Ninth Grade Mathematics: Review: Perimeters & Areas of 2D Figures
Mensuration is the branch of applied geometry dedicated to measuring geometric parameters such as lengths, perimeters, and two-dimensional surface areas. In this chapter, you will review the standard perimeter and area formulas for foundational plane figures—triangles, rectangles, parallelograms, rhombuses, trapezoids, and circles—establishing the computational baseline needed for advanced geometric analysis.
Foundational Formulas for Polygons
The perimeter of any polygon is the total linear distance around its outer boundary, obtained by summing the lengths of all its boundary segments. The area measures the total two-dimensional space enclosed within those boundaries.
Figure Perimeter Formula Area Formula
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Square (side s) P = 4 * s Area = s^2
Rectangle (l, w) P = 2 * (l + w) Area = l * w
Parallelogram (b, h) P = 2 * (side1 + side2) Area = b * h
Triangle (b, h) P = a + b + c Area = (1/2) * b * h
Rhombus (d1, d2) P = 4 * s Area = (1/2) * d1 * d2
Trapezoid (b1, b2, h) P = a + b1 + c + b2 Area = (1/2) * (b1 + b2) * h
RHOMBUS DIAGONAL AREA TRAPEZOID AREA
* b1
/|\ *----------*
/ | \ /| |\
d1 / | \ d2 h / | h h | \
*---+---* *--+----------+--*
\ | / b2
\ | /
* Area = (1/2) * (b1 + b2) * h
Area = (1/2) * d1 * d2
Notice that the area of a rhombus can be computed directly from its two perpendicular diagonals: each of the four right triangles formed by the diagonals has area (1/2)(d1/2)(d2/2) = (1/8)d1d2, and summing all four yields (1/2)d1d2.
Circles and Circular Rings
For a circle of radius r and diameter d = 2*r:
Circle Formulas:
Circumference: C = 2 * pi * r = pi * d
Area: A = pi * r^2
When two concentric circles share the same center with outer radius R and inner radius r, they enclose a circular ring (annulus):
Circular Ring (Annulus)
. - ~ ~ - .
. ' R ' .
/ . - ~ . \
/ / r \ \
; ; * ; ;
\ \ / /
\ ' . _ . ' /
. ' ' .
' - ~ ~ - .
Area of Ring = pi * (R^2 - r^2)
The area of the ring is simply the difference between the area of the larger circle and the area of the smaller circle: Area = pi * R^2 - pi * r^2 = pi * (R^2 - r^2).
Unit Conversions in Mensuration
Calculating areas requires converting all given dimensions into consistent physical units before performing arithmetic operations:
Unit Conversion Rules:
Length:
1 meter (m) = 100 centimeters (cm) = 1,000 millimeters (mm)
1 kilometer (km) = 1,000 meters (m)
Area (Square Units):
1 m^2 = (100 cm) * (100 cm) = 10,000 cm^2
1 cm^2 = (10 mm) * (10 mm) = 100 mm^2
1 hectare (ha) = 100 m * 100 m = 10,000 m^2
1 km^2 = 1,000,000 m^2 = 100 hectares
A common student mistake is converting 1 square meter to 100 square centimeters; remember that because area is two-dimensional, the conversion factor must be squared: 100 * 100 = 10,000 cm^2.
Chapter Exercises
Exercise 1: Find the area and perimeter of a rhombus whose diagonals have lengths 24 cm and 10 cm.
Exercise 2: The parallel sides of a trapezoid are 25 cm and 13 cm, and its non-parallel sides are equal, each measuring 10 cm. Calculate the perpendicular height and the exact area of the trapezoid.
Exercise 3: A circular racetrack has an inner circumference of 352 meters and an outer circumference of 396 meters. Find the width of the track and the total surface area of the running track (use 22/7 for pi).
Solutions and Explanations
Solution 1: The area of a rhombus is (1/2) * d_1 * d_2. Given diagonals d_1 = 24 cm and d_2 = 10 cm: Area = (1/2) * 24 * 10 = 120 square cm. The diagonals of a rhombus bisect each other at right angles, forming four congruent right-angled triangles with legs d_1 / 2 = 12 cm and d_2 / 2 = 5 cm. By the Pythagorean theorem, the side length s is the hypotenuse: s = sqrt(12^2 + 5^2) = sqrt(144 + 25) = sqrt(169) = 13 cm. The perimeter of the rhombus is 4 * s = 4 * 13 = 52 cm.
Solution 2: In the trapezoid, the parallel bases are b_1 = 25 cm and b_2 = 13 cm. Drop perpendicular altitudes from the ends of the shorter base to the longer base, forming a central rectangle of width 13 cm and two congruent right triangles on either side. The combined base length of the two right triangles is 25 - 13 = 12 cm, so each right triangle has a base of 12 / 2 = 6 cm. The hypotenuse of each right triangle is the non-parallel leg (10 cm). By the Pythagorean theorem, the perpendicular height is h = sqrt(10^2 - 6^2) = sqrt(100 - 36) = sqrt(64) = 8 cm. Now compute the area: Area = (1/2) * (b_1 + b_2) * h = (1/2) * (25 + 13) * 8 = (1/2) * 38 * 8 = 152 square cm.
Solution 3: Let the inner radius be r and outer radius be R. The inner circumference is 2 * pi * r = 352 meters, so 2 * (22/7) * r = 352. Multiply by 7/44: r = (352 * 7) / 44 = 8 * 7 = 56 meters. The outer circumference is 2 * pi * R = 396 meters, so 2 * (22/7) * R = 396. Multiply by 7/44: R = (396 * 7) / 44 = 9 * 7 = 63 meters. The width of the track is the difference in radii: Width = R - r = 63 - 56 = 7 meters. The area of the track is the area of the circular ring: Area = pi * (R^2 - r^2) = (22/7) * (63^2 - 56^2). Factor using difference of squares: (63 - 56) * (63 + 56) = 7 * 119. Thus, Area = (22/7) * 7 * 119 = 22 * 119 = 2,618 square meters.