Question 1

Find the diagonal of a square with sidelength 5 cm.

Accepted Answer

The diagonal d of a square with side length a is given by the Pythagoras theorem: d = √(a² + a²) = √(2a²) = a√2. Here, a = 5 cm, so d = 5√2 cm ≈ 7.07 cm. — Using the Pythagoras theorem in the right triangle formed by two adjacent sides and the diagonal, diagonal = √(5² + 5²) = √(25 + 25) = √50 = 5√2 cm.

Question 2

Find the missing sidelengths in the following right triangles:

[The textbook shows several right triangles with some sides missing.]

Accepted Answer

Use the Pythagoras theorem a² + b² = c² where c is the hypotenuse.

For each triangle:
- Triangle with sides 4, 7, and missing side 10: Check if 4² + 7² = 10²? 16 + 49 = 65, 10²=100, no. So missing side must be calculated accordingly.
- Triangle with sides 9, 10, and missing side 41: Since 41 is large, likely hypotenuse.
- Triangle with sides 40, 45, and missing side 27: Calculate missing side using Pythagoras theorem.
- Triangle with sides 10, 150, and missing side 3: Check which is hypotenuse and calculate missing side.

Calculate each missing side by rearranging the Pythagoras theorem accordingly. — For each triangle, identify the hypotenuse (largest side). Then use the formula:
- If hypotenuse c is missing: c = √(a² + b²)
- If a leg is missing: missing side = √(c² - known leg²)
Calculate each missing side step-by-step.

Question 3

Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.

Accepted Answer

The diagonals of a rhombus bisect each other at right angles. Each side of the rhombus is the hypotenuse of a right triangle with legs half the diagonals.

Half diagonals: 24/2 = 12 units, 70/2 = 35 units.

Side length = √(12² + 35²) = √(144 + 1225) = √1369 = 37 units. — Using the property of rhombus diagonals intersecting at right angles and bisecting each other, each side is the hypotenuse of a right triangle with legs half the diagonals. Apply Pythagoras theorem to find the side.

Question 4

Is the hypotenuse the longest side of a right triangle? Justify your answer.

Accepted Answer

Yes, the hypotenuse is always the longest side of a right triangle. This is because, according to the Pythagoras theorem, the square of the hypotenuse equals the sum of the squares of the other two sides, so it must be longer than either of them. — If the hypotenuse were not the longest side, then one of the legs would be longer, contradicting the Pythagoras theorem. Hence, the hypotenuse is always the longest side.

Question 5

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

Accepted Answer

True. Every Baudhāyana triple (a Pythagorean triple) is either a primitive triple (where the three numbers are coprime) or a scaled version (multiple) of a primitive triple. — By definition, primitive triples have no common divisor other than 1. Any other triple can be obtained by multiplying a primitive triple by a common factor. Hence the statement is true.

Question 6

Give 5 examples of rectangles whose sidelengths and diagonals are all integers.

Accepted Answer

Examples of rectangles with integer sides and integer diagonals are those whose sides form Pythagorean triples:
1) 3, 4, 5
2) 6, 8, 10
3) 5, 12, 13
4) 9, 12, 15
5) 8, 15, 17

Each set represents the lengths of the two sides and the diagonal respectively. — Since the diagonal of a rectangle forms the hypotenuse of a right triangle with the sides as legs, the sides and diagonal must satisfy the Pythagoras theorem with integer values. Using known Pythagorean triples gives such rectangles.

Question 7

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

Accepted Answer

Area of square with side 7 = 7² = 49 units²
Area of square with side 5 = 5² = 25 units²
Difference = 49 - 25 = 24 units²

Construct a square with area 24 units². The side length of this square = √24 = 2√6 units. — The difference of the areas of the two squares is 24. To construct a square with this area, find the side length as the square root of 24. The side length is irrational, so the construction involves approximating 2√6 units.

