NCERT Solutions Ganita Prakash (Part 2) Chapter 7 – 152Section 7.1 Rectangle and Squares — Figure it Out
Book page 150 Updated on2026-09-05
Q1.
Identify the missing sidelengths. [Figure (i): a pinwheel of four rectangles of areas 28 in², 21 in², 35 in² and 14 in², with 4 in, 7 in, 3 in and 2 in marked and one side marked “? in”.] [Figure (ii): a rectangle of area 50 m² whose top strip of height 4 m has area 29 m², with a further 11 m² rectangle attached on its right; three sidelengths are marked “?”.]
Answer
(i) ? = 2 in. Each rectangle hands you a length that the next one needs.
Four rectangles round one point. Each one shares a full side with the next, so the chain of deductions never breaks.
The right edge of the 28 in² rectangle = 4 in (given) + 3 in (the height of the 21 in² rectangle) = 7 in
So its other side = 28 ÷ 7 = 4 in
Top of the 35 in² rectangle = 3 in (overhang) + 4 in = 7 in
So its height = 35 ÷ 7 = 5 in
Left edge of the 14 in² rectangle = 5 in + 2 in (the stub below) = 7 in
So ? = 14 ÷ 7 = 2 in
(The 21 in² rectangle starts the chain: its width is the given 7 in, so its height is 21 ÷ 7 = 3 in.)
(ii) The top strip has height 4 m throughout, so divide each area by 4.
Width of the 29 m² part = 29 ÷ 4 = 7.25 m
Width of the 11 m² part = 11 ÷ 4 = 2.75 m
The bold rectangle has area 50 m², and its top part is 29 m²,
so its lower part = 50 − 29 = 21 m²
Its width is the same 7.25 m, so
height of the lower part = 21 ÷ 7.25 = 84/29 = 2.90 m (approx.)
Check it yourself: the whole top strip is 7.25 + 2.75 = 10 m long and 4 m high, giving 40 m² = 29 + 11. ✓
Why it happens: whenever two rectangles sit side by side sharing a full edge, that edge is a common factor of both areas. Dividing an area by the shared side is what recovers the unknown side.
Q2.
The figure shows a path (the shaded portion) laid around a rectangular park EFGH. (i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area. An example of a formula — Area of a rectangle = length × width. [Hint: There is a relation between the areas of EFGH, the path, and ABCD.] (ii) If the width of the path along each side is given, can you find its area? If not, what other measurements do you need? Assign values of your choice to these measurements and find the area of the path. Give a formula for the area using these measurements. [Hint: Break the path into rectangles.] (iii) Does the area of the path change when the outer rectangle is moved while keeping the inner rectangular park EFGH inside it, as shown?
Answer
(i) Four measurements are enough: the two sidelengths of the outer rectangle ABCD and the two sidelengths of the park EFGH.
Area of ABCD = Area of EFGH + Area of the path
so Area of the path = Area of ABCD − Area of EFGH Area of the path = (AB × BC) − (EF × FG)
Taking AB = 20 m, BC = 15 m, EF = 16 m, FG = 11 m:
Area of the path = (20 × 15) − (16 × 11) = 300 − 176 = 124 m²
(ii) The four widths alone are not enough — you also need the length and width of the park. With the park l by b, and the path of width p on the left, q on the right, t on top and d at the bottom, break the path into eight rectangles: four strips and four corners.
top and bottom strips: l(t + d)
left and right strips: b(p + q)
the four corner rectangles: pt + qt + pd + qd = (p + q)(t + d)
Area of the path = l(t + d) + b(p + q) + (p + q)(t + d)
With l = 16 m, b = 11 m and every width 2 m (so p = q = t = d = 2):
= 16 × 4 + 11 × 4 + 4 × 4 = 64 + 44 + 16 = 124 m²
Same answer as (i) — as it must be, since the outer rectangle is then 20 m by 15 m.
(iii) No, it does not change.
Why it happens: however the outer rectangle is shifted, its size is unchanged and the park's size is unchanged. The path is exactly "outer minus inner", so its area stays Area(ABCD) − Area(EFGH). What changes is only the shape of the path — wide on one side, narrow on the other. Area is preserved; the shape is not.
Q3.
The figure shows a plot with sides 14m and 12m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose. [Math Talk]
Answer
You need the widths of the two strips — nothing else.
Let the plot be 14 m long and 12 m wide. Let the strip that runs across the length have width a, and the strip that runs across the width have width b.
Horizontal strip = 14 × a
Vertical strip = 12 × b
But the little square where they cross has been counted twice, and it measures a × b.
