Chapter 6: Perimeter and Area (Mensuration)

a. $2\times (l+2)=14\Rightarrow l+2=7\Rightarrow l=$ 5 cm

b. $4\times s=20\Rightarrow s=$ 5 cm

c. $2\times (3+b)=12\Rightarrow 3+b=6\Rightarrow b=$ 3 m (it is a square!)

a. 5 cm b. 5 cm c. 3 m

Length of wire = perimeter of the rectangle $=2\times (5+3)=16$ cm. The square has the same perimeter, so its side $=16÷4=$ 4 cm.

4 cm

Third side $=55-(20+14)=55-34=$ 21 cm.

21 cm

Perimeter $=2\times (150+120)=540$ m. Cost $=540\times 40=$ ₹21,600.

₹21,600

a. $36÷4=$ 9 cm b. $36÷3=$ 12 cm c. $36÷6=$ 6 cm

a. 9 cm b. 12 cm c. 6 cm

One round = perimeter $=2\times (230+160)=780$ m. Three rounds $=3\times 780=$ 2340 m.

2340 m

One round $=2\times (70+40)=220$ m, so 5 rounds $=5\times 220=$ 1100 m.

1100 m

One round $=2\times (60+30)=180$ m, so 7 rounds $=7\times 180=$ 1260 m. Toshi ran the longer distance (1260 m > 1100 m), even though her track is smaller.

Toshi: 1260 m, more than Akshi's 1100 m.

Both girls start at the bottom-right corner of their own track and run clockwise (first along the bottom, then up the left side).

• a. A: 250 = 220 + 30, so 30 m along the bottom from her start.
• b. B: 500 = 2 × 220 + 60, so 60 m along the bottom (10 m before the bottom-left corner).
• c. 1000 = 4 × 220 + 120: 4 full rounds; then 70 m (bottom) + 40 m (left side) + 10 m along the top: C is 10 m from the top-left corner.
• d. X: 250 = 180 + 70: 60 m along the bottom and 10 m up the left side.
• e. Y: 500 = 2 × 180 + 140: 60 + 30 + 50, so 50 m along the top from the top-left corner.
• f. 1000 = 5 × 180 + 100: 5 full rounds; then 60 + 30 + 10, so Z is 10 m along the top from the top-left corner.

A: 30 m along the bottom; B: 60 m along the bottom; C: 4 full rounds, 10 m along the top; X: 10 m up the left side; Y: 50 m along the top; Z: 5 full rounds, 10 m along the top.

Work backwards 350 m from the finishing line (the middle of the bottom side), against the running direction:

• Inner track (A): 50 m to the corner + 100 m + 100 m + 100 m = 350 m. So A is at the corner at the end of the side opposite the finish (the bottom-right corner when running anticlockwise): A runs 100 + 100 + 100 + 50 = 350 m.
• Outer track (B): 75 m to the corner + 150 m (side) = 225 m; 125 m more along the top side. So B is on the top side, 125 m from the top-left corner: B runs 125 + 150 + 75 = 350 m.

A: at the corner of the inner track (100 + 100 + 100 + 50 m). B: on the top side of the outer track, 125 m from the corner (125 + 150 + 75 m).

Toshi is right. The perimeter is 6 straight units + 3 diagonal units (6s + 3d). A diagonal of a unit square is longer than its side (it is the slanted line across the square), so 3 diagonal units are more than 3 straight units. The perimeter is more than 9 units.

Toshi is right: each diagonal unit is longer than a straight unit, so 6s + 3d is more than 9 units.

Count the straight (horizontal or vertical) segments and the slanted (diagonal) segments between neighbouring dots:

8s + 2d, 4s + 6d, 12s + 6d and 18s + 6d (from left to right).

