Multiply a Triple by Any Number and It Still Works

It does — and so does every multiple. Doubling each number gives $6,8,10$, and

$$6^2+8^2=36+64=100=10^2$$

So one known triple hands you an endless supply. This page covers everything in the CBSE Class 8 Mathematics chapter's second part: testing a triple, generating new ones, the converse of the theorem, and real-life problems.

A Baudhayana-Pythagoras triple is a set of three positive integers $a$, $b$, $c$ satisfying

$$a^2+b^2=c^2$$

To test one, identify the largest number as $c$, then check whether the squares of the other two add to its square.

Worked tests.

$(3,4,5)$: $9+16=25=5^2$. A triple.

$(5,12,13)$: $25+144=169=13^2$. A triple.

$(8,15,17)$: $64+225=289=17^2$. A triple.

$(7,24,25)$: $49+576=625=25^2$. A triple.

$(4,5,6)$: $16+25=41$, but $6^2=36$. Not a triple.

$(6,7,9)$: $36+49=85$, but $9^2=81$. Not a triple.

The four basic triples worth memorising are $(3,4,5)$, $(5,12,13)$, $(8,15,17)$ and $(7,24,25)$, because exam questions are built from them and their multiples.

The step that must come first is picking out the largest number, and getting it wrong wrecks the test. For $(5,12,13)$, testing $5^2+13^2$ against $12^2$ compares the wrong pair and would reject a perfectly good triple — so only the largest can ever play the role of the hypotenuse.

Multiply all three numbers by the same whole number. The relation survives, because every term is scaled by the same factor.

From $(3,4,5)$:

- $\times 2$ gives $(6,8,10)$, and $36+64=100$
- $\times 3$ gives $(9,12,15)$, and $81+144=225=15^2$
- $\times 4$ gives $(12,16,20)$, and $144+256=400=20^2$
- $\times 5$ gives $(15,20,25)$
- $\times 10$ gives $(30,40,50)$

From $(5,12,13)$:

- $\times 2$ gives $(10,24,26)$, and $100+576=676=26^2$
- $\times 3$ gives $(15,36,39)$

From $(8,15,17)$: $\times 2$ gives $(16,30,34)$.

The reason multiplying works can be shown in one line. If $a^2+b^2=c^2$ and each side is multiplied by $k$, then

$$(ka{)}^2+(kb{)}^2=k^2{a}^2+k^2{b}^2=k^2(a^2+b^2)=k^2{c}^2=(kc{)}^2$$

So the new set satisfies the same relation, for any whole number $k$.

A triple with no common factor, such as $(3,4,5)$ or $(5,12,13)$, is called a primitive triple, and the others are its multiples. Note that dividing works only when a common factor exists: $(6,8,10)$ divides by 2 back to $(3,4,5)$, but $(3,4,5)$ cannot be reduced any further — and multiplying by a fraction would leave the numbers non-integer, so it would no longer be a triple.

The converse says that if the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled — and the right angle lies opposite the longest side.

So the theorem and its converse run in opposite directions:

- Theorem — given a right angle, the sides satisfy $a^2+b^2=c^2$
- Converse — given $a^2+b^2=c^2$, there is a right angle

Worked tests.

$(9,12,15)$: $81+144=225=15^2$. Right-angled, with the right angle opposite the 15 cm side.

$(5,6,8)$: $25+36=61$, but $8^2=64$. Not right-angled.

$(10,24,26)$: $100+576=676=26^2$. Right-angled.

$(6,8,11)$: $36+64=100$, but $11^2=121$. Not right-angled.

The comparison tells you more than yes or no:

- Two smaller squares adding to less than the largest — the angle opposite the longest side is obtuse, as in $(5,6,8)$
- Adding to more — that angle is acute
- Adding to exactly the largest — right

A mason checking a corner uses this directly: measuring 3 units along one wall and 4 along the other, the corner is square only if the distance between those marks is exactly 5.

The order of the check matters, and it is the same trap as before. Always square the largest side on its own and compare with the sum of the other two — testing the wrong pair can make a right-angled triangle look ordinary or the reverse.

Sketch the situation, mark the right angle, and decide which length is the hypotenuse — always the one opposite that angle.

Ladder against a wall. A ladder $13$ m long has its foot $5$ m from the wall. The wall is vertical and the ground horizontal, so the right angle is at the base and the ladder is the hypotenuse:

$$\text{height}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\text{ m}$$

Wire from a pole. A pole is $24$ m tall and a wire runs from its top to a point $7$ m from its base:

$$\text{wire}=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25\text{ m}$$

Diagonal of a rectangle. A rectangular field is $12$ m by $5$ m, so the diagonal is

$$\sqrt{144+25}=\sqrt{169}=13\text{ m}$$

Straight-line distance. A man walks $9$ m east, then $12$ m north. He has covered $21$ m of ground, but his distance from the start is

$$\sqrt{81+144}=\sqrt{225}=15\text{ m}$$

Height of a broken tree. A tree snaps so that its top touches the ground $8$ m from the base, and the broken piece is $10$ m long. The standing part is $\sqrt{100-64}=6$ m, so the tree was $6+10=16$ m tall.

Spotting the hypotenuse is the move that decides these questions, and it is always the slanting length — the ladder rather than the wall, the wire rather than the pole, the diagonal rather than a side, the direct route rather than either leg of the walk.