Q: Which has a greater area — an equilateral triangle or a square of the same side length as the triangle? Which has a greater area: two identical equilateral triangles together or a square of the same side length as the triangle? Give reasons.

Area of equilateral triangle = $\frac{\sqrt{3}}{4} a^2$ Area of square = a 2 ⇒ $\frac{\sqrt{3}}{4} a^2$

Question 1

Identify the missing sidelengths.

Accepted Answer

(i) After naming the figure
In rectangle ABCD,
Area of rectangle = Length × Breadth
7 × BC = 21
⇒ BC = 3 in
∴ AF = AD + DF
= 3 in + 4 in
= 7 in
In the rectangle EFAG,
EF × AF = 28 in
2
⇒ EF × 7 in = 28 in
2
⇒ EF = 4 in
∴ HA = HG + GA
= 3 in + 4 in
= 7 in
In rectangle HIJA,
Area = HA × AJ
⇒ 35 in
2
= 7 in × AJ
⇒ AJ = 5 in
∴ AK = AJ + JK
= 5 in + 2 in
= 7 in
In rectangle KLMA,
Area = KL × LM
⇒ x in × 7 in = 14 in
2
⇒ x = 2 in
Thus, the missing sidelength = 2 in
(ii) After naming the figure,
In rectangle ABGH,
Area = AB × AH
AB × 4 m = 29 m²
AB = $\frac {29}{4}$ m or 7.25 m
Area of rectangle HGDC = Area of rectangle ABDC – Area of rectangle ABGH
= 50 m
2
– 29 m
2
= 21 m
2
In rectangle HGDC,
CD × GD = 21 m
2
⇒ $\frac {29}{4}$m × GD = 21 m
2
⇒ GD = $\frac {84}{29}$ m or 2.9 m
In rectangle BEFG,
BG × BE = 11 m
2
⇒ 4 m × M = 11 m
2
⇒ BE = $\frac {11}{4}$m or 2.75 m
Thus, AB = $\frac {29}{4}$ m; BE = $\frac {11}{4}$ m and GD = $\frac {84}{29}$ m

Question 2

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

Accepted Answer

(i) Measurements needed:
Length of outer rectangle ABCD = A
Width of outer rectangle ABCD = B
Length of inner rectangle EFGH = a
Width of inner rectangle EFGH = b
Let A = 10, B = 8, a = 6, b = 4.
Calculation: Area of path = Area of ABCD – Area of EFGH
= 10 × 8 – 6 × 4
= 80 – 24
= 56 m
2
Formula:
Area of path = (A × B) – (a × b)
(ii) Yes, if the width of the path is uniform along each side, we can find its area, but we also need the dimensions of either the outer or inner rectangle.
If the width of the path d = 2 m (Uniform on all sides)
Length of inner path EFGH = l = 16 m
Width of inner path EFGH = w = 11 m
Length of outer rectangle = l + 2d
= 16 + 2(2)
= 20 m
Width of outer rectangle = w + 2d
= 11 + 2(2)
= 15 m
Breaking the path into rectangles
Now there are 4 rectangles
Left rectangle = w × d
= 11 × 2
= 22 m
2
Right rectangle = w × d
= 11 × 2
= 22 m
2
Top rectangle = (l + 2d) d
= 20 × 2
= 40 m
2
Bottom rectangle = (l + 2d) d
= 20 × 2
= 40 m
2
Total area of path = 22 + 22 + 40 + 40 = 124 m
2
Formula:
Area of path = 2d(l + w) + 4d
where d – width of path
l – length of inner path
w – width of inner path
(iii) No, the area of the path does not change.
Reason: The area of the path depends only on:
Area of outer rectangle ABCD.
Area of inner rectangle EFGH.

Question 3

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.

Accepted Answer

Measurements needed:
Length of plot = 14 m
Width of plot = 12 m
Width of horizontal path = x
1
Width of vertical path = x
2
Assign values:
Let x
1
= 2 m
x
2
= 2 m
Now area of horizontal path = 14 × 2 = 28 m
Area of vertical path = 12 × 2 = 24 m
Area of overlapping square = 2 × 2 = 4 m
2
∴ Area of cross path = 28 + 24 – 4 = 48 m
2
Formula = Area of cross path = (L × w
1
) + (W × w
2
) – (w
1
× w
2
)
Here, L = length of plot
W = Width of plot
w
1
= Width of horizontal path
w
2
= Width of vertical path

Question 4

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?

Accepted Answer

After naming the figure,
The area of the spiral tube = Area of the rectangle, ABEC + Area of the rectangle, DEGF + Area of the rectangle, GHIJ + Area of the rectangle, JKML + Area of the rectangle, NOPL + Area of the rectangle, PQRS + Area of the rectangle, STUV + Area of the rectangle, VWYX + Area of the rectangle, XZA
1
B
1
= AC × AB + EG × DE + IH × JI + LJ × LM + NO × NL + PQ × PS + UT × ST + VX × VW + ZA
1
× A
1
B
1
= 20 × 1 + 18 × 1 + 20 × 1 + 13 × 1 + 15 × 1 + 8 × 1 + 10 × 1 + 3 × 1 + 5 × 1
= 20 + 18 + 20 + 13 + 15 + 8 + 10 + 3 + 5
= 112 sq. units
Thus, the area of the spiral tube is 112 sq. units.
Let the length of the straight tube be x.
The area of the bent tube on the left = Area of rectangle, BACD + Area of rectangle, BGFE
= AC × CD + BG × BE
= 5 × 1 + 4 × 1
= 9 sq. units
Area of straight tube = x × 1
Area of bent tube = 9 sq. units
According to the question, both are the same.
So x × 1 = 9
⇒ x = 9 units.

Question 5

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.

Accepted Answer

Let ‘a’ be the side of the square.
Area of triangle, DCB (region 3)
Thus, the increase in the areas of regions 1, 2, and 3 is 4 times.
Reason: If the sidelength of the square is doubled, then the area becomes 4 times.

Question 6

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.

Accepted Answer

1. Let us take a square of cardboard (8 cm × 8 cm).
2. Draw two perpendicular lines inside the square (not through the center), dividing it into 4 rectangular pieces.
3. Cut along these lines to get 4 pieces.
4. Rearrange these 4 pieces.
Place them at the four corners of a larger imaginary square.
The pieces should be arranged so that they form a square with a hole in the middle.
Triangles
Figure It Out (Pages 157-159)

Question 7

If M and N are the midpoints of XY and XZ, what fraction of the area of ∆XYZ is the area of ∆XMN? [Hint: Join XY]

Accepted Answer

Let O be the midpoint of YZ, then join M to O and N to O.
According to mid point theorem,
MN = $\frac {1}{2}$ YZ, and MN is parallel to YZ.
The triangle XYZ is divided into four equal triangles.
So, Area of ∆XMN = $\frac {1}{4}$ × Area of ∆XYZ.

Question 8

Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.

Accepted Answer

Let P be the point on the bank of the river, then the shortest path from H to P to T is created by the point P such that ∆HPT has minimum area.
Shortest path
House (H) → P (Point on river) → Water tank
Area of a Polygon
Figure It Out (Page 160)

Question 9

Which has a greater area — an equilateral triangle or a square of the same side length as the triangle? Which has a greater area: two identical equilateral triangles together or a square of the same side length as the triangle? Give reasons.

Accepted Answer

Area of equilateral triangle = $\frac{\sqrt{3}}{4} a^2$
Area of square = a
2
⇒ $\frac{\sqrt{3}}{4} a^2$