8s + 2d, 4s + 6d, 12s + 6d, 18s + 6d

Both are regular polygons: all their sides are equal and all their angles are equal. So the perimeter is simply (number of sides) × (length of a side): square 4 × side, equilateral triangle 3 × side, and in general:

$$\text{Perimeter of a regular polygon}=\text{number of sides}\times \text{length of a side}$$

(e.g. a regular hexagon of side 5 cm has perimeter 6 × 5 = 30 cm).

Both have all sides and angles equal; perimeter of a regular polygon = number of sides × side.

Each piece has perimeter $2\times (6+2)=16$ cm, so both together have 32 cm of edge. When the pieces touch, the touching part is not on the boundary, and it is lost from both pieces:

$$\text{Perimeter}=32-2\times (\text{length of the touching edge})$$

• b. (L-shape) touching edge 2 cm: $32-4=$ 28 cm
• c. (T-shape) touching edge 2 cm: $32-4=$ 28 cm
• d. (side by side, shifted by 3 cm) touching edge 6 − 3 = 3 cm: $32-6=$ 26 cm

For 22 cm, the touching edge must be $(32-22)÷2=5$ cm: put the strips side by side along their long sides, shifted by 1 cm.

b. 28 cm c. 28 cm d. 26 cm. For 22 cm, place the strips side by side, shifted by 1 cm (touching along 5 cm).

Width $=300÷25=$ 12 m.

12 m

Area $=500\times 200=1,00,000$ sq m = 1000 hundreds of sq m. Cost $=1000\times 8=$ ₹8000.

₹8000

Area $=100\times 50=5000$ sq m. Number of trees $=5000÷25=$ 200.

200 trees

a. The staircase figure is 7 m wide and 7 m tall. Cut it with horizontal lines into four rectangles:

b. The figure is a 5 × 3 rectangle with a 3 × 2 rectangle cut out of the bottom middle (the two legs are 1 m wide). Area $=5\times 3-3\times 2=15-6=$ 9 sq m. (Or split it: top strip 5 × 1 = 5, two legs 1 × 2 = 2 each: 5 + 2 + 2 = 9.)

a. 28 sq m b. 9 sq m

• A and B (the two large triangles) have the same area.
• C and E (the two small triangles) have the same area.
• D, F and G have the same area: each can be covered exactly by the two small triangles C and E.

A = B; C = E; and D, F and G are equal (each equals C + E).

D is 2 times as big as C, because D can be covered exactly by C and E, and C and E are equal: D = C + E (= 2 × C).

D is twice C; D = C + E.

Neither: they are equal. Both can be covered exactly by the two small triangles C and E, so each has twice the area of C.

Equal: each is made of two small triangles.

Equal. F and G can each be covered exactly by the two small triangles C and E (try placing them over each other), so they have the same area.

Equal: both are twice the area of C.

A (a large triangle) can be covered by two of the 2-unit pieces (for example D and two small triangles, or the square and two small triangles), so A = 4 × C, while G = 2 × C. Shape A is twice as big as Shape G (not four times).

A is twice as big as G (A = 4 × C, G = 2 × C).

Measuring everything in units of C:

$$A+B+C+D+E+F+G=4+4+1+2+1+2+2=16$$

The big square has an area of 16 times the area of Shape C.

16 times the area of C.

Still 16 times the area of C. The rectangle is made of exactly the same seven pieces, without gaps or overlaps, so its area is the sum of the areas of the pieces, which does not change when we rearrange them.

16 times the area of C, because the same pieces are used.

They are different. When the pieces are rearranged, different edges come to the outside, so the boundary changes even though the area stays the same. The rectangle is longer and thinner than the square, so its perimeter is larger (of all rectangles with the same area, the square has the smallest perimeter).

Different: the area stays the same but the boundary changes; the rectangle has a larger perimeter than the square.

Counting full squares and halves (two half squares make one square unit):

about 4, 9, 10 and 11 sq units (from left to right).

About 4, 9, 10 and 11 square units.

