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12. Plane Geometry

CSS GSA — Paper 7, Part-II, Section VI, Sub-topic 12 (Plane Geometry)

Past Paper Weightage

  • Very high — angles, triangles, and circle theorems are among the most reliably recurring GSA quant questions (marked as a focus area).

Angles

  • Complementary angles sum to 90°; Supplementary angles sum to 180°
  • Angles on a straight line sum to 180°; angles around a point sum to 360°
  • Vertically opposite angles are equal; when a transversal crosses parallel lines — alternate angles equal, corresponding angles equal, co-interior (allied) angles sum to 180°

Triangles

  • Sum of interior angles = 180°
  • Types: Equilateral (all sides/angles equal, each angle 60°), Isosceles (two equal sides/angles), Scalene (all different), Right-angled (one 90° angle)
  • Pythagorean theorem (right triangle):
a2+b2=c2(c=hypotenuse)a^2 + b^2 = c^2 \quad (c = \text{hypotenuse})
  • Area:
A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}

for a right triangle, the two legs can serve as base and height

  • Similar triangles: same shape, proportional sides, equal corresponding angles; Congruent triangles: identical in shape and size (SSS, SAS, ASA, RHS criteria)

Circles

Circumference=2πrArea=πr2\text{Circumference} = 2\pi r \qquad \text{Area} = \pi r^2
  • Angle in a semicircle = 90°; angle at the center = 2× angle at the circumference (same arc)
  • Tangent to a circle is perpendicular to the radius at the point of contact

Quadrilaterals

  • Sum of interior angles of any quadrilateral = 360°
  • Parallelogram: opposite sides equal and parallel, opposite angles equal, diagonals bisect each other

Past Paper Style Numericals (Solved)

  1. Two angles are complementary and one is 20° more than the other. Find both angles.

    Answer: x+(x+20)=90 → x=35 → 35° and 55°

  2. In an isosceles triangle, the vertex angle is 40°. Find each base angle.

    Answer: (180−40)/2 = 70° each

  3. A ladder 13 m long leans against a wall, its foot 5 m from the wall base. How high up the wall does it reach?

    Answer: √(13²−5²) = √(169−25) = √144 = 12 m

  4. Find the area of a triangle with base 14 cm and height 9 cm.

    Answer: ½×14×9 = 63 cm²

  5. Find the circumference and area of a circle with radius 21 cm (use π≈22/7).

    Answer: Circumference=2×22/7×21=132 cm; Area=22/7×441=1386 cm²

MCQ Traps

  • Sum of angles: triangle = 180°, quadrilateral = 360°, pentagon = 540° — general formula for an n-sided polygon:
S=(n−2)×180∘S = (n-2) \times 180^\circ
  • Similar ≠ congruent: similar triangles need not be the same size, only the same shape

Self-Test Questions

  1. In a right triangle, the legs are 6 cm and 8 cm. Find the hypotenuse.
  2. Find the sum of interior angles of a hexagon.
  3. Two parallel lines are cut by a transversal; one angle is 65°. Find its corresponding angle and its co-interior angle.

Past-Paper MCQs (FPSC)

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Taken from the FPSC papers you shared (2015–2025). FPSC does not publish MCQ keys, so the answer is worked out by me.

  1. (CSS 2024, Q1.5) If the radius of a circle is increased three times then its area is increased: (A) Three times (B) Six times (C) Nine times (D) None of these
  • ✅ Answer key
    1. (C) Nine times, since area ∝ r², so (3r)² = 9r²

Past-Paper Questions with Answers (FPSC)

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Every question below appeared in the FPSC papers you shared; the year and question number are in each title. Answers are worked out by me.

  • CSS 2025 Q7(c): Rectangle with sides (3x + 6) m, (x + 1) m and diagonal (4x + 1) m. Area and perimeter. (5)

    1. Pythagoras gives x² − 5x − 6 = 0, so x = 6
    2. Sides 24 m and 7 m; diagonal 25 m
    3. Area = 24 × 7 = 168 m²; perimeter = 2(24 + 7) = 62 m

  • CSS 2024 Q8(c): A 10 m tree stands 20 m from a 100 m-high building wall; its shadow reaches the top of the wall. How far from the wall is the light source? (5)

    1. Let D = distance of the source from the wall; the tree is D − 20 from the source
    2. Similar triangles: 10 ÷ (D − 20) = 100 ÷ D
    3. 10D = 100D − 2,000 → 90D = 2,000
    4. D ≈ 22.22 m

  • CSS 2023 Q6(d): Ahmad (1.6 m) stands 2 m in front of a spotlight, facing a wall 10.5 m from him. How tall is his shadow BC on the wall? (5)

    1. Distance from spotlight to wall = 2 + 10.5 = 12.5 m
    2. Similar triangles: BC ÷ 12.5 = 1.6 ÷ 2
    3. BC = 10 m

  • CSS 2023 Q7(a): Ali stands 10 m from a tree; his eyes are 1.5 m above the ground and 15 m from the treetop. Find the tree's height. (5)

