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**Hint**: To find area of rhombus we first calculate half of given diagonal and then applying Pythagoras theorem to find third side of a right triangle so formed after intersection of diagonals of rhombus, using this to find length of diagonals and finally substituting values in formula to get required area of rhombus.

Formulas used: Area of rhombus when diagonals are known $ \dfrac{1}{2} \times {d_1} \times {d_2} $ , where $ {d_1} $ and $ {d_2} $ are diagonals of rhombus. Pythagoras theorem $ {\left( {Hpy.} \right)^2} = {\left( {Perp.} \right)^2} + {\left( {Base} \right)^2} $

**:**

__Complete step-by-step answer__In given rhombus PQRS, side PQ = $ 17 $ cm and diagonal PR = $ 16 $ cm.

We know that in rhombus diagonals bisects each other.

Let O is the point where two diagonals of rhombus PQRS meet.

Therefore, point O bisects both diagonals PR and QS of rhombus PQRS.

But PR = $ 16 $ cm given.

Therefore length of OP will be given as $ \dfrac{1}{2} $ of PR.

Hence, $ OP = \dfrac{1}{2}\left( {PR} \right) = \dfrac{1}{2}\left( {16} \right) $

$ \Rightarrow OP = 8cm $

Also, in rhombus we know that diagonals are perpendicular to each other.

Hence, $ \Delta OPR $ so formed after intersection of diagonals is a right angle triangle.

In $ \Delta OPR $ , two sides are known to find the third side; we use Pythagora's theorem. Which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides of the triangle.

$ \Rightarrow {\left( {PQ} \right)^2} = {\left( {OP} \right)^2} + {\left( {OQ} \right)^2} $

Substituting values of $ PR = 17 $ and $ OP = 8 $ to find $ OQ $

$ \Rightarrow {\left( {17} \right)^2} = {\left( 8 \right)^2} + {\left( {OQ} \right)^2} $

$

\Rightarrow 289 = 64 + {\left( {OQ} \right)^2} \\

\Rightarrow 289 - 64 = {\left( {OQ} \right)^2} \\

\Rightarrow {\left( {OQ} \right)^2} = 225 \\

\Rightarrow {\left( {OQ} \right)^2} = {\left( {15} \right)^2} \\

\Rightarrow \left( {OQ} \right) = 15 \\

$

Hence, from above we see that length $ OQ = 15cm $ .

Therefore, the length of SQ will be given as $ 2\left( {OQ} \right) $ , because O is the midpoint of QS.

$

\Rightarrow QS = 2\left( {OQ} \right) \\

\Rightarrow QS = 2\left( {15} \right) \\

\Rightarrow QS = 30 \\

$

From above we see that the length of diagonals of rhombus PQRS are 16 and 30 respectively.

Also, we know that area of the rhombus is given as $ \dfrac{1}{2} \times {d_1} \times {d_2} $ , where $ {d_1}\,and\,{d_2} $ are diagonals of rhombus.

On substituting values we have

Area of rhombus PQRS = $ \dfrac{1}{2} \times 16 \times 30 $

Area of rhombus = $ 8 \times 30 $

Area of rhombus = $ 240c{m^2} $

Hence, the required area of the rhombus of a given dimension is $ 240c{m^2} $ .

**Note**: In case of rhombus there are two ways to calculate its area. In the first case the base and altitude of the rhombus are given. In this case area of rhombus is given as $ Area = base \times altitude $ .

In seconds case instead of altitude diagonals are given and area in this case is given as: $ Area = \dfrac{1}{2} \times {d_1} \times {d_2} $ where $ {d_1} $ and $ {d_2} $ are diagonals of rhombus.

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