FEUCAT Mathematics — Perimeter, Area, Volume & Equation of a LineStudy Notes

Perimeter is the total distance around the boundary of a two-dimensional figure. Think of it as the length of fence needed to enclose a plot of land or the distance you'd walk around the edge of a basketball court. **Key Formulas:** • Rectangle: P = 2(l + w) where l = length, w = width • Square: P = 4s where s = side length • Triangle: P = a + b + c where a, b, c are the three sides • Regular polygon: P = ns where n = number of sides, s = side length • Circle (circumference): C = 2πr = πd where r = radius, d = diameter **Step-by-Step Problem Solving Approach:** 1. Identify the shape you're dealing with 2. List the given measurements 3. Choose the appropriate formula 4. Substitute the values 5. Calculate and include proper units (always linear units like cm, m) **Worked Example 1: Rectangle** Problem: A rectangular lot measures 25 meters by 15 meters. What is its perimeter? Step 1: Identify - This is a rectangle Step 2: Given - length = 25 m, width = 15 m Step 3: Formula - P = 2(l + w) Step 4: Substitute - P = 2(25 + 15) Step 5: Calculate - P = 2(40) = 80 meters **Worked Example 2: Circle** Problem: A circular garden has a diameter of 14 meters. Find its circumference. Step 1: Identify - This is a circle Step 2: Given - diameter = 14 m, so radius = 7 m Step 3: Formula - C = 2πr or C = πd Step 4: Substitute - C = π(14) = 14π Step 5: Calculate - C = 14 × 3.14159 ≈ 43.98 meters **Common Mistakes to Avoid:** • Forgetting to double the length and width in rectangles • Using diameter instead of radius (or vice versa) in circle problems • Mixing up units (always convert to the same unit first) • Not including units in your final answer

Area measures the amount of surface space inside a two-dimensional figure. Imagine painting a wall - the area tells you how much paint you need to cover the entire surface. **Key Formulas:** • Rectangle: A = lw (length × width) • Square: A = s² (side squared) • Triangle: A = ½bh (½ × base × height) • Parallelogram: A = bh (base × height) • Trapezoid: A = ½(b₁ + b₂)h where b₁, b₂ are parallel sides • Circle: A = πr² (π × radius squared) • Rhombus: A = ½d₁d₂ (½ × diagonal₁ × diagonal₂) **Special Case - Heron's Formula for Triangles:** When you know all three sides but not the height: A = √[s(s-a)(s-b)(s-c)] where s = (a+b+c)/2 **Step-by-Step Problem Solving:** 1. Identify the shape 2. Determine what measurements you have 3. Choose the appropriate formula 4. Substitute carefully 5. Calculate and use squared units (cm², m²) **Worked Example 1: Rectangle** Problem: A rectangular rice field is 120 meters long and 80 meters wide. What is its area? Step 1: Shape - Rectangle Step 2: Given - length = 120 m, width = 80 m Step 3: Formula - A = lw Step 4: Substitute - A = 120 × 80 Step 5: Calculate - A = 9,600 m² **Worked Example 2: Triangle** Problem: A triangular garden plot has a base of 16 meters and height of 12 meters. Find its area. Step 1: Shape - Triangle Step 2: Given - base = 16 m, height = 12 m Step 3: Formula - A = ½bh Step 4: Substitute - A = ½ × 16 × 12 Step 5: Calculate - A = ½ × 192 = 96 m² **Worked Example 3: Circle** Problem: A circular swimming pool has a radius of 5 meters. What is its area? Step 1: Shape - Circle Step 2: Given - radius = 5 m Step 3: Formula - A = πr² Step 4: Substitute - A = π × 5² Step 5: Calculate - A = 25π ≈ 78.54 m² **Composite Figures Strategy:** For complex shapes, break them into familiar pieces: 1. Divide the figure into rectangles, triangles, circles, etc. 2. Calculate each area separately 3. Add or subtract as needed **Common Mistakes:** • Forgetting to square the radius in circle area formula • Using perimeter formulas instead of area formulas • Not converting diameter to radius • Forgetting the ½ in triangle and trapezoid formulas

