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Aptitude & Reasoning · Question Set
Geometry, Mensuration & Coordinate Basics interview questions for placements and exams.
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18
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Subject
Aptitude & Reasoning
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Go through each question and its explanation. Use this set as a focused practice pack for Aptitude & Reasoning.
The interior angles of any triangle sum to 180°. Subtract the known angles: 180° − (50° + 60°) = 180° − 110° = 70°. This follows directly from the Triangle Angle Sum Theorem.
∠C = 180° − (50° + 60°) = 70°
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The total surface area of a sphere is 4πr². For r = 7, TSA = 4π×49 = 196π cm² (≈ 615.75 cm² if π ≈ 3.1416).
TSA = 4πr² = 4π·49 = 196π
For a right cone, the slant height l is the hypotenuse of a right triangle formed by radius and height: l = √(r² + h²) = √(9² + 12²) = √(81 + 144) = √225 = 15 cm. The curved surface area (lateral area) is πrl = π×9×15 = 135π cm².
l=√(r²+h²)=√(81+144)=15; CSA=πrl=π·9·15=135π
By the Pythagorean Theorem, hypotenuse² = 6² + 8² = 36 + 64 = 100. Taking square root gives 10 cm.
c = √(6² + 8²) = √100 = 10
The area of a triangle is half the product of base and height. So, Area = ½ × 12 × 7 = 42 cm².
A = (1/2)×b×h = (1/2)×12×7 = 42
When only the three sides are known, use **Heron’s formula**. First compute the semi–perimeter s = (7 + 8 + 9)/2 = 12. Then area = √[s(s − a)(s − b)(s − c)] = √[12 × 5 × 4 × 3] = √720 = 12√5 ≈ 26.83 cm². This avoids needing a height.
s=12 ⇒ A=√(12·5·4·3)=√720=12√5
Circumference C = 2πr. Substituting r = 7 and π = 22/7 gives C = 2 × (22/7) × 7 = 44 cm.
C = 2πr = 2×(22/7)×7 = 44
Area A = πr² = (22/7) × 14² = (22/7) × 196 = 22 × 28 = 616 cm².
A = πr² = (22/7)×196 = 616
The area of a sector is the corresponding fraction of the circle’s area: (θ/360°)×πr². With θ = 60°, r = 6, the fraction is 60/360 = 1/6, so area = (1/6)×π×36 = 6π cm² (≈ 18.85 cm²).
A_sector = (60/360)π(6²) = 6π
For an n–sided regular polygon, each interior angle = [(n − 2)×180°]/n. For n = 8, we get [(8 − 2)×180°]/8 = (6×180°)/8 = 1080°/8 = 135°.
[(n−2)·180°]/n ⇒ (6·180)/8 = 135°
Volume of a cylinder is V = πr²h. Substituting r = 5, h = 14: V = (22/7)×25×14 = (22/7)×350 = 22×50 = 1100 cm³.
V = πr²h = (22/7)·25·14 = 1100
Distance between (x₁,y₁) and (x₂,y₂) is √[(x₂−x₁)² + (y₂−y₁)²]. Here, Δx = 5 − (−3) = 8 and Δy = −2 − 4 = −6. So distance = √(8² + (−6)²) = √(64 + 36) = √100 = 10.
d=√[(5+3)²+(-2-4)²]=√(8²+(-6)²)=10
Midpoint M = ((x₁ + x₂)/2, (y₁ + y₂)/2). So M = ((2 + 8)/2, (−1 + 5)/2) = (10/2, 4/2) = (5, 2).
M=((2+8)/2,(−1+5)/2)=(5,2)
Use point–slope form: y − y₁ = m(x − x₁). With m = 4 and (x₁,y₁) = (3, −2), we get y + 2 = 4(x − 3) ⇒ y + 2 = 4x − 12 ⇒ y = 4x − 14.
y−(−2)=4(x−3) ⇒ y=4x−14
If P divides AB internally in ratio m:n (AP:PB = m:n), then P = ((m x_B + n x_A)/(m+n), (m y_B + n y_A)/(m+n)). With m=2, n=1, A(−1,2), B(5,8): x_P = (2×5 + 1×(−1))/3 = 9/3 = 3; y_P = (2×8 + 1×2)/3 = 18/3 = 6.
P=((2·5+1·(−1))/3,(2·8+1·2)/3)=(3,6)
Use the coordinate (shoelace) formula: Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|. Substituting A(1,2), B(4,6), C(6,2): Area = ½|1(6−2) + 4(2−2) + 6(2−6)| = ½|4 + 0 − 24| = ½·20 = 10.
A=½|1(6−2)+4(2−2)+6(2−6)|=10
The standard form is (x − h)² + (y − k)² = r² for center (h, k) and radius r. Here h = −2 and k = 3, so (x − (−2))² + (y − 3)² = (5)² ⇒ (x + 2)² + (y − 3)² = 25.
(x−h)²+(y−k)²=r² ⇒ (x+2)²+(y−3)²=25
An exterior angle of a triangle equals the sum of the two remote interior angles. Let the unknown remote interior angle be x. Then x + 45° = 110°, so x = 65°.
x + 45° = 110° ⇒ x = 65°