Question 1

Cylinder melted into a cone (same radius)
A cylinder of radius 4 cm and height 8 cm is melted and recast into a cone of the same radius. Find the height of the cone (in cm).

Accepted Answer

Introduction / Context: Melting and recasting conserves volume. With equal radii, we equate the cylinder’s volume to the cone’s volume and solve for the cone’s height directly. Given Data / Assumptions: Cylinder radius r = 4 cm, height H_cyl = 8 cm Cone radius r = 4 cm, height H_cone = ? Volumes: V_cyl = π r^2 H_cyl; V_cone = (1/3) π r^2 H_cone Concept / Approach: Set π r^2 H_cyl = (1/3) π r^2 H_cone. Since r is the same, π r^2 cancels, yielding a simple proportionality between heights. Step-by-Step Solution: π * 4^2 * 8 = (1/3) * π * 4^2 * H_cone 128π = (16π/3) * H_cone ⇒ H_cone = 128 * 3 / 16 = 24 cm Verification / Alternative check: Compute volumes numerically: cylinder = 128π; cone with H = 24 gives (1/3)*16π*24 = 128π—exactly equal. Why Other Options Are Wrong: 12, 36, and 48 correspond to one-half, 1.5x, or 2x the correct value; only 24 preserves equal volume with same radius. Common Pitfalls: Forgetting the 1/3 factor in cone volume or not cancelling common factors leads to algebra mistakes. Final Answer: 24 cm

Question 2

Power of a point: tangent–secant relation
In a circle, chord AB is extended to meet the tangent DE at an external point D. If AB = 12 cm and the tangent length DE = 8 cm, find the length of AD (in cm).

Accepted Answer

Introduction / Context: This uses the power of a point theorem: from an external point, the square of the tangent length equals the product of the external and total lengths of a secant through that point. Adjusting the query to AD (the external segment) aligns with standard statements and the given options. Given Data / Assumptions: Chord AB = 12 cm; extended line meets tangent DE at D. Tangent length from D: DE = 8 cm. Let AD be the external segment; DB = AD + AB. Concept / Approach: Power of a point: (tangent)^2 = (external secant) * (entire secant). Hence, DE^2 = AD * (AD + AB). Solve for AD using a quadratic relation. Step-by-Step Solution: DE^2 = AD * (AD + AB) ⇒ 8^2 = AD * (AD + 12) 64 = AD^2 + 12AD ⇒ AD^2 + 12AD − 64 = 0 Discriminant = 12^2 + 4 * 64 = 400 ⇒ AD = (−12 + 20)/2 = 4 cm (positive root) Verification / Alternative check: If AD = 4, then DB = 4 + 12 = 16 and 4 * 16 = 64 = DE^2, confirming the theorem. Why Other Options Are Wrong: 5, 6, and 7 fail the equation AD(AD + 12) = 64. Only 4 satisfies it. Common Pitfalls: Assuming chord length enters directly without forming the total secant (external + chord) or misapplying the tangent–secant relation leads to incorrect values. Final Answer: 4 cm

Question 3

Arc length with central angle 30° and radius 21 cm
Find the length of the arc (in cm) subtended by a central angle of 30° in a circle of radius 21 cm.

Accepted Answer

Introduction / Context: Arc length for central angle θ (in degrees) uses s = (θ/360) * 2πr. Substituting θ = 30° simplifies to one-twelfth of the circumference, which is convenient for mental checks too. Given Data / Assumptions: θ = 30°, r = 21 cm Use π = 22/7 (or 3.14) for evaluation. Concept / Approach: Compute s = (30/360) * 2πr = (1/12) * 2πr = (1/6) * πr. With r = 21, this becomes (1/6)*π*21 = 3.5π, which is close to 11 when π ≈ 22/7. Step-by-Step Solution: s = (30/360) * 2 * π * 21 = (1/12) * 42π = 3.5π With π = 22/7 ⇒ s = 3.5 * (22/7) = 11 cm Verification / Alternative check: Using π ≈ 3.14 ⇒ 3.5 * 3.14 ≈ 10.99 cm, which rounds to 11 cm, matching the choice. Why Other Options Are Wrong: 16.5 and 22 assume larger angles; 28 is a radius multiple not tied to 30°. Only 11 cm fits exactly. Common Pitfalls: Mixing degrees with radians or forgetting to divide by 360 when θ is in degrees is common. Final Answer: 11 cm

Question 4

Area change with scaled sides of a rectangle
The length and breadth of a rectangle increase by 40% and 70%, respectively. By what percentage does the area increase?

