Visual models of identities

The correct answer is A. A square with side \(a+b\) has area \((a+b)^2\). Divide each side into parts of lengths \(a\) and \(b\). The large square is then divided into an \(a\) by \(a\) square with area \(a^2\), two rectangles each with area \(ab\), and a \(b\) by \(b\) square with area \(b^2\). Adding these non-overlapping parts gives \((a+b)^2=a^2+ab+ab+b^2=a^2+2ab+b^2\). Option B describes the difference-square identity and has a negative middle term. Option C describes the difference of two squares, represented by a rectangle with sides \(a+b\) and \(a-b\), not this square. Option D is not a valid identity in general; addition and multiplication are different operations. Memory cue: a plus sign in the side produces a plus middle term and two equal \(ab\) rectangles.

The correct answer is B, \(x^2+6x+9\). A square with side \(x+3\) has area \((x+3)^2\). In its area model, the parts are an \(x\) by \(x\) square of area \(x^2\), two rectangles of dimensions \(x\) and 3, each having area \(3x\), and a 3 by 3 square of area 9. Adding them gives \(x^2+3x+3x+9=x^2+6x+9\). Option A counts only one of the two \(3x\) rectangles. Option C has the correct constant term but the wrong sign for the middle term; a plus inside the bracket produces a plus middle term. Option D places the coefficients incorrectly and does not follow the identity. Use \((a+b)^2=a^2+2ab+b^2\) with \(a=x\), \(b=3\): the middle term is \(2\times x\times3=6x\).

The correct answer is C, \(a^2-b^2\). The area of the large square is side times side, so it is \(a^2\). The removed square has side \(b\), so its area is \(b^2\). Remaining area means the original area minus the removed area: \(a^2-b^2\). Option A adds the two areas, which would describe the total of separate squares rather than the part left after removal. Option B, \(2ab\), is the area of two rectangles in some square identities, not the difference between these two square areas. Option D, \(a-b\), is a length, not an area; the remaining region’s area cannot be found by subtracting side lengths alone. The expression can also be factored as \(a^2-b^2=(a+b)(a-b)\), but the direct area description is \(a^2-b^2\). Memory cue: removed area is always subtracted from original area.

The correct answer is D, \(a^2-b^2\). The area of a rectangle is length multiplied by breadth, so here it is \((a+b)(a-b)\). Expand carefully: \(a(a-b)+b(a-b)=a^2-ab+ab-b^2\). The middle terms \(-ab\) and \(+ab\) cancel, leaving \(a^2-b^2\). Option A is the square-of-a-sum result, \((a+b)^2\), not a product of a sum and a difference. Option B is the square-of-a-difference result, \((a-b)^2\). Option C, \(2a+2b\), is related to a perimeter expression in a different setting, not the area of this rectangle. A useful memory cue is “sum times difference gives difference of squares”: \((a+b)(a-b)=a^2-b^2\).

The correct answer is A, \(x^2-4x+4\). Apply the identity \((a-b)^2=a^2-2ab+b^2\) with \(a=x\) and \(b=2\). This gives \((x-2)^2=x^2-2(x)(2)+2^2=x^2-4x+4\). In the area model, the original \(x^2\) square loses two rectangular strips, each of area \(2x\), and the small corner square of area 4 is accounted for positively, producing the same result. Option B uses the plus identity and therefore has the wrong sign for the middle term. Option C forgets the factor 2 in the two equal strips, so it has \(-2x\) instead of \(-4x\). Option D has both an incorrect sign pattern and an incorrect constant term. Memory cue: for \((x-2)^2\), the middle term is negative and equals twice \(x\) times 2.

Direct answer: option B, \(p^2+2pq+q^2\). The total area of a figure made from non-overlapping pieces is found by adding the area of every piece exactly once. Here the four pieces have areas \(p^2\), \(pq\), \(pq\), and \(q^2\). Therefore, total area \(=p^2+pq+pq+q^2=p^2+2pq+q^2\). This is the area form of the identity \((p+q)^2=p^2+2pq+q^2\): one square has area \(p^2\), two middle rectangles have total area \(2pq\), and the other square has area \(q^2\). Option A is incomplete because it omits both \(pq\) parts. Option C belongs to the difference identity \((p-q)^2\), where the middle term is negative, but here both given rectangular areas are positive. Option D is a perimeter-like expression, not an area sum, and it also does not include the products \(p^2\) and \(q^2\). Memory cue: two equal middle rectangles mean \(pq+pq=2pq\).

