inadvertently searching for that number in the options, or forgetting to multiply the sum of length and breadth by 2 for the perimeter. Final Answer: Therefore, the correct answer is None of these.

Question 9

A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?

Accepted Answer

Concept & Formula: The total length of the fence is the perimeter of the rectangle. For a closed continuous shape (like a fenced rectangle), the number of poles equals the total perimeter divided by the distance between consecutive poles. Perimeter = 2(L + B) Number of poles = \frac{Total Perimeter}{Spacing} Step-by-Step Solution: • Given: Length = 90 m, Breadth = 50 m, Spacing = 5 m. • Step 1: Calculate the perimeter of the plot. Perimeter = 2 × (90 + 50) Perimeter = 2 × 140 = 280 m • Step 2: Calculate the number of poles. Poles = (280/5) Poles = 56 Exam Strategy & Shortcut: To divide any number by 5 quickly, double the number and divide by 10. Here, 280 × 2 = 560, and 560 / 10 = 56. Common Pitfall: A common mistake is treating this like a straight line problem and adding 1 to the final count (i.e., calculating 56 + 1 = 57). Remember that in a closed loop, the starting pole and ending pole are the exact same pole, so you do not add 1. Final Answer: Therefore, the correct answer is 56.

Question 10

The area of a rectangle is $252	ext{ cm}^2$ and its length and breadth are in the ratio of 9 : 7 respectively. What is its perimeter?  (Bank. P.O., 2009)

Accepted Answer

Concept & Logic: When dimensions are given as a ratio, introduce a common multiplier (like x) to express the exact length and breadth. Set up an equation using the area to solve for x. Area = Length × Breadth Step-by-Step Solution: • Given: Area = 252 cm^2, Ratio of Length to Breadth = 9:7. • Step 1: Express dimensions using variable x. Let Length (L) = 9x Let Breadth (B) = 7x • Step 2: Set up the area equation and solve for x. 9x × 7x = 252 63x^2 = 252 x^2 = (252/63) x^2 = 4 x = 2 • Step 3: Find actual dimensions. L = 9 × 2 = 18 cm B = 7 × 2 = 14 cm • Step 4: Calculate the perimeter. Perimeter = 2(L + B) Perimeter = 2(18 + 14) = 2(32) = 64 cm Exam Strategy & Shortcut: Notice the ratio sum is 9 + 7 = 16. The semi-perimeter is 16x, so the full perimeter is 32x. This tells you immediately that the perimeter must be a multiple of 32. Looking at the options (64, 68, 96, 128), 64, 96, and 128 are multiples of 32. Testing x=2 gives a perimeter of 64 and an area of (9×2)×(7×2) = 18 × 14 = 252, which matches perfectly. Common Pitfall: A frequent error is forgetting that multiplying 9x and 7x yields an x^2 term, not just x. Solving 63x = 252 \implies x = 4 will lead to completely incorrect dimensions. Final Answer: Therefore, the correct answer is 64 cm.

Question 11

A carpenter is designing a table. The table will be in the form of a rectangle whose length is 4 feet more than its width. How long should the table be if the carpenter wants the area of the table to be 45 sq ft? (J.M.E.T., 2010)

Accepted Answer

Concept & Algebra: Translate the given word problem into algebraic equations. We need to find the dimensions using the area of a rectangle. Area = length × width Step-by-Step Solution: • Let the width of the table be w feet. • The length is 4 feet more than the width, so length l = w + 4. • The area is given as 45 sq ft, which yields the equation: w(w + 4) = 45 • Expanding this gives a quadratic equation: w^2 + 4w - 45 = 0 • Factoring the quadratic equation: (w + 9)(w - 5) = 0 • Since a dimension cannot be negative, w = 5 feet. • The question asks for the length of the table: l = w + 4 = 5 + 4 = 9 feet. Exam Strategy & Shortcut: Instead of solving the quadratic equation, use the options! The question asks for the length. Test Option (b): If length = 9, then width = 9 - 4 = 5. Area = 9 × 5 = 45. This matches the condition perfectly. Common Pitfall: A common mistake is solving for the width (w = 5) and assuming that is the final answer. Always re-read the question to check what specific value is being asked for (in this case, "How long", meaning length). Final Answer: Therefore, the correct answer is 9 ft.

