Find Area by Composing and Decomposing
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6th Grade Math › Find Area by Composing and Decomposing
A bulletin board is shaped like a trapezoid that you decompose into a rectangle and a triangle.
- The shorter base is $6\text{ cm}$, the longer base is $10\text{ cm}$, and the height is $4\text{ cm}$.
If you decompose it into:
- a rectangle $6\text{ cm}\times 4\text{ cm}$ and
- a right triangle with base $4\text{ cm}$ and height $4\text{ cm}$,
what is the total area?
$32\text{ cm}^2$
$40\text{ cm}^2$
$24\text{ cm}^2$
$28\text{ cm}^2$
Explanation
This question tests finding the area of a trapezoid by decomposing into a rectangle and a right triangle, using A = lw and (1/2)bh, summing to total. Strategies: rectangle 6 cm × 4 cm = 24 cm², triangle (1/2)×4 cm×4 cm = 8 cm², total 32 cm². Formulas: rectangle A = lw, triangle (1/2)bh as half rectangle; example, similar trapezoid bases 6 and 10, height 4: 24+8=32. Like triangle half 24=12, L-shape 30+12=42. The correct area is 24 + 8 = 32 cm² using decomposition. Errors include no 1/2 for triangle getting 24+16=40, or wrong base like 10×4=40. Decomposing: (1) identify parts, (2) divide, (3) calculate and sum; real-world for boards like covering. Mistakes: arithmetic like 24+8=30, forgetting squared units.
An irregular patio can be decomposed into two rectangles that do not overlap: Rectangle 1 is $5\text{ m} \times 6\text{ m}$ and Rectangle 2 is $3\text{ m} \times 4\text{ m}$. What is the total area of the patio?
$34\text{ m}^2$
$54\text{ m}^2$
$46\text{ m}^2$
$42\text{ m}^2$
Explanation
This question tests finding the area of polygons by composing into rectangles (large minus cutout) or decomposing into triangles/rectangles (sum parts), using A=(1/2)bh for triangles, A=lw for rectangles. Strategies include decomposing by breaking into simpler shapes like a triangle as half a rectangle with A=(1/2)bh, L-shape as two rectangles summing areas, or composing by enclosing in a rectangle and subtracting outside parts, such as a right triangle in a 6×4 rectangle giving (1/2)×24=12. Formulas are rectangle A=lw (length×width), triangle A=(1/2)bh (base×height divided by 2—half of a rectangle); example: trapezoid bases 6 and 10, height 4, decompose into rectangle 6×4=24 plus triangle base 4, height 4: (1/2)×4×4=8, total 32. For this patio, decompose into two rectangles: 5×6=30 m² and 3×4=12 m², summing to 42 m². A common error is adding wrong, like 30+12=40, or treating as one shape without decomposing. To decompose, identify simpler shapes like rectangles and triangles, draw dividing lines, calculate each (rectangle: lw, triangle: (1/2)bh), and sum; for composing, enclose in rectangle, identify cutouts, calculate, and subtract. Real-world: patio area for paving; mistakes include arithmetic errors or forgetting non-overlapping parts.
A triangular pennant is a right triangle with base $6\text{ cm}$ and height $4\text{ cm}$. What is its area? (Use $A=\tfrac{1}{2}bh$.)
$24\text{ cm}^2$
$10\text{ cm}^2$
$12\text{ cm}^2$
$20\text{ cm}^2$
Explanation
This question tests finding the area of a right triangle by using A = (1/2)bh, composing it as half of a rectangle. Strategies include directly applying the formula for the triangle with base 6 cm and height 4 cm, or composing a 6×4 rectangle of 24 cm² and taking half for 12 cm². Formulas: triangle A = (1/2)bh, which is half of a rectangle's A = lw with same base and height; example, trapezoid bases 6 and 10, height 4, decomposes to rectangle 24 plus triangle 8 for 32. For example, this right triangle base 6, height 4 is (1/2)×6×4=12, or half of 24=12; an L-shape might be 48 minus 6=42 by composition. The correct area is (1/2)×6×4 = 12 cm² using the formula. A common error is forgetting the 1/2, calculating bh=24 instead of 12, or mixing with rectangle formula. For triangles: always use (1/2)bh; real-world like pennant fabric area. Steps: identify base and height, multiply and halve; mistakes include arithmetic errors or unsquared units.