Direct answer: option C, \(m^2\). In the area model for \((m+n)^2\), the four parts are a square of area \(m^2\), two rectangles of area \(mn\) each, and a square of area \(n^2\). Among these, the square-shaped parts are \(m^2\) and \(n^2\). If \(m>n\), then squaring preserves the order for positive lengths, so \(m^2>n^2\); consequently, the \(m^2\) square is the larger square part. The wording “can be” is important because no relation between \(m\) and \(n\) is stated: if \(n>m\), then \(n^2\) would be larger, and if they are equal, the two square parts would be equal. Option A is the area of one rectangle, not a square. Option B is the combined area of two rectangles, not one square part. Option D is a length, while the question asks for a square part described by its area. Memory cue: in \((m+n)^2\), the square pieces are \(m^2\) and \(n^2\); \(mn\) pieces are rectangles.

Direct answer: option D, \((x+5)^2\). Find the area of each piece. The \(x\) by \(x\) square has area \(x^2\). Each \(x\) by \(5\) rectangle has area \(5x\), so the two rectangles together have area \(10x\). The \(5\) by \(5\) square has area \(25"). Adding gives \(x^2+10x+25\). By the identity \(a^2+2ab+b^2=(a+b)^2\), with \(a=x\) and \(b=5\), this becomes \((x+5)^2\). Option A has a negative middle term and would require subtracting the two rectangles, which is not what the combined positive pieces show. Option B is a difference of squares and has no middle term. Option C contains only \(5x\), but the two rectangles contribute \(5x+5x=10x\), so it is incomplete. Memory cue: two positive equal rectangles create the \(+2ab\) term.

Direct answer: option A, \((y-4)^2=y^2-8y+16\). Start with the square of side \(y\), whose area is \(y^2\). Removing two rectangles of area \(4y\) removes \(4y+4y=8y\). Adding the small square of side 4 contributes \(4^2=16\). Thus the resulting expression is \(y^2-8y+16\). Using \((a-b)^2=a^2-2ab+b^2\), take \(a=y\) and \(b=4\), giving \((y-4)^2=y^2-2(y)(4)+4^2=y^2-8y+16\). Option B has a positive middle term and represents addition. Option C is the difference-of-squares identity and has no \(-8y\) term. Option D incorrectly equates a difference of squares with a perfect-square expansion. Memory cue: subtracting two equal rectangles means the middle term is \(-2ab\), not zero.

Direct answer: option B, \(36+24+4\). A square with side \(6+2\) has area \((6+2)^2\). Divide the side into lengths 6 and 2. The area model then contains a \(6\) by \(6\) square with area \(36"); two rectangles, each \(6\) by \(2\), with total area \(12+12=24\); and a \(2\) by \(2\) square with area 4. Adding all non-overlapping parts gives \(36+24+4=64\). This is the numerical form of \((a+b)^2=a^2+2ab+b^2\). Option A omits the two middle rectangles. Option C adds lengths or partial products incorrectly and is not the complete area. Option D happens to equal 60, not the actual area 64, and it does not represent the partition. Memory cue: for \((6+2)^2\), remember square, two rectangles, square: \(36+24+4\).

Direct answer: option C, \((a+b)^2\). In the area model, split both sides of a square of side \(a+b\) into lengths \(a\) and \(b\). This creates an \(a\) by \(a\) square of area \(a^2\), two rectangles measuring \(a\) by \(b\), each with area \(ab\), and a \(b\) by \(b\) square of area \(b^2\). Therefore, the total area is \(a^2+ab+ab+b^2=a^2+2ab+b^2=(a+b)^2\). Option A and option B describe the difference-of-squares arrangement; their total expression is \(a^2-b^2\), not a square with two positive middle rectangles. Option D is only a length, not an area model with four parts. The key clue is the two positive equal rectangles: they produce the middle term \(+2ab\). Memory cue: “plus square” means \(a^2\), two \(ab\) rectangles, and \(b^2\).

Direct answer: option A, \(r^2+rs\). The area of a rectangle is length multiplied by breadth. Split the length \(r+s\) into two parts, \(r\) and \(s\), while keeping the breadth \(r\). The first smaller rectangle has dimensions \(r\) by \(r\), so its area is \(r^2\). The second has dimensions \(s\) by \(r\), so its area is \(rs\). Adding the two non-overlapping areas gives \(r^2+rs\). Equivalently, the distributive law gives \(r(r+s)=r\cdot r+r\cdot s=r^2+rs\). Option B incorrectly uses \(s^2\), which would require a square with side \(s\). Option C doubles the product and does not represent the given rectangle. Option D is only the length, not an area. Memory cue: multiply the unchanged breadth by each split part: \(r\cdot r+r\cdot s\).