Question 12

The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is

Accepted Answer

Concept & Distance: Cycling along the boundary of the park means covering the perimeter. First, calculate the distance covered using speed and time, which gives the perimeter. Distance = Speed × Time Step-by-Step Solution: • Convert the time into hours to match the speed units: Time = (8/60) hours = (2/15) hours • Calculate the distance (perimeter) covered in km: Perimeter = 12 km/hr × (2/15) hr = (24/15) km = 1.6 km • Convert the perimeter to meters (since the final area needs to be in sq. m): 1.6 km = 1600 m • Use the ratio 3:2 to find the dimensions. Let length be 3x and breadth be 2x. 2(3x + 2x) = 1600 2(5x) = 1600 \implies 10x = 1600 \implies x = 160 • Calculate the actual sides: • Length = 3 × 160 = 480 m • Breadth = 2 × 160 = 320 m • Calculate the area: Area = 480 × 320 = 153600 sq. m Exam Strategy & Shortcut: To quickly convert 12 km/hr to m/s, multiply by 5/18, yielding 10/3 m/s. The time is 8 minutes = 480 s. Perimeter = (10/3) × 480 = 1600 m. The ratio is 3:2, so perimeter is 10 parts. 1 part = 160. Area = 3 parts × 2 parts = 6 parts^2 = 6 × (160)^2 = 6 × 25600 = 153600. Common Pitfall: Failing to convert the speed from km/hr to m/min or time from minutes to hours, leading to a massive miscalculation in the perimeter size. Final Answer: Therefore, the correct answer is 153600 sq. m.

Question 13

The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field?

Accepted Answer

Concept & Algebra: Express the relationship between the length and breadth algebraically, then substitute it into the area formula to solve for the missing dimension. Area = l × b Step-by-Step Solution: • Let the breadth of the rectangular field be b. • The length is 15% more than the breadth, meaning: l = b + 0.15b = 1.15b • We are given the area: l × b = 460 • Substitute l: (1.15b) × b = 460 1.15b^2 = 460 • Solve for b^2: b^2 = (460/1.15) = (46000/115) = 400 • Solve for b: b = \sqrt{400} = 20 metres • The breadth is 20 metres. This value does not match options a, b, or c. Exam Strategy & Shortcut: 15% is (3/20). So if breadth is 20x, length is 23x. Area = 20x × 23x = 460x^2. Setting this to 460 gives x^2 = 1 \implies x = 1. The breadth is 20 × 1 = 20. This completely bypasses decimal division. Common Pitfall: Incorrectly setting up the "15% more" condition as just 0.15b instead of 1.15b. Alternatively, finding the value of 20 and hastily selecting "Cannot be determined" when they don't see it in the first few options, instead of "None of these". Final Answer: Therefore, the correct answer is None of these.

Question 14

A farmer wishes to start a 100 sq. m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is:

Accepted Answer

Concept & Optimization: The problem involves finding dimensions that satisfy two constraints simultaneously: a fixed area and a fixed length for three sides. Area = l × b = 100 Fencing = l + 2b = 30 (or 2l + b = 30 ) Step-by-Step Solution: • Let the length be l and the breadth be b. The fencing covers three sides. • This means the fencing length is either l + 2b = 30 (if the wall is on the length side) or 2l + b = 30 (if the wall is on the breadth side). • Test Option (a): 15 × 6.67. Area \approx 100. Fencing could be 15 + 2(6.67) = 28.34 or 2(15) + 6.67 = 36.67. Neither is 30. • Test Option (b): 20 × 5. Area = 20 × 5 = 100. Fencing could be 20 + 2(5) = 30. This perfectly matches the 30 m of wire available! • Test Option (c): 30 × 3.33. Area \approx 100. Fencing could be 30 + 2(3.33) = 36.66. Does not match. • Test Option (d): 40 × 2.5. Area = 100. Fencing could be 40 + 2(2.5) = 45 or 2(40) + 2.5 = 82.5. Does not match. Exam Strategy & Shortcut: For dual-constraint problems with options provided, back-solving is infinitely faster than setting up a quadratic equation (l(15 - 0.5l) = 100). Simply plug the dimensions into L + 2B to see which one equals 30. Common Pitfall: Attempting to solve this by assuming a square or a specific ratio first, rather than letting the two given conditions (Area = 100, Fencing = 30) guide the selection. Final Answer: Therefore, the correct answer is 20 m × 5 m.

