Measuring Space: Perimeter and Area | EOT

Question 1

Identities in algebra can sometimes be shown as area relationships. For example:

The figure shown corresponds to the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Do you see how?

Draw figures corresponding to the identities (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2 and (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.

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Solution
Understand the Question
  • Algebraic identities can be interpreted geometrically by comparing the area of a large geometric figure with the sum of the areas of its component parts.
  • For (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, a square of side (a+b)(a + b) is divided into four regions: two squares of areas a2a^2 and b2b^2, and two rectangles each of area abab.
  • We can geometrically represent (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2 and (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca by constructing appropriate squares and rectangles.

(i) Draw a figure corresponding to the identity (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2.

Step 1 · Geometric Representation of (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2

Consider a square of side length aa with total area a2a^2.

Remove a smaller square of side length bb from one corner: Area of remaining region=a2b2\text{Area of remaining region} = a^2 - b^2

Divide the remaining L-shaped region into two rectangles:

  • Rectangle 1 with length aa and breadth (ab)(a - b), having area a(ab)a(a - b)
  • Rectangle 2 with length bb and breadth (ab)(a - b), having area b(ab)b(a - b)

Combining these two rectangles side-by-side forms a single rectangle of length (a+b)(a + b) and breadth (ab)(a - b): Total Area=a(ab)+b(ab)=(a+b)(ab)\text{Total Area} = a(a - b) + b(a - b) = (a + b)(a - b)

Equating the two expressions for the area: (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2

Answer

(i) (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2

(ii) Draw a figure corresponding to the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.

Step 1 · Geometric Representation of (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Consider a square of side length (a+b+c)(a + b + c): Total Area=(a+b+c)2\text{Total Area} = (a + b + c)^2

Divide each side into three segments of lengths aa, bb, and cc. This splits the large square into 99 smaller regions:

  • 33 squares of areas a2a^2, b2b^2, and c2c^2
  • 22 rectangles of dimensions a×ba \times b, each of area abab
  • 22 rectangles of dimensions b×cb \times c, each of area bcbc
  • 22 rectangles of dimensions c×ac \times a, each of area caca

Summing the areas of all 99 parts: Total Area=a2+b2+c2+2ab+2bc+2ca\text{Total Area} = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Therefore: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Answer

(ii) (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Common Mistakes
  • Missing Cross-Terms: Forgetting the rectangular regions (2ab2ab, 2bc2bc, 2ca2ca) and mistakenly assuming (a+b+c)2=a2+b2+c2(a + b + c)^2 = a^2 + b^2 + c^2.
  • Incorrect Dimension Splitting: In (a+b)(ab)(a + b)(a - b), failing to notice that the common breadth (ab)(a - b) allows the two sub-rectangles to combine into one rectangle of length (a+b)(a + b).

More questions in EOT

Q1

Identities in algebra can sometimes be shown as area relationships. For example:

The figure shown corresponds to the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Do you see how?

Draw figures corresponding to the identities (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2 and (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.

Q2

An isosceles triangle has perimeter 40 cm40\text{ cm}; the equal sides are 15 cm15\text{ cm} each. Find the area of the triangle.

Q3

An isosceles triangle has base 10 cm10\text{ cm}, and its area is 60 cm260\text{ cm}^2. What are the lengths of the equal sides?

Q4

The area of a right-angled triangle is 54 sq. cm54\text{ sq. cm}. One of its legs has length 12 cm12\text{ cm}. Find its perimeter.

Q5

The sides of a triangle are in the ratio 2:3:42 : 3 : 4, and its perimeter is 45 cm45\text{ cm}. Find its area.

Q6

The sides of a triangle have lengths 7 cm7\text{ cm}, 24 cm24\text{ cm}, 25 cm25\text{ cm}. Find the area of the triangle in two different ways.

Q7

If the wheel of a bicycle has a diameter of 60 cm60\text{ cm}, find how far a cyclist will have travelled after the wheel has rotated 100100 times.

Q8

Find the area of a quadrant of a circle whose circumference is 66 cm66 \text{ cm}.

Q9

The wheel of a car has an outer radius of 28 cm28\text{ cm}. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km1\text{ km}.

Q10

Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Q11

You know that the area of a parallelogram is base ×\times height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides ×\times height, i.e., 12(a+b)h\dfrac{1}{2}(a + b)h.

Q12

By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).

Q13

Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?

Q14

Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.

Q15

Three problems about fitting congruent shapes together:

(i) Rectangle ABCD has sides aa, bb, and rectangle PQRS has sides 2a2a, 2b2b. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? Check and see!

(ii) ΔABC\Delta ABC has sides aa, bb, cc, and ΔPQR\Delta PQR has sides 2a2a, 2b2b, 2c2c. Show that ΔPQR\Delta PQR has 4 times the area of ΔABC\Delta ABC. Does this mean that 4 copies of ΔABC\Delta ABC will fit into ΔPQR\Delta PQR? Check and see!

(iii) ΔABC\Delta ABC has sides aa, bb, cc, and ΔPQR\Delta PQR has sides 3a3a, 3b3b, 3c3c. Show that ΔPQR\Delta PQR has 9 times the area of ΔABC\Delta ABC. Does this mean that 9 copies of ΔABC\Delta ABC will fit into ΔPQR\Delta PQR? Check and see!

Q16
  1. Find the fraction of the shaded region in each of the following figures:

(i) Fig. 6.43: What fraction of the triangle is shaded? (ii) Fig. 6.44: What fraction of the square is shaded?

Q17

Find the fraction of the rectangle covered by the circles in each of the following figures:

(i) Fig. 6.45: What fraction of the rectangle is covered by the circles?

(ii) Fig. 6.46: What fraction of the rectangle is covered by the circles?

Q18

Use the above to make a conjecture about the area occupied by circles fitted into a rectangle in the manner shown. Test your conjecture for particular cases: 10 circles; 20 circles; 50 circles. Then prove your conjecture!

Q19

*19. The figure shows nine identical rectangles fitted together to make a large rectangle whose area is 72 cm272\text{ cm}^2. Find the perimeter of each small rectangle.

Q20

Show that the areas of the shaded blue triangle and the shaded red triangle are equal.

Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.

Q21

The figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions AA and BB.

Show that AA and BB have equal area.

Q22

In Fig. 6.50, four semicircles have been drawn within the given square whose side is 2 units2 \text{ units}. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.

Q23

In Fig. 6.51 we see two concentric circles with a common centre OO. A chord BCBC of the larger circle is drawn, touching the smaller circle at AA. The length of BCBC is ll. Show that the area of the green region enclosed between the two circles is 14πl2\dfrac{1}{4} \pi l^2.

Q24

In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A)+Area (B)=Area (C)\text{Area (A)} + \text{Area (B)} = \text{Area (C)}.

Q25

Fig. 6.53 shows two circles passing through each other's centres. Find the area of the region enclosed by the two circles in terms of the common radius rr.

Q26

In Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A,B,CA, B, C, as marked. Show that the area of the rectangle is

2(A+C)(B+C)C\dfrac{2(A + C)(B + C)}{C}

Q27

In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle.

Show that the areas of the two shaded regions are equal.

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