A pentagon-shaped sign can be decomposed into a rectangle and a triangle that sit on top of it.
- The rectangle is $8\text{ ft}$ wide and $5\text{ ft}$ tall.
- The triangle on top has the same base as the rectangle ($8\text{ ft}$) and a height of $3\text{ ft}$.
What is the area of the pentagon?
$64\text{ ft}^2$
$40\text{ ft}^2$
$52\text{ ft}^2$
$28\text{ ft}^2$
Explanation
This question tests finding the area of polygons by composing into rectangles (large minus cutout) or decomposing into triangles/rectangles (sum parts), using A=(1/2)bh for triangles, A=lw for rectangles. Strategies: decompose (break into simpler shapes: triangle as half rectangle with A=(1/2)bh, L-shape as two rectangles summing areas), compose (enclose in rectangle, subtract outside parts: right triangle in 6×4 rectangle is (1/2)×24=12). Formulas: rectangle A=lw (length×width), triangle A=(1/2)bh (base×height divided by 2—half of rectangle). Example: trapezoid bases 6 and 10, height 4, decompose into rectangle 6×4=24 plus triangle with base 4, height 4: (1/2)×4×4=8, total 24+8=32. Example: right triangle base 6, height 4, fits in rectangle 6×4=24, triangle is half: (1/2)×24=12, or directly (1/2)×6×4=12; or L-shape: large rectangle 8×6=48 minus cutout 3×2=6 gives 48-6=42, or decompose into rectangles 5×6=30 and 3×4=12, sum 30+12=42 (both methods work). The correct area is found by decomposing: rectangle 8 ft × 5 ft = 40 ft² plus triangle (1/2) × 8 ft × 3 ft = 12 ft², total 52 ft². A common error is forgetting the 1/2 for the triangle (40 + 24 = 64 ft²), or using rectangle height for triangle base, or arithmetic like 40 + 12 = 54. Decomposing: (1) identify simpler shapes (can polygon be split into rectangles and triangles?), (2) draw lines dividing (dotted lines showing decomposition), (3) calculate each part (rectangle: lw, triangle: (1/2)bh), (4) sum (total=part₁+part₂+...). Composing: (1) enclose in rectangle (smallest rectangle containing polygon), (2) identify cutouts (triangles or rectangles outside polygon), (3) calculate rectangle and cutouts, (4) subtract (rectangle area minus cutout areas). Triangle key: A=(1/2)bh always (half of rectangle with same base and height). Real-world: room area for flooring (L-shaped room composed/decomposed), garden area for fencing, window area for glass. Mistakes: forgetting (1/2) for triangles, wrong decomposition (incorrect shapes), arithmetic errors, units not squared.
A trapezoid-shaped window has parallel sides $6\text{ in}$ and $10\text{ in}$ and height $4\text{ in}$. A student computes $(6+10)\cdot 4=64\text{ in}^2$. What is the correct area?
$32\text{ in}^2$
$40\text{ in}^2$
$64\text{ in}^2$
$16\text{ in}^2$
Explanation
This question tests finding the area of polygons by composing into rectangles (large minus cutout) or decomposing into triangles/rectangles (sum parts), using A=(1/2)bh for triangles, A=lw for rectangles. Strategies include decomposing by breaking into simpler shapes like a triangle as half a rectangle with A=(1/2)bh, L-shape as two rectangles summing areas, or composing by enclosing in a rectangle and subtracting outside parts, such as a right triangle in a 6×4 rectangle giving (1/2)×24=12. Formulas are rectangle A=lw (length×width), triangle A=(1/2)bh (base×height divided by 2—half of a rectangle); example: trapezoid bases 6 and 10, height 4, decompose into rectangle 6×4=24 plus triangle base 4, height 4: (1/2)×4×4=8, total 32 in². The student computed (6+10)×4=64, but the correct area is (6+10)/2 ×4=32 in², forgetting to divide by 2 for the average base. A common error is halving wrong, like (6+10)×(4/2)=32 but misplaced, or arithmetic like (16/2)×4=30. To decompose, identify simpler shapes, draw lines, calculate parts, and sum; for composing, enclose in rectangle, identify cutouts, and subtract. Real-world: window area for glass; mistakes include forgetting to average bases or units not squared.