Accepted Answer

Introduction / Context: Area scales multiplicatively with each dimension. Percent increases become factors (1 + r). Multiply the factors to get the new area factor, then convert back to a percent increase over the original. Given Data / Assumptions: Length factor = 1.40 (40% increase) Breadth factor = 1.70 (70% increase) Concept / Approach: New area factor = 1.40 * 1.70 = 2.38. Percentage increase = (2.38 − 1) * 100% = 138% over the original area. Step-by-Step Solution: Area(new) / Area(old) = 1.4 * 1.7 = 2.38 Increase% = (2.38 − 1) * 100% = 138% Verification / Alternative check: Pick a = 10 by 10 ⇒ area 100. New sides 14 and 17 ⇒ area 238 ⇒ increase 138%, confirming the factor method. Why Other Options Are Wrong: 118, 110, and 128 arise from additive or partial scaling errors rather than multiplicative scaling. Common Pitfalls: Adding 40% and 70% to get 110% is incorrect because area depends on both dimensions multiplicatively. Final Answer: 138

Question 5

Slope of a line parallel to the line through (4, −2) and (−3, 5)
What is the slope of any line parallel to the line joining (4, −2) and (−3, 5)?

Accepted Answer

Introduction / Context: Parallel lines share the same slope. Compute the slope of the given line from two points, then use that value directly for any parallel line. Given Data / Assumptions: Points: (x1, y1) = (4, −2), (x2, y2) = (−3, 5) Slope m = (y2 − y1) / (x2 − x1) Concept / Approach: Compute m once; all parallels have identical m. Take care with signs in numerator and denominator for accuracy. Step-by-Step Solution: m = (5 − (−2)) / (−3 − 4) = 7 / (−7) = −1 Verification / Alternative check: Reversing point order gives the same slope: (−2 − 5)/(4 − (−3)) = −7/7 = −1, confirming consistency. Why Other Options Are Wrong: 3/7 and −3/7 reflect partial differences; 1 is the negative inverse (perpendicular), not parallel. Common Pitfalls: Sign mistakes in the denominator are common; always subtract coordinates in the same order for x and y. Final Answer: −1

Question 6

Volume of a right circular cylinder
Find the volume (in cm³) of a right circular cylinder with diameter 21 cm and height 10 cm.

Accepted Answer

Introduction / Context: The volume of a cylinder is V = π r^2 h. With a given diameter, we first halve it to get the radius, square it, and multiply by height and π. Given Data / Assumptions: Diameter = 21 cm ⇒ radius r = 10.5 cm Height h = 10 cm Use π = 22/7 for an exact integer outcome. Concept / Approach: Compute r^2, then apply V = π r^2 h. Keeping π as 22/7 simplifies to an exact value when r is a half-integer compatible with 7. Step-by-Step Solution: r^2 = 10.5^2 = 110.25 V = (22/7) * 110.25 * 10 = (22/7) * 1102.5 = 3465 cm^3 Verification / Alternative check: Using π ≈ 3.14 yields ≈ 3462 cm³, which rounds to the exact 3465 obtained with 22/7—options list the exact value. Why Other Options Are Wrong: 3118, 3275, 3621 do not match π r^2 h with r = 10.5 and h = 10; only 3465 does. Common Pitfalls: Using diameter in place of radius (forgetting to halve) or squaring 21 instead of 10.5 leads to large errors. Final Answer: 3465

Question 7

Combinatorics of a pentagonal prism
A pentagonal prism has 15 edges. How many vertices does it have?

Accepted Answer

Introduction / Context: For a right or oblique prism based on an n-gon, counts are standard: vertices = 2n, edges = 3n, faces = n + 2. Recognizing this pattern avoids lengthy enumeration. Given Data / Assumptions: Pentagonal prism ⇒ n = 5 Edges = 3n = 15 (consistent) Concept / Approach: Use the prism formula for vertices: V = 2n. Substitute n = 5 to get the vertex count directly. These relations can also be derived from Euler’s formula V − E + F = 2 with known face counts. Step-by-Step Solution: V = 2n = 2 * 5 = 10 Verification / Alternative check: Faces F = n + 2 = 7. Check Euler: V − E + F = 10 − 15 + 7 = 2 (valid), confirming counts. Why Other Options Are Wrong: 12, 15, and 20 correspond to different polyhedra or misapplied formulas; only 10 satisfies the prism identities. Common Pitfalls: Confusing prisms with pyramids (where counts differ) or forgetting that there are two congruent n-gon bases leads to errors. Final Answer: 10

Question 8

y-intercept of a line
At what point does the line 4x − 3y = −6 cut the y-axis?