Direct answer: option C, \(x^2+5x+6\). Divide one side of the rectangle into lengths \(x\) and 2, and the other side into \(x\) and 3. The four smaller rectangles have areas \(x\cdot x=x^2\), \(x\cdot3=3x\), \(2\cdot x=2x\), and \(2\cdot3=6\). Add them: \(x^2+3x+2x+6=x^2+5x+6\). This is also obtained by distribution: \((x+2)(x+3)=x^2+3x+2x+6\). Option A leaves out the two middle products and the constant product. Option B incorrectly uses \(6x\) as the middle term; 6 is the product of 2 and 3, not the sum of the two \(x\)-terms. Option D does not include all four products. Memory cue: multiply every term on the first side by every term on the second side, then combine like terms.

Direct answer: option D, \(t^2+2t+1\). The area of a square is the square of its side, so the total area is \((t+1)^2\). In the visual model, split both sides into lengths \(t\) and 1. The four pieces are a \(t\) by \(t\) square with area \(t^2\), two \(t\) by 1 rectangles with total area \(t+t=2t\), and a 1 by 1 square with area 1. Thus the total is \(t^2+2t+1\). This is the identity \((a+b)^2=a^2+2ab+b^2\). Option A omits both middle rectangles. Option B has a negative middle term, which belongs to a difference such as \((t-1)^2\). Option C uses the wrong coefficient for the middle term; two rectangles give 2t, not 3t. Memory cue: square, two equal rectangles, square gives \(t^2+2t+1\).

The direct answer is option A. A square with side \(a\) has area \(a^2\), and a square with side \(b\) has area \(b^2\). If the smaller square is removed from the larger one, the remaining area is \(a^2-b^2\). In the visual rearrangement, the remaining pieces can be joined to make a rectangle whose side lengths are \(a+b\) and \(a-b\). Its area is therefore \((a+b)(a-b)\), proving the difference-of-squares identity. Option B is wrong because \((a+b)^2=a^2+2ab+b^2\), so it contains an extra \(2ab\). Option C is wrong because \((a-b)^2=a^2-2ab+b^2\), not the expression shown. Option D is not a valid general area identity. Memory cue: removing one square from another means subtracting squares; rearrangement gives the two factors.

The direct answer is option B. The square has side \(10-3\), so its area is \((10-3)^2\). Use the identity \((a-b)^2=a^2-2ab+b^2\), with \(a=10\) and \(b=3\). This gives \(10^2-2(10)(3)+3^2=100-60+9\). The value is also \(49\), which agrees with \(7^2\). Option A, \(100-9\), is the difference of two squares, not the square of a difference. Option C has the wrong sign for the middle term; it belongs to a sum pattern. Option D uses \(30\) instead of \(60\), because it forgets the factor 2 in \(2ab\). A useful check is to calculate the side first: \(10-3=7\), so the area must be 49.

The direct answer is option C. Imagine dividing each side of the large square into lengths \(a\), \(b\), and \(c\). The small regions then have two shapes. A matching length and width produce square regions such as \(a^2\), \(b^2\), and \(c^2\). Two different lengths produce rectangular regions such as \(ab\), \(ac\), and \(bc\); some occur twice because the grid is symmetric. Thus the expansion of \((a+b+c)^2\) contains square terms and rectangle terms. Option A is incomplete because it ignores mixed products. Option B is incomplete because it ignores the three square regions. Option D is wrong because the model describes areas, not lengths. Memory cue: equal factors make squares; unlike factors make rectangles.

The direct answer is option B. In an area model, multiply the two total side lengths: \((x+4)(x+2)\). Distributing gives \(x\cdot x=x^2\), \(x\cdot2=2x\), \(4\cdot x=4x\), and \(4\cdot2=8\). These are exactly the four listed regions, so the rectangle can have dimensions \(x+4\) and \(x+2\). Option A would give \(x^2+9x+8\), so its middle parts do not match. Option C expands to \(x^2-6x+8\), containing negative mixed terms rather than the stated positive ones. Option D gives different coefficients and constant term. The corner region is a helpful clue: the constants 4 and 2 multiply to 8.