Circles cannot be packed together without gaps, so a region filled with circles still has uncovered parts; the count depends on how they are arranged (the same rectangle holds 42 circles one way and 44 another). Triangles and rectangles can fill a space without gaps, but squares are the best because:

• they fit together with no gaps and no overlaps, in rows and columns;
• all squares of the same size are identical, whichever way they are turned;
• a square has equal length and breadth, so counting rows × columns gives the area directly (length × breadth);
• they match the units of length (a 1 cm square has area 1 sq cm).

Circles leave gaps; squares fill a region exactly in rows and columns, have equal sides and match the length units, so they are the best unit of area.

a. 1 × 24 (perimeter 50). b. 4 × 6 (perimeter 20).

c. Greatest: 1 × 32 cm (66 cm); least: 4 × 8 cm (24 cm).

Prediction: the longest, thinnest rectangle (breadth 1) has the greatest perimeter, and the rectangle whose sides are closest to each other (closest to a square) has the least perimeter. For example, for area 36 the 6 × 6 square has perimeter 24, the least.

For 24: greatest 1 × 24 (50), least 4 × 6 (20). For 32: greatest 1 × 32 (66 cm), least 4 × 8 (24 cm). Thinnest gives the greatest perimeter; closest to a square gives the least.

The two triangles cut along a diagonal overlap exactly, so each triangle has half the area of the rectangle. For triangle ABE (whose top corner E is on side DC), the line EF splits the rectangle into two rectangles AFED and BFEC, and each part of the triangle is half of one of them, so again:

$$\text{Area of triangle ABE}=\frac{1}{2}\times \text{area of rectangle ABCD}$$

Conclusion: the area of a triangle is half the area of a rectangle that has the same base and the same height as the triangle.

Each triangle (BAD and ABE) has half the area of rectangle ABCD: a triangle's area is half that of the rectangle with the same base and height.

a. 24 b. 30 c. 48 d. 16 e. 12 square units

Each square has 4 sides (36 in all); each place where two squares touch hides 2 of these sides. So perimeter = 36 − 2 × (number of touching edges).

1. 12 units: the 3 × 3 square has the most touching edges (12).
2. 20 units: when the squares touch the fewest times (8 touches, e.g. a straight row of 9).
3. 18 units: 9 touching edges, e.g. a row of 7 squares with 2 squares on top of one end (one 2 × 2 block).

4. Perimeter 12: only one shape (the 3 × 3 square). Perimeters 20 and 18: many shapes: for 20, any shape without a 2 × 2 block (a row, an L, a T, a zig-zag…); for 18, any shape with exactly one 2 × 2 block.

1. 12 (3 × 3 square only) 2. 20 (e.g. a row of 9) 3. e.g. a row of 7 with 2 squares on one end 4. Only the 3 × 3 gives 12; many shapes give 18 or 20.

A new square adds its 4 sides but hides every side where it touches the figure (each touch removes 1 side of the new square and 1 side of the figure):

• touches 1 side: perimeter increases by 2 (24 → 26);
• touches 2 sides (in a corner): perimeter stays the same (24);
• touches 3 sides (in a notch): perimeter decreases by 2 (24 → 22).

Touching 1 side: +2 (increases); 2 sides: no change; 3 sides (a notch): −2 (decreases).

Width of the plot = master bedroom 15 + toilet 5 + kitchen 15 = 35 ft.

b. Area of the house (the whole plot) $=35\times 30=$ 1050 sq ft. (The rooms add up to 1025 sq ft; the remaining 25 sq ft is the 5 ft × 5 ft passage below the toilet.)

Small bedroom 15 × 12 (180), utility 15 × 3 (45), garden 20 × 3 (60), parking 15 × 3 (45), hall 20 × 12 (240); house area 35 ft × 30 ft = 1050 sq ft.

Depth of the house = master bedroom 15 + small bedroom 10 = 25 ft.

b. Area $=42\times 25=$ 1050 sq ft (check: 180 + 120 + 50 + 180 + 70 + 345 + 105 = 1050).