Accepted Answer

Introduction / Context: Intercepts are found by setting the other coordinate to zero. The y-intercept occurs where x = 0. Solve for y from the given linear equation to get the intercept point explicitly. Given Data / Assumptions: Equation: 4x − 3y = −6 At y-intercept, x = 0 Concept / Approach: Substitute x = 0 to get −3y = −6 ⇒ y = 2. The y-intercept point is (0, 2). Step-by-Step Solution: 4(0) − 3y = −6 ⇒ −3y = −6 ⇒ y = 2 Point = (0, 2) Verification / Alternative check: Rearrange to slope–intercept form: −3y = −4x − 6 ⇒ y = (4/3)x + 2. Intercept is 2, confirming (0, 2). Why Other Options Are Wrong: (0, 3/2) corresponds to y = 1.5, not satisfying the equation; (2, 0) and (3/2, 0) are x-intercepts candidate points, not y-intercepts here. Common Pitfalls: Confusing x- and y-intercepts or sign slips when isolating y are typical mistakes. Final Answer: (0,2)

Question 9

Area of a circle (radius = 10.5 cm)
Calculate the area (in cm²) of a circle with radius 10.5 cm.

Accepted Answer

Introduction / Context: The area of a circle is A = π r^2. With r = 10.5, using π = 22/7 provides an exact decimal outcome because 10.5 is 21/2 and simplifies with 7 conveniently. Given Data / Assumptions: r = 10.5 cm Use π = 22/7 for exactness. Concept / Approach: Compute r^2 and multiply by π. Keep units in cm². Check with π ≈ 3.14 for a rough confirmation. Step-by-Step Solution: r^2 = 10.5^2 = 110.25 A = (22/7) * 110.25 = 346.5 cm^2 Verification / Alternative check: Using π ≈ 3.14: 3.14 * 110.25 ≈ 346.185, close to 346.5; the 22/7 form gives the exact option value. Why Other Options Are Wrong: 693 and 315 are multiples wrong by factors of 2 or 0.9; 157.5 is far too small for r = 10.5 cm. Common Pitfalls: Using diameter instead of radius or forgetting to square the radius commonly produces errors. Final Answer: 346.5

Question 10

Area of a rhombus from its diagonals
Find the area (in cm²) of a rhombus whose diagonals are 25 cm and 20 cm.

Accepted Answer

Introduction / Context: Rhombus area equals half the product of its diagonals—one of the most direct area formulas for this quadrilateral. This avoids computing side or height. Given Data / Assumptions: d1 = 25 cm d2 = 20 cm Concept / Approach: Area = (d1 * d2) / 2. Substitute and compute exactly; no trigonometry is needed. Step-by-Step Solution: Area = (25 * 20)/2 = 500/2 = 250 cm^2 Verification / Alternative check: If needed, one can confirm by splitting the rhombus into four right triangles formed by the diagonals (which are perpendicular), summing their areas. Why Other Options Are Wrong: 500 doubles the correct area; 125 halves it; 200 is unrelated to (d1 * d2)/2 here. Common Pitfalls: Forgetting the 1/2 factor is the most common mistake with this formula. Final Answer: 250

Question 11

Effect on area when radius is reduced to 25%
If the radius of a circle becomes 25% of its original value, by what percentage does the area decrease?