Question 15

A rectangle of certain dimensions is chopped off from one corner of a larger rectangle as shown. $AB = 8$ cm and $BC = 4$ cm. The perimeter of the figure $ABCPQRA$ (in cm) is  <img    src="/api/images/rectangle-chopped-off-corner-larger-rectangle-ab…

Accepted Answer

Concept & Geometric Transformation: When a rectangular corner is removed from a larger rectangle, the perimeter of the resulting shape remains exactly the same as the original un-chopped rectangle. Step-by-Step Solution: • The total perimeter of the figure is the sum of its outer boundary: AB + BC + CP + PQ + QR + RA. • Notice the horizontal segments: The sum of the chopped horizontal segment CP and the inner horizontal segment QR is exactly equal to the full bottom length AB. CP + QR = AB = 8 cm • Notice the vertical segments: The sum of the inner vertical segment PQ and the chopped vertical segment RA is exactly equal to the full right-side length BC. PQ + RA = BC = 4 cm • Now, substitute these sums into the perimeter equation: Perimeter = AB + BC + (CP + QR) + (PQ + RA) Perimeter = 8 + 4 + 8 + 4 Perimeter = 24 cm Exam Strategy & Shortcut: Visualize "pushing" the inner line segments PQ and QR outward to fill the gap. PQ slides right to complete the right edge, and QR slides up to complete the top edge. The shape becomes a perfect 8 × 4 rectangle. Perimeter = 2(8 + 4) = 24. Common Pitfall: Assuming that removing area must also reduce the perimeter, leading test-takers to guess lower numbers without analyzing the lengths of the new edges created by the cut. Final Answer: Therefore, the correct answer is 24.

Question 16

A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?

Accepted Answer

Concept & Linear Equations: Translate the word problem into a system of two linear equations. You need to express the total area, the average area, and the difference between the two parts. Step-by-Step Solution: • Let the areas of the two parts be x and y, where x is the larger part. • The total area is 700 hectares: x + y = 700 • The average of the two areas is: Average = (x + y/2) = (700/2) = 350 • The problem states the difference (x - y) is one-fifth of the average: x - y = (1/5) × 350 x - y = 70 • Now we have a simple system of equations: 1) x + y = 700 2) x - y = 70 • We want to find the smaller area (y). Subtract equation (2) from equation (1): (x + y) - (x - y) = 700 - 70 2y = 630 y = 315 Exam Strategy & Shortcut: You know the total is 700, so x + y = 700. The difference x - y = (1/5) × ((700/2)) = 70. To find the smaller number when given a sum and a difference, subtract the difference from the sum and divide by 2: (700 - 70/2) = (630/2) = 315. Common Pitfall: A common error is taking one-fifth of the *total* area (700) instead of the *average* area (350), which would incorrectly make the difference 140, leading to a wrong smaller area of 280 (which is a trap option). Final Answer: Therefore, the correct answer is 315.

Question 17

An artist has completed one-fourth of a rectangular oil painting. When he will paint another 100 square centimetres of the painting, he would complete three-quarters of the painting. If the height of the oil painting is 10 cm, determine the length (…

Accepted Answer

Concept & Fraction Logic: This problem is solved by translating word fractions into a simple algebraic equation to find the total area, and then using the standard area formula for a rectangle to find the missing dimension. Area = Length × Height (Breadth) Step-by-Step Solution: 1. Let the total area of the rectangular oil painting be A square centimeters. 2. The artist has already completed (1/4) of the painting, which is (A/4). 3. Painting another 100 cm^2 brings the total completed portion to (3/4) of the painting. We can set up the equation: (A/4) + 100 = (3A/4) 4. Isolate the A terms on one side to solve for the total area: 100 = (3A/4) - (A/4) 100 = (2A/4) 100 = (A/2) 5. Multiply by 2: A = 200 cm^2 6. We are given that the height (which acts as the breadth of the rectangle) is 10 cm. Let the length be L. L × 10 = 200 L = (200/10) = 20 cm Exam Strategy & Shortcut: You can solve this mentally using parts. The difference between completing (3/4) and (1/4) of the painting is (2/4), which simplifies to (1/2) (half of the painting). If half the painting is 100 cm^2, the full painting is 200 cm^2. Divide by the given height of 10 cm to immediately get a length of 20 cm. Common Pitfall: A potential pitfall is solving for the total area (200 cm^2) and mistakenly selecting an option matching that number if it were available, rather than taking the final step to find the specific dimension (length) requested. Final Answer: Therefore, the correct answer is 20.

Question 18

A courtyard 25 m long and 16 m broad is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is