Volume measures the amount of space inside a three-dimensional object. Think of it as how much water a container can hold or how much concrete is needed to fill a foundation. **Key Formulas:** • Cube: V = s³ (side cubed) • Rectangular prism: V = lwh (length × width × height) • Cylinder: V = πr²h (π × radius² × height) • Cone: V = ⅓πr²h (⅓ × π × radius² × height) • Sphere: V = ⅔πr³ (⅔ × π × radius cubed) • Pyramid (rectangular base): V = ⅓lwh (⅓ × length × width × height) **Memory Tip:** Notice that cones and pyramids have the ⅓ factor - they hold exactly one-third the volume of a cylinder or prism with the same base and height. **Step-by-Step Problem Solving:** 1. Identify the 3D shape 2. List all given dimensions 3. Choose the correct volume formula 4. Substitute values carefully 5. Calculate and use cubic units (cm³, m³, liters) **Worked Example 1: Rectangular Water Tank** Problem: A water tank measures 4m long, 3m wide, and 2m high. What is its volume? Step 1: Shape - Rectangular prism Step 2: Given - length = 4m, width = 3m, height = 2m Step 3: Formula - V = lwh Step 4: Substitute - V = 4 × 3 × 2 Step 5: Calculate - V = 24 m³ **Worked Example 2: Cylindrical Tank** Problem: A cylindrical grain silo has a radius of 6 meters and height of 15 meters. Find its volume. Step 1: Shape - Cylinder Step 2: Given - radius = 6m, height = 15m Step 3: Formula - V = πr²h Step 4: Substitute - V = π × 6² × 15 Step 5: Calculate - V = π × 36 × 15 = 540π ≈ 1,696.46 m³ **Worked Example 3: Cone** Problem: An ice cream cone has a radius of 3 cm and height of 10 cm. What is its volume? Step 1: Shape - Cone Step 2: Given - radius = 3cm, height = 10cm Step 3: Formula - V = ⅓πr²h Step 4: Substitute - V = ⅓ × π × 3² × 10 Step 5: Calculate - V = ⅓ × π × 9 × 10 = 30π ≈ 94.25 cm³ **Real-World Applications:** • Water storage tanks (rectangular prism, cylinder) • Concrete for foundations (rectangular prism) • Grain silos (cylinder) • Traffic cones (cone) • Ball volume (sphere) **Common Mistakes:** • Forgetting the ⅓ factor for cones and pyramids • Using diameter instead of radius • Confusing area formulas with volume formulas • Not cubing all dimensions properly

Linear equations describe straight lines on a coordinate plane. UPCAT problems often combine geometry with coordinate systems, so knowing all three forms is essential. **Form 1: Slope-Intercept Form** y = mx + b • m = slope (steepness and direction) • b = y-intercept (where line crosses y-axis) • Easiest form for graphing **Form 2: Point-Slope Form** y - y₁ = m(x - x₁) • Use when you know one point (x₁, y₁) and the slope m • Most useful for finding equations **Form 3: Standard Form** Ax + By = C • A, B, C are integers • Useful for finding intercepts • Can represent any line **Finding Slope Between Two Points:** m = (y₂ - y₁)/(x₂ - x₁) Slope is 'rise over run' - vertical change over horizontal change **Step-by-Step: Finding Line Equation** **Worked Example 1: Given Two Points** Problem: Find the equation of the line passing through (2, 3) and (6, 11). Step 1: Find the slope m = (11 - 3)/(6 - 2) = 8/4 = 2 Step 2: Use point-slope form with either point Using (2, 3): y - 3 = 2(x - 2) Step 3: Simplify to slope-intercept form y - 3 = 2x - 4 y = 2x - 1 **Worked Example 2: Given Point and Slope** Problem: Find the equation of the line with slope -3 passing through (4, 5). Step 1: Use point-slope form directly y - 5 = -3(x - 4) Step 2: Expand and simplify y - 5 = -3x + 12 y = -3x + 17 **Understanding Slope:** • Positive slope: line rises from left to right • Negative slope: line falls from left to right • Zero slope: horizontal line • Undefined slope: vertical line **Parallel and Perpendicular Lines:** • Parallel lines have equal slopes: m₁ = m₂ • Perpendicular lines have slopes that multiply to -1: m₁ × m₂ = -1 • If one line has slope ⅔, a perpendicular line has slope -3/2 **Worked Example 3: Parallel Line** Problem: Find the equation of the line parallel to y = 2x + 5 passing through (3, 1). Step 1: Parallel lines have same slope Slope = 2 (from the given line) Step 2: Use point-slope form y - 1 = 2(x - 3) y - 1 = 2x - 6 y = 2x - 5