Accepted Answer

Introduction / Context: Area of a circle scales with the square of its radius. If the radius is scaled by a factor k, the area scales by k^2. Here, reducing radius to 25% means k = 0.25, so the area greatly diminishes. Given Data / Assumptions: New radius r2 = 0.25 * r1 Area scales as A ∝ r^2 Concept / Approach: Compute the new-to-old area ratio: (r2/r1)^2 = (0.25)^2 = 0.0625. That means the new area is 6.25% of the original, so the decrease is 100% − 6.25% = 93.75%. Step-by-Step Solution: Area factor = (0.25)^2 = 0.0625 Decrease% = (1 − 0.0625) * 100% = 93.75% Verification / Alternative check: Take a concrete example: original radius = 4 ⇒ area = 16π. New radius = 1 ⇒ area = π. Decrease = 15π out of 16π ≈ 93.75%, confirming the proportion. Why Other Options Are Wrong: 25% and 50% ignore the square scaling; 43.75% is half of the correct decrease; only 93.75% matches k^2 scaling. Common Pitfalls: Applying linear rather than quadratic scaling to area leads to significant underestimation of the decrease. Final Answer: 93.75%

Question 12

Rectangle from perimeter and breadth:
The perimeter of a rectangle is 60 cm and its breadth is 14 cm. Find the area of the rectangle (in cm²).

Accepted Answer

Introduction / Context: This problem combines the rectangle perimeter formula with the area formula. Once one dimension (breadth) and the perimeter are known, the other dimension (length) is immediate and area follows as length * breadth. Given Data / Assumptions: Perimeter P = 60 cm. Breadth b = 14 cm. All measures are in centimetres. Concept / Approach: For a rectangle, P = 2(L + B). Hence L + B = P/2. Knowing B, compute L. Then Area = L * B. No special cases apply here because the numbers are integral and consistent. Step-by-Step Solution: L + B = 60/2 = 30.Given B = 14 ⇒ L = 30 − 14 = 16.Area = L * B = 16 * 14 = 224 cm². Verification / Alternative check: Check the perimeter using L = 16, B = 14: 2*(16 + 14) = 2*30 = 60 (matches). Therefore the dimensions are consistent, and the area 224 cm² is correct. Why Other Options Are Wrong: 112, 336, 448 arise from partial products or doubling mistakes (e.g., mixing length with perimeter or forgetting the factor 2 in perimeter). Common Pitfalls: Using P = L + B instead of P = 2(L + B); using breadth directly with perimeter for area without first finding the correct length. Final Answer: 224

Question 13

Square to diagonal:
The area of a square is 100 cm². Find the exact length of its diagonal (in cm).

Accepted Answer

Introduction / Context: Relating a square’s area to its side and diagonal is routine geometry. If the area is given, the side is the square root of the area, and the diagonal equals side * √2 by the Pythagorean relation in the square’s right triangle. Given Data / Assumptions: Area A = 100 cm². Let side = s; diagonal = d. Concept / Approach: Area of a square: A = s^2 ⇒ s = √A. Diagonal: d = s√2. Keep the exact radical form (no decimals) because the option set is exact with √2. Step-by-Step Solution: s = √(100) = 10 cm.d = s√2 = 10√2 cm. Verification / Alternative check: If s = 10, the square’s right triangle has legs 10 and 10, so d^2 = 10^2 + 10^2 = 200 ⇒ d = √200 = 10√2, confirming the result. Why Other Options Are Wrong: 10 cm is the side length, not the diagonal; 20 cm and 20√2 cm are too large relative to an area of 100 cm². Common Pitfalls: Confusing side with diagonal; rationalizing √200 incorrectly. Final Answer: 10√2 cm

Question 14

Circle – radius increased by 14%:
If the radius of a circle increases by 14%, by what percentage does its area increase?

Accepted Answer

Introduction / Context: Area of a circle is proportional to the square of its radius. Therefore, any percentage change in radius compounds (squares) when translated to area. A 14% radius increase needs squaring of the scale factor 1.14. Given Data / Assumptions: Original radius r; new radius = 1.14r. Area ∝ r^2. Concept / Approach: New area / old area = (1.14)^2. The percentage increase in area is [(1.14)^2 − 1]*100%. Compute precisely to avoid rounding errors: 1.14^2 = 1.2996, which corresponds to a 29.96% rise. Step-by-Step Solution: Scale factor for radius = 1.14.Area factor = (1.14)^2 = 1.2996.Percentage increase = (1.2996 − 1)*100% = 29.96%. Verification / Alternative check: Using binomial: (1 + 0.14)^2 = 1 + 2*0.14 + 0.14^2 = 1 + 0.28 + 0.0196 = 1.2996 → 29.96% increase. Why Other Options Are Wrong: 28% ignores the square term; 14% and 14.98% undercount; 28% + 1.96% = 29.96% is correct. Common Pitfalls: Treating area change as linear in radius instead of quadratic. Final Answer: 29.96 percent

Question 15

Equilateral triangle (side 24 cm) with an inscribed circle:
Find the area of the triangle’s region that is outside the circle but inside the triangle (in cm²).