Area of the crosspath = 14a + 12b − ab
Choosing a = 2 m and b = 1 m:
= 14 × 2 + 12 × 1 − 2 × 1
= 28 + 12 − 2 = 38 m²
Check it yourself: the crosspath leaves four rectangles of grass. Along the 14 m side the path takes away 1 m, leaving 13 m; along the 12 m side it takes away 2 m, leaving 10 m. So the grass measures 13 × 10 = 130 m², and 130 + 38 = 168 m² = 14 × 12. ✓
Why it happens: the crossing square belongs to both strips. Adding the two strip areas counts it twice, so it must be subtracted once. This is the same "add, then remove the overlap" idea used whenever two regions meet.
Q4.
Find the area of the spiral tube shown in the figure. The tube has the same width throughout. [Hint: There are different ways of finding the area. Here is one method.] What should be the length of the straight tube if it is to have the same area as the bent tube on the left?
Answer
Area of the spiral = 112 sq. units.
First settle the hint. The bent tube is an L whose two outer arms are 5 and 5, and whose width is 1.
Area of the L = 5 × 1 + 5 × 1 − 1 × 1 = 9 sq. units
(the corner square lies in both arms, so it is subtracted once)
A straight tube of width 1 and length ℓ has area ℓ × 1.
So ℓ = 9 units.
The same idea straightens the whole spiral. Every bend costs one unit square. Add all nine outer arms, then subtract 1 for each of the 8 bends.
Arm (measured along the outer edge)
1
2
3
4
5
6
7
8
9
Total
Length
20
20
20
15
15
10
10
5
5
120
Straightened length = 120 − 8 = 112 units
Width = 1 unit Area = 112 × 1 = 112 sq. units
Why it happens: unrolling the tube is a dissection — cut it at the bends and lay the pieces end to end. Nothing is added or thrown away, so the area is unchanged. Each bend is a square of side 1 shared by the two arms meeting there, which is why it is counted once instead of twice.
Check it yourself: the inner edge of the spiral measures 19 + 18 + 18 + 13 + 13 + 8 + 8 + 3 + 4 = 104 units. The average of the outer and inner edges is (120 + 104) ÷ 2 = 112 — the length of the middle line of the tube, and the same answer again.
Q5.
In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.
Answer
Every region becomes 4 times as big, so each increases by 3 times its original area.
In the figure the diagonal from the bottom-left corner to the top-right corner splits the square in half. Region 3 is that lower-right half. The segment from the top-left corner to the centre of the square then splits the upper half into regions 1 and 2. Let the sidelength be a.
Region
Side a
Side 2a
Increase
1
a²/4
a²
3a²/4
2
a²/4
a²
3a²/4
3
a²/2
2a²
3a²/2
Region 3 = half the square = a²/2 → (2a)²/2 = 4a²/2 = 2a²
The centre is the midpoint of the diagonal, so the segment from the top-left corner is a median of the upper triangle
Region 1 = Region 2 = ½ × a²/2 = a²/4 → a²
Why it happens: doubling the sidelength doubles every length in the figure, because all the cuts are described by the corners and the centre. A region that was l by w becomes 2l by 2w, so its area is multiplied by 2 × 2 = 4. Areas scale by the square of the length factor, so "twice as long" means "four times the area" — never twice.
Tip: the same rule answers many questions at once. Triple the sidelength and every area is multiplied by 9; halve it and every area is divided by 4.
Q6.
Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure. Rearrange the pieces to get a larger square, with a hole inside. You can try this activity by constructing the square using cardboard, thick chart paper, or similar materials. [Math Talk]
Answer
Draw the two cuts as in the figure: one line from a point on the left side to a point on the right side, and the other perpendicular to it, from a point on the top side to a point on the bottom side. Cut out the four pieces.
Now slide the four pieces outwards, keeping each one's slant direction, until each cut edge of one piece lies against the matching cut edge of the next. They close up into a bigger square with a square hole in the middle.
The four pieces are only moved, never stretched — so the hole is exactly the extra area of the bigger square.
Area of the four pieces = area of the original square = s² (unchanged)
Area of the big square = area of the pieces + area of the hole
so area of the hole = L² − s²
The side L of the new square is the length of each cut, and the side of the hole is the offset d — the distance between the two points where one cut meets the pair of opposite sides. So L² = s² + d², exactly the Baudhāyana–Pythagoras relation.
Why it happens: a dissection can never create or destroy area — that is the one thing which stays invariant when pieces are rearranged. The puzzle only looks paradoxical because the outline has grown; the growth is precisely accounted for by the hole. If you make the two cuts pass exactly through the centre, all four pieces are congruent and the hole is a neat square in the middle.
Try This: make the two cuts nearly parallel to the sides. The offset d becomes tiny, and so does the hole. Make them steep and the hole grows.