The direct answer is option A. A large square can be divided into one square of side \(a\), with area \(a^2\), one square of side \(b\), with area \(b^2\), and two rectangles each having area \(ab\). Adding the four regions gives \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Their outer boundary is a square whose side is \(a+b\), so the same total area is \((a+b)^2\). Option B is wrong because the model has two rectangles, not one. Option C is wrong because \(ab\) is only the area of one mixed rectangle. Option D is wrong because the variables need not be zero. Memory cue: one square, two equal rectangles, and one square form the “square-plus-double-product” pattern.

The direct answer is option B. The area of a square is side multiplied by itself. Therefore a square with side \(p+q\) has area \((p+q)(p+q)=(p+q)^2\). If the square is divided into parts, those parts have areas \(p^2\), \(pq\), \(pq\), and \(q^2\), giving \(p^2+2pq+q^2\). Option A has area \((p+q)(p-q)=p^2-q^2\), the difference-of-squares expression. Option C has area \((p-q)^2\), a different identity with a negative middle term. Option D has area only \(pq\). The key distinction is that a squared bracket means the entire bracket is used as both dimensions.

The direct answer is option C. Add the area of every separate region in the diagram: the first square contributes \(a^2\), each of the two rectangles contributes \(ab\), and the second square contributes \(b^2\). Thus the total is \(a^2+ab+ab+b^2=a^2+2ab+b^2\). This is also the identity \((a+b)^2\), because the four regions can form a square with side \(a+b\). Option A omits both rectangular regions. Option B subtracts one rectangle even though all listed regions are being added. Option D subtracts both rectangles and represents \((a-b)^2\), not this diagram. A reliable method is to write each small area separately before combining like terms.

The direct answer is option D. The area of the square is \((z-1)^2\). Applying \((a-b)^2=a^2-2ab+b^2\), with \(a=z\) and \(b=1\), gives \(z^2-2z+1\). The square term \(z^2\) is positive, the mixed term is \(-2z\), and the constant square \(1^2\\) is positive. Among the listed choices, the positive term asked for is \(1\). Option A is negative because it is the middle term. Option B incorrectly changes the sign of \(z^2\); squaring does not make this term negative. Option C is the negative of the constant square, while the constant contribution is positive. Memory cue: in \((z-1)^2\), the signs are plus, minus, plus.

The direct answer is option A. The square contributes area \(a\times a=a^2\), and the rectangle contributes area \(a\times b=ab\). Adding them gives \(a^2+ab\). Both terms contain the common factor \(a\), so factorising gives \(a(a+b)\). Geometrically, the two pieces share the same width \(a\); their lengths \(a\) and \(b\) combine to make a total length \(a+b\). Therefore the joined shape has area \(a(a+b)\). Option B is only a length, not an area. Option C incorrectly treats the sum as two separate squares and omits the mixed product. Option D has the wrong powers and does not equal \(a^2+ab\) in general. Memory cue: common width times combined length gives the factored form.

The direct answer is option B. In a rectangle model, each small area comes from multiplying one segment on one side by one segment on the other side. The four regions can be identified as \(x\cdot x=x^2\), \(x
e5=5x\), \(1
e x=x\), and \(1
e5=5\). Thus the two complete side lengths are \(x+5\) and \(x+1\), giving \((x+5)(x+1)\). Its expansion is \(x^2+5x+x+5=x^2+6x+5\), exactly the sum of the four parts. Option A would create \(x^2+10x+25\). Option C creates negative mixed terms. Option D uses side segments 6 and 5 and gives different regions. Memory cue: read the two outer side labels, not just one interior piece.

The direct answer is A, the area model. Begin with the key idea: the area of one fixed figure does not change merely because we calculate it in a different way. In an area model, a rectangle or square is divided into smaller parts. We can find the whole area by multiplying its complete length and breadth, or by adding the areas of all the smaller pieces. Because both calculations describe the same figure, the two algebraic expressions must be equal. For example, a square of side a+b has area (a+b)^2. Dividing it into four parts gives a^2+ab+ab+b^2, so (a+b)^2=a^2+2ab+b^2. A is correct because it directly uses equal areas to prove an identity. B represents positions and distances on a line, not usually a partitioned-area proof. C may show time relationships but is not the standard model here. D organizes values in rows and columns but does not itself show that two expressions represent the same geometric area. Memory cue: same shape, same area, two calculations means an area proof.