Question 8

(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. unit?
(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Accepted Answer

(i) Yes, by choosing appropriate vertices on the grid, squares with areas 2, 3, 4, and 5 sq. units can be constructed. For example:
- Area 2: A square rotated so that its vertices lie on grid points can have area 2.
- Area 3: Similarly, a square with area 3 can be constructed using lattice points.
- Area 4: A normal square with side 2 units.
- Area 5: A square with side √5 units.

(ii) The possible integer-valued areas correspond to sums of two squares, i.e., numbers that can be expressed as a² + b² where a and b are integers. This is because the square's side squared equals the sum of squares of horizontal and vertical components between vertices on the grid. — Squares on a lattice grid can be oriented such that their sides are not parallel to the axes. The area of such squares is given by the sum of squares of the horizontal and vertical distances between points. Hence, all integer areas that are sums of two squares are possible.

Question 9

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

In an equilateral triangle of side 6 units, the altitude bisects the opposite side into two segments of 3 units each.
Using Pythagoras theorem, height h = √(6² - 3²) = √(36 - 9) = √27 = 3√3 units.
Area = (1/2) × base × height = (1/2) × 6 × 3√3 = 9√3 square units. — The altitude divides the equilateral triangle into two right triangles. Using Pythagoras theorem, calculate the height. Then use the formula for area of triangle.

Question 10

Find the Colours! There are 3 closed boxes — one containing only red balls, the second containing only blue balls and the third containing only green balls. The boxes are labelled RED, BLUE and GREEN such that ‘no’ box has the correct label. We need to find which label goes with which box. How can this be done if we are allowed to open only one box?

Accepted Answer

Open the box labelled 'RED'. Since no box has the correct label, this box cannot contain red balls. It must contain either blue or green balls. Suppose it contains blue balls. Then the box labelled 'BLUE' cannot contain blue balls (wrong label), so it must contain green balls. The remaining box labelled 'GREEN' must contain red balls.

Thus, by opening the box labelled 'RED' and seeing the color inside, we can deduce the correct labels for all boxes. — Because all labels are wrong, the box labelled 'RED' cannot have red balls. Opening it reveals its actual contents, which helps deduce the contents of the other boxes by elimination.

Question 11

In Baudhāyana’s Śulba-Sūtra, how can one construct a square having double the area of a given square?

Accepted Answer

By constructing a square on the diagonal of the original square — Doubling the side length of a square quadruples its area, not doubles it. Baudhāyana’s Śulba-Sūtra states that the square constructed on the diagonal of the original square has double the area of the original square.

Question 12

Why does the square constructed on the diagonal of a square have double the area of the original square?

Accepted Answer

The square on the diagonal has double the area because the diagonal divides the original square into two congruent right triangles. When a square is constructed on the diagonal, it contains four such congruent triangles, doubling the area of the original square. — The original square can be divided into two congruent right triangles. The new square on the diagonal contains four such triangles, which is twice the number of triangles in the original square, hence its area is double.

Question 13

If the side length of a square is doubled, what is the factor by which the area increases?

Accepted Answer

4 times — Area of a square = side × side. Doubling the side length multiplies the area by 2 × 2 = 4 times.

Question 14

How can you construct a square whose area is half the area of a given square?

Accepted Answer

To construct a square with half the area of a given square, draw a smaller square inside the original square tilted such that its vertices lie on the midpoints of the sides of the original square. This smaller square has half the area of the original. — By joining the midpoints of the sides of the original square, a smaller square is formed inside it. The area of this smaller square is half the area of the original square, as shown by geometric reasoning and congruent triangles.

Question 15

In an isosceles right triangle with equal sides of length 1 unit, what is the length of the hypotenuse?

Accepted Answer

\sqrt{2} units — Using Pythagoras theorem, hypotenuse c satisfies c² = 1² + 1² = 2. So, c = \sqrt{2}.

Pythagoras Theorem

Pythagoras Theorem — Study Notes

Introduction

Pythagoras Theorem

Proof of Pythagoras Theorem

Practice Questions — Pythagoras Theorem

All 7 Chapters in Ganita Prakash Part-II