Accepted Answer

Concept & Surface Area Paving: To find the total number of items (like bricks or tiles) needed to cover a surface, you divide the total area of the surface by the area of a single item. It is critical that both areas are in the exact same unit of measurement. Number of Bricks = \frac{Total Area of Courtyard}{Area of One Brick} Step-by-Step Solution: 1. Find the dimensions of the courtyard in centimeters: Since the brick dimensions are in cm, it is easiest to convert the courtyard dimensions from meters to centimeters (1 m = 100 cm). Length = 25 m = 25 × 100 = 2500 cm Breadth = 16 m = 16 × 100 = 1600 cm 2. Calculate the area of the courtyard: Area of Courtyard = 2500 × 1600 = 4,000,000 cm^2 3. Calculate the area of a single brick: Area of Brick = 20 cm × 10 cm = 200 cm^2 4. Calculate the total number of bricks required: Number of Bricks = (4,000,000/200) Number of Bricks = (40,000/2) = 20000 Exam Strategy & Shortcut: To avoid large numbers and multiple zeros which can lead to calculation errors, keep the factors unmultiplied until the final step. Number of bricks = (2500 × 1600/20 × 10). Cancel out the zeros and simplify: (250 × 160/2) = 250 × 80 = 20000. Common Pitfall: The most standard mistake is failing to convert the units. Dividing the courtyard area in m^2 (400) by the brick area in cm^2 (200) will give an absurdly small answer of 2 bricks. Always homogenize units before calculation. Final Answer: Therefore, the correct answer is 20000.

Question 19

The diagonal of a rectangle is 10 cms and is twice the length of one of the sides. What is the area of the rectangle in sq. cm? (R.R.B. 2006)

Accepted Answer

Concept & Right Triangle Properties: This problem requires using a known diagonal and its proportional relationship to one side to find the second side via the Pythagorean theorem. Area = Length × Breadth Step-by-Step Solution: 1. We are given the diagonal (d) = 10 cm. 2. The diagonal is twice the length of one of the sides. Let this known side be w (width). d = 2w 10 = 2w \implies w = 5 cm 3. Let the other side (length) be l. Use the Pythagorean theorem to find l: d^2 = l^2 + w^2 10^2 = l^2 + 5^2 100 = l^2 + 25 4. Isolate l^2: l^2 = 100 - 25 = 75 5. Take the square root to find l: l = \sqrt{75} = \sqrt{25 × 3} = 5\sqrt{3} cm 6. Calculate the area of the rectangle: Area = l × w Area = 5\sqrt{3} × 5 = 25\sqrt{3} sq. cm Exam Strategy & Shortcut: If a right triangle has a hypotenuse that is exactly twice the length of one leg, it is a special 30^\circ-60^\circ-90^\circ triangle. The sides of such a triangle are in the ratio 1 : \sqrt{3} : 2. Since the hypotenuse is 10 (representing the 2 part), the shorter side is 5 (the 1 part), and the longer side is 5\sqrt{3} (the \sqrt{3} part). Area = 5 × 5\sqrt{3} = 25\sqrt{3}. Common Pitfall: A common mistake is stopping after finding the unknown side (5\sqrt{3}) and assuming this is the final answer, neglecting the final step of multiplying the sides together to find the area. Final Answer: Therefore, the correct answer is 25\sqrt{3}.

Question 20

The diagonal of a rectangular field is 15 metres and the difference between its length and width is 3 metres. The area of the rectangular field is (M.A.T., 2007)

Accepted Answer

Concept & Quadratic Geometry: This question combines the Pythagorean theorem with a linear constraint between the two sides, leading to a quadratic equation. Let the sides be expressed in terms of a single variable to solve for their exact values. Step-by-Step Solution: 1. Let the width of the rectangular field be w. 2. The difference between the length (l) and width is 3 m, which means l = w + 3. 3. We are given the diagonal (d) = 15 m. Apply the Pythagorean theorem: l^2 + w^2 = d^2 (w + 3)^2 + w^2 = 15^2 4. Expand the squared binomial: (w^2 + 6w + 9) + w^2 = 225 2w^2 + 6w + 9 = 225 5. Form a standard quadratic equation by setting it to zero: 2w^2 + 6w + 9 - 225 = 0 2w^2 + 6w - 216 = 0 6. Simplify by dividing the entire equation by 2: w^2 + 3w - 108 = 0 7. Factor the quadratic equation. We need two numbers that multiply to -108 and add to 3. Those numbers are 12 and -9. (w + 12)(w - 9) = 0 Since width cannot be negative, w = 9 m. 8. Find the length: l = w + 3 = 9 + 3 = 12 m 9. Calculate the area of the rectangular field: Area = l × w = 12 × 9 = 108 m^2 Exam Strategy & Shortcut: Instead of solving quadratics, look at the diagonal 15. Recognize that 15 is a multiple of the famous 3-4-5 Pythagorean triple (multiplied by 3). The sides would be 3 × 3 = 9 and 4 × 3 = 12. Check the condition: is the difference between these sides equal to 3? Yes (12 - 9 = 3). The sides are definitively 12 and 9. Area = 12 × 9 = 108. Solved in 5 seconds. Common Pitfall: A recurring algebraic error is expanding (w + 3)^2 as w^2 + 9 instead of w^2 + 6w + 9. Missing the middle term completely changes the quadratic and leads to incorrect side lengths. Final Answer: Therefore, the correct answer is 108 m^2.

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