Question 1

Earlier, we saw a method to create a square with double the area of a given square piece of paper. There is another method to do this in which two identical square papers are cut in the following way.
Can you arrange these pieces to create a square with double the area of either square?

Accepted Answer

Given: Two identical squares
These two are cut diagonally, forming two equal triangles, as shown in the figure.
Thus, we have four identical triangles from two identical squares.
Area of two triangles = Area of the given square
Area of four triangles = double the area of the given square.

Question 2

The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.
(i) 3
(ii) 4
(iii) 6
(iv) 8
(v) 9

Accepted Answer

If a is the length of two equal sides of an isosceles right triangle, then
hypotenuse = $\sqrt{a^2+a^2}=\sqrt{2 a^2}=a \sqrt{2}$
(i) a = 3
Hypotenuse = a√2 = 3√2 = $\sqrt{3 \times 3 \times 2}$ = √18
√18 lies between √16 and √25
√16

Question 3

The hypotenuse of an isosceles right triangle is 10. What are its other two side lengths?
[Hint: Find the area of the square composed of two such right triangles.]

Accepted Answer

If a is the length of two equal sides of an isosceles right triangle, then
a
2
+ a
2
= 102
⇒ 2a
2
= 100
⇒ a
2
= 50
⇒ a = √50
⇒ a = 5√2
2.4 Combining Two Different Squares
Figure It Out (Page 47)

Question 4

Let a, b, and c denote the lengths of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in each of the following cases:
(i) a = 5, b = 7
(ii) a = 8, b = 12
(iii) a = 9, c = 15
(iv) a = 7, b = 12
(v) a = 1.5, b = 3.5

Accepted Answer

Here, c
2
= a
2
+ b
2
(i) Now, c
2
= 5
2
+ 7
2
= 25 + 49
= 74
⇒ c = √74
(ii) Now, c
2
= 8
2
+ 12
2
= 64 + 144
= 208
⇒ c = √208
= $\sqrt{2 \times 2 \times 2 \times 2 \times 13}$
= 4√13
(iii) Here, 15
2
= 9
2
+ b
2
b
2
= 15
2
– 9
2
= 225 – 81
= 144
⇒ b = √144
= $\sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 3}$
= 12
(iv) c
2
= 7
2
+ 12
2
= 49 + 144
= 193
⇒ c = √193
(v) c
2
= 1.5
2
+ 3.5
2
= 2.25 + 12.25
= 14.5
⇒ c = √14.5
2.5 Right-Triangles Having Integer Sidelengths
Figure It Out (Page 50)

Question 5

True or False — Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple.

Accepted Answer

Let (a, b, c) be a Baudhayana triple.
Then HCF (a, b, c) = 1 or HCF (a, b, c) ≠ 1
(3, 4, 5): HCF is 1
(10, 24, 26): HCF = 2
If HCF is 1, then the triple is primitive.
If HCF is a number other than 1, then the triple is scaled.

Question 6

Construct a square whose area is equal to the difference of the areas of squares of side lengths 5 units and 7 units.

Accepted Answer

Area of square = 7
2
– 5
2
= 49 – 25
= 24 sq. units
1. Construct a square ABCD with a side of 2 cm.
Then DB = 2√2 units
2. Draw BE ⊥ DB at B such that BE = 4 units
3. Join DE.
DE
2
= (2√2)
2
+ 4
2
= 8 + 16
= 24
⇒ DE = √24
4. Draw a square with side DE (DEFG).
Area of DEFG = (√24)
2
= 24 square units.

Question 7

(i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. units?
(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Accepted Answer

(i) (a) 2 = 1
2
+ 1
2
Mark dots A, B, C, and D as shown.
Join AB, BC, CD, and DA.
Then ABCD is a square and area ABCD = 2 sq. units
(b) Square with area 3 units is not possible as 3 ≠ a
2
+ a
2
for any integer ‘a’.
(c) 4 = 2 × 2
Mark dots A, B, C, D as shown.
Join AB, BC, CD, DA.
Then ABCD is a square.
and ar ABCD = 2 × 2 = 4 sq. units
(d) (i)
Mark dots A, B, C, and D as shown.
Join A, B, C, and D
AB
2
= 2
2
+ 1
2
= 5
AB = √5 units
Hence, ABCD is a square with an area of 5 sq units.
(ii) Let the given value of area be x, where ‘x’ is an integer.
Then, x = a
2
+ b
2
, where ‘a’ and ‘b’ are integers or x is a perfect square, we can create squares with vertices as dots of the grid.

Question 8

Find the area of an equilateral triangle with sidelength 6 units.
[Hint: Show that an altitude bisects the opposite side. Use this to find the height.]

Accepted Answer

Let ΔABC be an equilateral triangle.
AB = BC = CA = 6 cm
Let AD be perpendicular to BC.
Then ∠1 = ∠2 (each = 90°)
AB = AC (each = 6 cm)
AD = AD (common)
ΔADB ≅ ΔADC (RHS)
BD = DC (CPCT)
∴ BD = DC = $\frac {1}{2}$ × 6 cm = 3 cm
In ΔADC,
h
2
+ 3
2
= 6
2
(Baudhayana’s triple)
⇒ h
2
= 36 – 9 = 27
⇒ h = $\sqrt{3 \times 3 \times 3}$
⇒ h = 3√3 cm
Ar ABC = $\frac {1}{2}$ × BC × AD
= $\frac {1}{2}$ × 6 × 3√3 sq. units
= 9√3 sq. units

The Baudhayana-Pythagoras Theorem