A table top is shaped like a trapezoid with parallel sides $6\text{ cm}$ and $10\text{ cm}$ and height $4\text{ cm}$. What is the area of the trapezoid?
$16\text{ cm}^2$
$32\text{ cm}^2$
$24\text{ cm}^2$
$64\text{ cm}^2$
Explanation
This question tests finding the area of polygons by composing into rectangles (large minus cutout) or decomposing into triangles/rectangles (sum parts), using A=(1/2)bh for triangles, A=lw for rectangles. Strategies include decomposing by breaking into simpler shapes like a triangle as half a rectangle with A=(1/2)bh, L-shape as two rectangles summing areas, or composing by enclosing in a rectangle and subtracting outside parts, such as a right triangle in a 6×4 rectangle giving (1/2)×24=12. Formulas are rectangle A=lw (length×width), triangle A=(1/2)bh (base×height divided by 2—half of a rectangle); for this trapezoid with bases 6 and 10, height 4, decompose into rectangle 6×4=24 plus triangle base 4, height 4: (1/2)×4×4=8, total 32 cm², or use average base (6+10)/2×4=32. For example, an L-shape: large rectangle 8×6=48 minus cutout 3×2=6 gives 48-6=42, or decompose into 5×6=30 and 3×4=12 summing to 42. A common error is forgetting to average the bases, like (6+10)×4=64 without dividing by 2, or arithmetic like 24+8=30 instead of 32. To decompose, identify simpler shapes, draw lines dividing, calculate each part (rectangle: lw, triangle: (1/2)bh), and sum; for composing, enclose in rectangle, identify cutouts, calculate, and subtract. Triangle key: A=(1/2)bh always; real-world: table top area for covering, mistakes include wrong decomposition or units not squared.
A school garden is shaped like a right triangle with base $6\text{ m}$ and height $4\text{ m}$. What is the area of the garden?
$10\text{ m}^2$
$24\text{ m}^2$
$28\text{ m}^2$
$12\text{ m}^2$
Explanation
This question tests finding the area of polygons by composing into rectangles (large minus cutout) or decomposing into triangles/rectangles (sum parts), using A=(1/2)bh for triangles, A=lw for rectangles. Strategies include decomposing by breaking into simpler shapes like a triangle as half a rectangle with A=(1/2)bh, or composing by enclosing in a rectangle and subtracting outside parts, such as a right triangle in a 6×4 rectangle giving (1/2)×24=12. Formulas are rectangle A=lw (length×width), triangle A=(1/2)bh (base×height divided by 2—half of a rectangle); for example, a trapezoid with bases 6 and 10, height 4, can be decomposed into a rectangle 6×4=24 plus a triangle with base 4, height 4: (1/2)×4×4=8, total 24+8=32. For this right triangle with base 6 m and height 4 m, you can compose it into a 6×4 rectangle of area 24 and take half, or directly use (1/2)×6×4=12 m². A common error is treating it like a rectangle without dividing by 2, getting 6×4=24 m² instead of 12. To decompose, identify simpler shapes like triangles, draw dividing lines, calculate each part using (1/2)bh, and sum them up; for composing, enclose in a rectangle, identify cutouts, calculate areas, and subtract. Remember, a triangle's area is always (1/2)bh as it's half a rectangle with the same base and height; in real-world uses, like garden areas for planting, mistakes include forgetting the 1/2 for triangles or arithmetic errors like 6×4=26.
A right triangle banner has a base of $6\text{ ft}$ and a height of $4\text{ ft}$. What is its area?