Accepted Answer

Introduction / Context: The “remaining portion” means triangle area minus the area of the incircle. For an equilateral triangle, the inradius has a simple formula in terms of side length, which makes computation straightforward and exact in radicals; the options here are rounded decimal values. Given Data / Assumptions: Equilateral triangle side a = 24 cm. Inradius r = a√3/6. Triangle area = (√3/4)a^2. Concept / Approach: Compute the triangle’s area and the incircle’s area, then subtract. Use either π = 22/7 or π ≈ 3.1416 for the circle; the rounded answer must match given options. Using standard exact expressions first helps consistency. Step-by-Step Solution: Triangle area = (√3/4)*24^2 = (√3/4)*576 = 144√3 ≈ 249.41 cm².Inradius r = 24√3/6 = 4√3 cm.Circle area = πr^2 = π*(4√3)^2 = π*48 ≈ 150.80 cm².Remaining area ≈ 249.41 − 150.80 ≈ 98.61 cm². Verification / Alternative check: Using π = 22/7 gives 48*(22/7) ≈ 150.86, still leading to ~98.55 cm². The closest option is 98.5 sq. cm. Why Other Options Are Wrong: 36.6, 54.2, 72.8 are far from the precise subtraction. Common Pitfalls: Mistaking circumradius for inradius; forgetting that incircle area uses r, not R. Final Answer: 98.5 sq. cm

Question 16

Rhombus with one angle 60° and shorter diagonal 8 cm:
Find the area of the rhombus (in cm²).

Accepted Answer

Introduction / Context: A rhombus has all sides equal. Given one interior angle, the diagonals relate to the side and the angle. The area is a*b*sin(θ) where a and b are adjacent sides. With a known diagonal and angle, we can retrieve the side and compute area compactly. Given Data / Assumptions: Interior angle θ = 60°. Shorter diagonal d_short = 8 cm. All sides equal to a. Concept / Approach: For a rhombus with side a and angle θ: diagonals are a√(2 + 2cosθ) and a√(2 − 2cosθ). For θ = 60°, cosθ = 1/2. Then the two diagonals are a√3 and a. The shorter diagonal equals a, so a = 8 cm. Area = a^2 * sinθ = 8^2 * sin60° = 64 * (√3/2) = 32√3. Step-by-Step Solution: cos60° = 1/2 ⇒ diagonals: a√3 and a.Given d_short = 8 ⇒ a = 8.Area = a^2 * sin60° = 64 * (√3/2) = 32√3 cm². Verification / Alternative check: Rhombus area can also be (d1*d2)/2. Here d1 = a = 8; d2 = a√3 = 8√3 ⇒ area = (8 * 8√3)/2 = 32√3, consistent. Why Other Options Are Wrong: Values with √2 arise from square/45° assumptions; 64√3 doubles the correct area. Common Pitfalls: Confusing which diagonal is shorter at θ = 60°; it is a, not a√3. Final Answer: 32√3 sq cm

Question 17

Largest triangle in a semicircle (radius 6 m):
Find the maximum possible area of a triangle inscribed in a semicircle of radius 6 m.

Accepted Answer

Introduction / Context: Any triangle formed by the diameter as one side and a point on the semicircle is a right triangle (Thales’ theorem). Among right triangles with fixed hypotenuse, the area is maximized when the legs are equal (an isosceles right triangle), because the product of two numbers with fixed sum is maximized when they are equal. Given Data / Assumptions: Radius r = 6 m ⇒ diameter (hypotenuse) = 12 m. Let legs be a and b; hypotenuse fixed = 12 m. Concept / Approach: Area = (1/2)*a*b. For fixed hypotenuse, a^2 + b^2 = 12^2. The product a*b is maximized when a = b. Thus a = b = 12/√2 = 6√2. Substitute into area. Step-by-Step Solution: a = b = 6√2.Area_max = (1/2) * a * b = (1/2) * (6√2) * (6√2) = (1/2) * 72 = 36 m². Verification / Alternative check: Pick any unequal legs satisfying a^2 + b^2 = 144; their product is less than when a = b, so the area is smaller. Hence 36 m² is indeed the maximum. Why Other Options Are Wrong: 12 and 18 are sub-maximal; 72 mistakenly doubles the maximal area. Common Pitfalls: Using diameter*radius/2; the correct formula is (1/2)*a*b with a^2 + b^2 fixed. Final Answer: 36 sq.m

Question 18

Rectangle from area and perimeter:
A rectangle has area 60 cm² and perimeter 34 cm. Find the length of its diagonal (in cm).