Comparison: both houses have the same area (1050 sq ft), but different perimeters: Charan's $2\times (35+30)=130$ ft and Sharan's $2\times (42+25)=134$ ft. Sharan's house has the longer boundary.

Utility 7 × 10, toilet 5 × 10, hall 23 × 15 (345), entrance 7 × 15 (105), small bedroom 120 sq ft; area 42 × 25 = 1050 sq ft. Same area as Charan's, but a longer perimeter (134 ft vs 130 ft).

a. The left pieces 13 and 15 have the same width; the right pieces 26 and ? have the same width. Since 26 = 2 × 13, the right column is twice as wide, so ? = 2 × 15 = 30 sq cm.

b. Bottom rectangle: height 2 cm, area 10 sq cm → length 5 cm. The upright piece stands on the last 5 − 3 = 2 cm, so its height = 10 ÷ 2 = 5 cm. The pink square is 3 cm wide and 5 − 2 = 3 cm tall: ? = 3 × 3 = 9 sq cm.

c. Middle piece: 42 ÷ 6 = 7 cm wide. Bottom piece: 7 + 5 = 12 cm wide, so 60 ÷ 12 = 5 cm tall. Top piece: 15 − 6 − 5 = 4 cm tall and 7 − 3 = 4 cm wide: ? = 4 × 4 = 16 sq cm.

d. Small piece: 18 ÷ 5 = 3.6 cm tall. The big piece is 4 + 3.6 = 7.6 cm tall, so its width ? = 38 ÷ 7.6 = 5 cm.

a. 30 sq cm b. 9 sq cm c. 16 sq cm d. 5 cm

Total area $=50+14=64$ sq m. Possible rectangles: 8 m × 8 m, 16 m × 4 m, 32 m × 2 m or 64 m × 1 m.

Any rectangle of area 64 sq m, e.g. 8 m × 8 m or 16 m × 4 m.

Width $=1000÷50=$ 20 m.

20 m

$$5\times 4-3\times 3=20-9=11\text{sq m}$$

11 sq m

$$15\times 12-4\times (2\times 1)=180-8=172\text{sq m}$$

172 sq m

Take A = a 1 × 18 rectangle (area 18, perimeter 2 × 19 = 38 units) and B = a 4 × 5 rectangle (area 20, perimeter 2 × 9 = 18 units). A has the smaller area but the much longer perimeter. (Another pair: A = 2 × 9, perimeter 22; B = 4 × 5, perimeter 18.)

For example A = 1 × 18 (perimeter 38) and B = 4 × 5 (perimeter 18).

Measure your page first. If the page is $L$ cm tall and $B$ cm wide, the border is $(L-2)$ cm tall and $(B-3)$ cm wide, so its perimeter is $2\times (L+B)-10$ cm.

For example, for a page 28 cm × 20 cm: border = 26 cm × 17 cm, perimeter $=2\times (26+17)=$ 86 cm.

Perimeter = (perimeter of the page) − 10 cm; e.g. 86 cm for a 28 cm × 20 cm page.

Area of the outer rectangle $=12\times 8=96$ sq units, so the inner one must have area 48 sq units, and must be less than 12 × 8 in both directions. An 8 × 6 rectangle works, placed 2 units from the left and right and 1 unit from the top and bottom.

An 8 × 6 rectangle (48 sq units) placed in the middle of the 12 × 8 rectangle.

Let the side of the square be $s$. Each rectangle is $s\times \frac{s}{2}$.

• Perimeter of the square $=4s$.
• Perimeter of each rectangle $=2\times (s+\frac{s}{2})=3s$; both together $=6s=1\frac{1}{2}\times 4s$.

So (c) is always true. (a) is false: each rectangle is half the square. (b) is false: 4s < 6s. (d) is false: the two rectangles together have the same area as the square.

(c): together the rectangles have perimeter 6s, which is 1½ times the square's 4s.