Squares and Square Roots | FIO

Question 14

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

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Solution
Understand the Question
  • In a right-angled triangle, the side opposite the 9090^\circ angle is called the hypotenuse.
  • By the Pythagorean theorem, the square of the hypotenuse equals the sum of the squares of the two legs: c2=a2+b2c^2 = a^2 + b^2.
  • Because side lengths are always positive, c2c^2 is strictly greater than both a2a^2 and b2b^2, which proves c>ac > a and c>bc > b.

Step 1 · Set Up the Pythagorean Theorem

Let the lengths of the two legs of the right triangle be aa and bb, and let the length of the hypotenuse be cc.Diagram 1

By the Pythagorean theorem: a2+b2=c2a^2 + b^2 = c^2

Step 2 · Compare Side Lengths

From the Pythagorean theorem: c2=a2+b2c^2 = a^2 + b^2

Since side lengths must be positive (a>0a > 0 and b>0b > 0), their squares a2a^2 and b2b^2 are also positive.

Therefore: c2>a2c^2 > a^2 c2>b2c^2 > b^2

Taking the square root for positive values: c>ac > a c>bc > b

Thus, the hypotenuse cc is longer than both legs aa and bb.

Answer

Yes, the hypotenuse is the longest side of a right triangle as c>ac > a and c>bc > b.

Common Mistakes
  • Omitting the Positivity Condition: Forgetting to state that a,b,c>0a, b, c > 0, which is required when concluding c>ac > a from c2>a2c^2 > a^2.
  • Incomplete Justification: Stating that the hypotenuse is the longest side without using the Pythagorean theorem (c2=a2+b2c^2 = a^2 + b^2) or the angle-side relationship (side opposite the largest angle 9090^\circ is largest) to justify the claim.

More questions in FIO

Q1

Earlier, we saw a method to create a square with double the area of a given square 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?

Q2

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

Q3

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

Q4

If a right-angled triangle has shorter sides of lengths 5 cm5\text{ cm} and 12 cm12\text{ cm}, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhāyana's Theorem.

Q5

If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhāyana's Theorem.

Q6

Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudhāyana's Śulba-Sūtra, Verse 1.10)

Q7

Let aa, bb and cc denote the length of the sides of a right triangle, with cc being the length of the hypotenuse. Find the missing sidelength in each of the following cases:

(i) a=5a = 5, b=7b = 7

(ii) a=8a = 8, b=12b = 12

(iii) a=9a = 9, c=15c = 15

(iv) a=7a = 7, b=12b = 12

(v) a=1.5a = 1.5, b=3.5b = 3.5

Q8

Find 5 more Baudhāyana triples using this idea.

Q9

Does this method yield non-primitive Baudhāyana triples?

[Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

Q10

Are there primitive triples that cannot be obtained through this method? If yes, give examples.

Q11

Find the diagonal of a square with sidelength 5 cm5 \text{ cm}.

Q12

Find the missing sidelengths in the following right triangles:

Q13

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

Q14

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

Q15

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

Q16

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

Q17

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

Q18

(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?

Q19

Find the area of an equilateral triangle with sidelength 6 units6\text{ units}. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]

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