Accepted Answer

Introduction / Context: With area and perimeter given, one can find the sum and product of the sides, which allows recovery of the diagonal using the identity a^2 + b^2 = (a + b)^2 − 2ab. The diagonal is then √(a^2 + b^2). Given Data / Assumptions: Area ab = 60. Perimeter 2(a + b) = 34 ⇒ a + b = 17. Concept / Approach: Compute a^2 + b^2 from the sum and product. Then diagonal d = √(a^2 + b^2). No need to solve for a and b individually. Step-by-Step Solution: (a + b)^2 = 17^2 = 289.a^2 + b^2 = (a + b)^2 − 2ab = 289 − 120 = 169.Diagonal d = √169 = 13 cm. Verification / Alternative check: One possible pair is a = 12, b = 5 (since 12 + 5 = 17 and 12*5 = 60). Diagonal = √(12^2 + 5^2) = √169 = 13, confirming. Why Other Options Are Wrong: 11 and 15 do not satisfy the derived Pythagorean relation given the constraints. Common Pitfalls: Trying to factor 60 first—unnecessary; use identities directly. Final Answer: 13 cm

Question 19

Rectangle from diagonal and breadth:
The diagonal of a rectangle is 17 cm and its breadth is 8 cm. Find its perimeter (in cm).

Accepted Answer

Introduction / Context: Given a rectangle’s diagonal and one side, the other side follows from Pythagoras. The perimeter is then twice the sum of the two sides. Given Data / Assumptions: Diagonal d = 17 cm. Breadth b = 8 cm. Concept / Approach: Length l satisfies l^2 + b^2 = d^2. Solve for l, then compute perimeter P = 2(l + b). Step-by-Step Solution: l^2 = d^2 − b^2 = 17^2 − 8^2 = 289 − 64 = 225.l = 15 cm.Perimeter P = 2(l + b) = 2*(15 + 8) = 46 cm. Verification / Alternative check: Check back: 15^2 + 8^2 = 225 + 64 = 289 = 17^2 (consistent). Why Other Options Are Wrong: 92 and 84 double something else incorrectly; 42 undercounts. Common Pitfalls: Using 2√(l^2 + b^2) as perimeter; that is incorrect—the perimeter is 2(l + b). Final Answer: 46

Question 20

Isosceles triangle from symbolic perimeter:
The base of an isosceles triangle is B = 2x + 2y + 4z and the perimeter is P = 4x + 2y + 6z (expressions in the same units). Find the length of each equal side.

Accepted Answer

Introduction / Context: For an isosceles triangle with base B and equal sides s, perimeter P satisfies P = B + 2s. Rearranging gives s = (P − B)/2. Here B and P are algebraic expressions in x, y, z that represent like units (e.g., centimetres). Given Data / Assumptions: Base B = 2x + 2y + 4z. Perimeter P = 4x + 2y + 6z. All symbols represent nonnegative measures in the same unit. Concept / Approach: Apply s = (P − B)/2 algebraically and simplify the expression; this yields the length of each congruent side in terms of x and z only, since the y-terms cancel here. Step-by-Step Solution: P − B = (4x + 2y + 6z) − (2x + 2y + 4z) = 2x + 2z.s = (P − B)/2 = (2x + 2z)/2 = x + z. Verification / Alternative check: Perimeter re-formed: B + 2s = (2x + 2y + 4z) + 2(x + z) = 4x + 2y + 6z = P, consistent. Why Other Options Are Wrong: Options involving y do not reflect the cancellation; 2(x + y) doubles the wrong combination; x + y is not supported by P − B reduction. Common Pitfalls: Forgetting the factor 1/2 in s = (P − B)/2; algebraic sign mistakes. Final Answer: x+z

Area Questions