Answer

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**Hint:**Here, we will use Pythagoras theorem to solve this. In Pythagoras theorem, the sum of the squares on the legs of the right triangle is equal to the square on the hypotenuse.

**Formula used**

${(hypotenuse)^2} = {(base)^2} + {(perpendicular)^2}$

**Complete step by step solution:**

Let $AC$be the perpendicular on the side $BD$.

Now, we will solve $\Delta ABC$,

Here, $AB = 30cm$

$BC = y$

By using Pythagoras theorem

${(hypotenuse)^2} = {(base)^2} + {(perpendicular)^2}$

${(AB)^2} = {(BC)^2} + {(AC)^2}$

${(30)^2} = {(y)^2} + {(AC)^2}$

$30 \times 30 = {y^2} + {(AC)^2}$

$900 - {y^2} = A{C^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....(i)$

We will take $\Delta ACD$

$AD = 25cm$

$CD = 7cm$

By using Pythagoras theorem

${(hypotenuse)^2} = {(base)^2} + {(perpendicular)^2}$

${(25)^2} = {(7)^2} + {(AC)^2}$

$25 \times 25 = 7 \times 7 + {(AC)^2}$

$625 = 49 + {(AC)^2}$

$625 - 49 = {(AC)^2}$

$576 = {(AC)^2}$

Square root both sides, we will get

\[\sqrt {576} = \sqrt {{{(AC)}^2}} \]

$\sqrt {576} = AC$

We will factorize the value $576$.

\[576 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3\]

Now, $\sqrt {576} $$ = AC$

$\sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3} = AC$

$\sqrt {{2^2} \times {2^2} \times {2^2} \times {3^2}} = AC$

$\sqrt {{{(24)}^2}} = AC$2

$24cm = AC$

Put the value of $AC$ in equation (i) we have

$900 - {y^2} = {(24)^2}$

$900 - {y^2} = {(24)^2}$

$900 - {y^2} = 576$

$900 - 576 = {y^2}$

$324 = {y^2}$

So, factorize the number $324$

$324 = 2 \times 2 \times 3 \times 3 \times 3 \times 3$

$ = {2^2} \times {3^2} \times {3^2}$

$ = {(2 \times 3 \times 3)^2} = {(18)^2}$

So, ${(18)^2} = {y^2}$

When the powers are same, we equate the base,

Hence, $y = 18$

**Note:**Students must first be able to identify the type of triangle they are dealing with and then apply the appropriate formula for solving the problem.Pythagoras formula can be applied only to right angled triangles

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