    1. Vertical rise above eye level = √(15² − 10²) = √125 ≈ 11.18 m
    2. Tree height = 11.18 + 1.5 ≈ 12.68 m

  • CSS 2023 Q7(c): Lines of symmetry in a regular hexagon, octagon and circle? (5)

    • A regular n-sided polygon has n lines of symmetry: hexagon 6, octagon 8
    • Circle: infinite (every diameter)
    • Draw each shape with its lines

  • CSS 2021 Q6(b): How many 20 cm² tiles are needed for a 1 m wide footpath around the outside of a 24 m by 14 m grass plot? (5)

    1. Outer rectangle = 26 × 16 = 416 m²; plot = 24 × 14 = 336 m²
    2. Footpath = 80 m² = 800,000 cm²
    3. Tiles = 800,000 ÷ 20 = 40,000

  • CSS 2022 Q8(b): Right-triangle tiles with legs 12 cm and 4 cm cost Rs. 15 each. Cost to cover an 8 m by 6 m lounge? (5)

    1. Tile area = ½ × 12 × 4 = 24 cm²; room = 800 × 600 = 480,000 cm²
    2. Tiles = 480,000 ÷ 24 = 20,000
    3. Cost = 20,000 × 15 = Rs. 300,000

  • CSS 2024 Q7(a): A man walks round a right-angled triangle with base 4 km and hypotenuse 5 km, then 6 km more, turns 90° and walks 8 km. Distance travelled and distance from start? (5)

    1. Third side = 3 km, so one round = 12 km
    2. Distance travelled = 12 + 6 + 8 = 26 km
    3. Final legs are at right angles: √(6² + 8²) = 10 km from the start

  • CSS 2016 Q10(a): Define and draw right-angled and equilateral triangles. (5)

    • Right-angled triangle: one angle is exactly 90°; the longest side is the hypotenuse (a² + b² = c²)
    • Equilateral triangle: all three sides equal and all three angles 60°
    • Draw: mark the right angle with a small square; mark equal sides with matching ticks

  • CSS 2023 Q6(c): Perimeter of a rectangle with sides (3x − y), (2x − 3), (2x + y) is 114 cm; find the area. (5)

    • Opposite sides of a rectangle are equal: 3x − y = 2x + y, so x = 2y
    • Perimeter = 2(3x − y) + 2(2x − 3) = 114 → 5x − y = 60
    • Substitute x = 2y: 9y = 60 → y = 20/3, so x = 40/3
    • Sides: 3x − y = 100/3 ≈ 33.33 cm and 2x − 3 = 71/3 ≈ 23.67 cm
    • Area = (100/3) × (71/3) = 7,100/9 ≈ 788.9 cm² (check: 2(33.33 + 23.67) = 114)
    • The values are not whole numbers; if your copy of the paper shows different labels, use the same method

  • CSS 2019 Q7(c): What is a polygon? Describe different types of regular polygon. (5)

    • Polygon: a closed plane figure bounded by three or more straight line segments
    • Regular polygon: all sides equal and all angles equal
    • Formulas for n sides
      • Sum of interior angles = (n − 2) × 180°
      • Each interior angle (regular) = (n − 2) × 180° ÷ n
      • Sum of exterior angles = 360°; each exterior angle = 360° ÷ n
      • Lines of symmetry = n
    Regular polygonSidesEach interior angle
    Equilateral triangle360°
    Square490°
    Pentagon5108°
    Hexagon6120°
    Heptagon7≈ 128.6°
    Octagon8135°
    Nonagon9140°
    Decagon10144°
    • Irregular polygons have unequal sides or angles; polygons may also be convex (no angle over 180°) or concave

  • CSS 2021 Q8(b), 8(c): Find the number of triangles in the two images; calculate the total area and perimeter of the given shape. (5 each)

    • The figures did not come through in the extracted text, so these answers depend on the diagrams; check them against your paper
    • 8(b)(ii) five-pointed star (pentagram): 10 triangles (5 small triangles at the points + 5 larger ones)
    • 8(b)(i) triangle with three lines through one inner point
      • If the lines run from each vertex to the opposite side (like medians): 16 triangles (6 small + 3 + 6 + 1 whole)
      • If the lines stop at the centre: 4 triangles
    • Method for 8(b): label every vertex and count systematically: single small triangles, then pairs, triples and so on
    • 8(c) composite shape
      • Split it into rectangles/squares and triangles; area = sum of the parts
      • Perimeter = sum of the outer edges only (not the dashed inner lines)
      • A slanted side of 15 cm beside a 12 cm height suggests a 9–12–15 right triangle; send me the exact dimensions and I will finish the numbers

CSS Exam Perspective

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  • Most repeated areas: similar triangles and shadows, Pythagoras, area of paths and tiling
  • Tip: draw the figure, label every length, and convert units (m² to cm²) before dividing