$10\text{ ft}^2$
$20\text{ ft}^2$
$24\text{ ft}^2$
$12\text{ ft}^2$
Explanation
This question tests finding the area of polygons by composing into rectangles (large minus cutout) or decomposing into triangles/rectangles (sum parts), using A=(1/2)bh for triangles, A=lw for rectangles. Strategies: decompose (break into simpler shapes: triangle as half rectangle with A=(1/2)bh, L-shape as two rectangles summing areas), compose (enclose in rectangle, subtract outside parts: right triangle in 6×4 rectangle is (1/2)×24=12). Formulas: rectangle A=lw (length×width), triangle A=(1/2)bh (base×height divided by 2—half of rectangle). Example: trapezoid bases 6 and 10, height 4, decompose into rectangle 6×4=24 plus triangle with base 4, height 4: (1/2)×4×4=8, total 24+8=32. Example: right triangle base 6, height 4, fits in rectangle 6×4=24, triangle is half: (1/2)×24=12, or directly (1/2)×6×4=12; or L-shape: large rectangle 8×6=48 minus cutout 3×2=6 gives 48-6=42, or decompose into rectangles 5×6=30 and 3×4=12, sum 30+12=42 (both methods work). The correct area is found using the triangle formula: (1/2) × 6 ft × 4 ft = 12 ft², or by composing into a 6 ft × 4 ft rectangle (24 ft²) and taking half. A common error is forgetting the 1/2 and calculating as a rectangle (6 × 4 = 24 ft²), or mixing up base and height, or arithmetic like (1/2) × 6 × 4 = 24 instead of 12. Decomposing: (1) identify simpler shapes (can polygon be split into rectangles and triangles?), (2) draw lines dividing (dotted lines showing decomposition), (3) calculate each part (rectangle: lw, triangle: (1/2)bh), (4) sum (total=part₁+part₂+...). Composing: (1) enclose in rectangle (smallest rectangle containing polygon), (2) identify cutouts (triangles or rectangles outside polygon), (3) calculate rectangle and cutouts, (4) subtract (rectangle area minus cutout areas). Triangle key: A=(1/2)bh always (half of rectangle with same base and height). Real-world: room area for flooring (L-shaped room composed/decomposed), garden area for fencing, window area for glass. Mistakes: forgetting (1/2) for triangles, wrong decomposition (incorrect shapes), arithmetic errors, units not squared.
A tabletop is shaped like a trapezoid. The parallel sides are $6\text{ ft}$ and $10\text{ ft}$, and the height is $4\text{ ft}$. What is the area of the tabletop? (You may use $A=\tfrac{1}{2}(b_1+b_2)h$.)
$32\text{ ft}^2$
$16\text{ ft}^2$
$64\text{ ft}^2$
$40\text{ ft}^2$
Explanation
This question tests finding the area of a trapezoid using A = (1/2)(b1 + b2)h, which relates to decomposing into a rectangle and triangles. Strategies include using the formula directly for bases 6 ft and 10 ft, height 4 ft, giving (1/2)×16×4=32 ft², or decomposing into a rectangle and triangle. Formulas: trapezoid as average bases times height, rectangle A=lw, triangle (1/2)bh; example, this trapezoid decomposes to 6×4=24 plus (1/2)×4×4=8 for 32. Like a right triangle in 6×4=24, half is 12; or L-shape 48-6=42. The correct area is (1/2)(6+10)×4 = 32 ft² using the formula. Errors include not halving, getting 16×4=64, or adding bases wrong like 6+10=18. Decomposing: (1) split into rectangle and triangle, (2) calculate each, (3) sum; real-world for tabletops like painting area. Mistakes: forgetting 1/2 in formula, arithmetic like 8×4=36 instead of 32, squared units.
A storage mat is an irregular shape that can be found by composing a large rectangle and subtracting a triangular cutout.
- The large rectangle is $12\text{ ft}$ by $7\text{ ft}$.
- A right-triangular cutout has base $6\text{ ft}$ and height $3\text{ ft}$.
What is the area of the mat?
$93\text{ ft}^2$
$75\text{ ft}^2$
$66\text{ ft}^2$
$84\text{ ft}^2$
Explanation
This question tests finding the area of an irregular mat by composing a large rectangle and subtracting a triangular cutout, using A = lw minus (1/2)bh. Strategies: large 12 ft × 7 ft = 84 ft², cutout (1/2)×6×3=9 ft², area 84-9=75 ft². Formulas: rectangle A = lw, triangle (1/2)bh; example, trapezoid 24 + 8 = 32. Like L-shape 48-6=42, triangle 12. The correct area is 84 - 9 = 75 ft² using composition. Errors include full triangle 18, 84-18=66, or no subtract 84. Composing: (1) enclose, (2) identify cutout, (3) calculate, (4) subtract; real-world for mats like cleaning area. Mistakes: forgetting 1/2, arithmetic like 84-9=74, unsquared units.