These formulas help solve coordinate geometry problems that often appear with mensuration questions in UPCAT. **Distance Formula:** d = √[(x₂ - x₁)² + (y₂ - y₁)²] This comes from the Pythagorean theorem. The distance between two points forms the hypotenuse of a right triangle. **Midpoint Formula:** Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2) The midpoint is simply the average of the x-coordinates and y-coordinates. **Worked Example 1: Distance** Problem: Find the distance between points A(1, 2) and B(5, 8). Step 1: Identify coordinates (x₁, y₁) = (1, 2) and (x₂, y₂) = (5, 8) Step 2: Apply distance formula d = √[(5-1)² + (8-2)²] d = √[4² + 6²] d = √[16 + 36] d = √52 = 2√13 ≈ 7.21 units **Worked Example 2: Midpoint** Problem: Find the midpoint of the segment connecting (-3, 4) and (7, -2). Step 1: Apply midpoint formula Midpoint = ((-3+7)/2, (4+(-2))/2) Midpoint = (4/2, 2/2) Midpoint = (2, 1) **Applications in Mensuration:** • Finding the diagonal of a rectangle on a coordinate plane • Determining if a triangle is isosceles or equilateral • Finding the center of a circle given endpoints of a diameter **Common Applications:** 1. Rectangle with vertices at (0,0), (a,0), (a,b), (0,b) has diagonal length √(a² + b²) 2. Circle with center (h,k) and point (x,y) on the circle has radius r = √[(x-h)² + (y-k)²]

Success in UPCAT mathematics requires not just knowing formulas, but developing systematic problem-solving approaches. **General Problem-Solving Strategy:** 1. **Read Carefully:** Identify what is given and what is asked 2. **Visualize:** Draw a diagram when possible 3. **Choose Formula:** Select the appropriate formula for the shape/situation 4. **Check Units:** Ensure all measurements are in the same units 5. **Substitute:** Plug values into the formula carefully 6. **Calculate:** Perform arithmetic step by step 7. **Verify:** Check if your answer makes sense **Composite Figure Strategy:** Many UPCAT problems involve complex shapes made from basic figures. **Worked Example: Composite Area** Problem: A playground consists of a rectangle 20m × 15m with a semicircle of diameter 15m attached to one of the shorter sides. Find the total area. Step 1: Identify components - Rectangle: 20m × 15m - Semicircle: diameter = 15m, so radius = 7.5m Step 2: Calculate rectangle area A₁ = 20 × 15 = 300 m² Step 3: Calculate semicircle area A₂ = ½ × π × (7.5)² = ½ × π × 56.25 = 28.125π ≈ 88.36 m² Step 4: Add areas Total Area = 300 + 88.36 = 388.36 m² **Mixed Problems Strategy:** Some problems combine mensuration with coordinate geometry. **Worked Example: Rectangle on Coordinate Plane** Problem: A rectangle has vertices at (1,2), (1,7), (6,7), and (6,2). Find its perimeter and area. Step 1: Find dimensions using distance formula or counting Length = 6 - 1 = 5 units Width = 7 - 2 = 5 units Step 2: Calculate perimeter P = 2(5 + 5) = 20 units Step 3: Calculate area A = 5 × 5 = 25 square units **Time Management Tips for UPCAT:** • Memorize basic formulas completely • Practice mental math for common calculations (π ≈ 3.14, √2 ≈ 1.41, etc.) • Skip difficult problems initially, return later • Always check units in your final answer • Use estimation to verify reasonableness **Common Trap Questions:** 1. Giving diameter when formula needs radius 2. Mixed units (cm and m in same problem) 3. Asking for perimeter but giving area information 4. Composite figures where you must subtract areas **Formula Sheet Strategy:** Create mental associations: • Perimeter = 'fence around' (linear units) • Area = 'paint needed' (square units) • Volume = 'water it holds' (cubic units) • Lines = 'slope and intercept' **Verification Methods:** • Does the perimeter make sense compared to the sides? • Is the area reasonable for the given dimensions? • Do the units match what's being asked? • Can you solve the problem a different way to check?