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A trapezium has vertices marked as $\text{P, Q, R}$ and $\text{S}$ (in that order anticlockwise). The side $\text{PQ}$ is parallel to side $\text{SR}.$

Further, it is given that, $\text{PQ = 11 cm, QR = 4 cm, RS = 6 cm}$ and $\text{SP = 3 cm.}$

What is the shortest distance between $\text{PQ}$ and $\text{SR (in cm)}?$

1. $1.80$
2. $2.40$
3. $4.20$
4. $5.76$

Let’s first draw the trapezium.

Let the distance between $\text{RS},$ and $\text{PQ}$ be $h\;\text{cm}.$

Here, $\text{AB = 6 cm} \Rightarrow {\color{Blue}{\boxed{\text{PA + BQ = 5 cm}}}} \quad \longrightarrow (1)$

The $\triangle \text{PAS, QBR}$ are right angle triangle, so we can apply the Pythagorean theorem.

Now, in $\triangle \text{PAS},$

${\color{Green}{\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2}}}$

$\Rightarrow (\text{PS})^{2} = (\text{AS})^{2} + (\text{PA})^{2}$

$\Rightarrow 3^{2} = h^{2} + (\text{PA})^{2}$

$\Rightarrow h^{2} + (\text{PA})^{2} = 9 \quad \longrightarrow (2)$

Now, in $\triangle \text{PAS},$

$\Rightarrow (\text{QR})^{2} = (\text{RB})^{2} + (\text{BQ})^{2}$

$\Rightarrow 4^{2} = h^{2} + (\text{BQ})^{2}$

$\Rightarrow h^{2} + (\text{BQ})^{2} = 16 \quad \longrightarrow (3)$

Subtract equation $(2)-(3).$

$\qquad \require{cancel}\begin{array}{} {\color{Red}{\cancel{h^{2}}}} + (\text{PA})^{2} = 9 \\ {\color{Red}{\cancel{h^{2}}}} + (\text{BQ})^{2} = 16 \\\; – \qquad \;\; – \qquad \quad – \\\hline (\text{PA})^{2} – (\text{BQ})^{2} = -7 \end{array}$

$\Rightarrow (\text{PA – BQ}) (\text{PA + BQ}) = -7 \quad [{\color{Green}{\because a^{2} – b^{2} = (a+b)(a-b)}}]$

$\Rightarrow (\text{PA – BQ}) \cdot 5 = -7$

$\Rightarrow {\color{Blue}{\boxed{\text{PA – BQ} = \frac{-7}{5}\;\text{cm}}}}\quad \longrightarrow (4)$

Adding the equation $(1)\; \& \;(4),$ we get.

$\text{PA + BQ + PA – BQ} = 5 \;– \frac{7}{5}$

$\Rightarrow \text{2PA} = \frac{25-7}{5}$

$\Rightarrow \text{2PA} = \frac{18}{5}$

$\Rightarrow {\color{Blue}{\boxed{\text{PA} = \frac{9}{5}\;\text{cm}}}}$

Put the value of $\text{PA}$ in the equation $(1).$

$h^{2} + (\text{PA})^{2} = 9$

$\Rightarrow h^{2} + \left(\frac{9}{5}\right)^{2} = 9$

$\Rightarrow h^{2} + \frac{81}{25} = 9$

$\Rightarrow h^{2} = 9 – \frac{81}{25}$

$\Rightarrow h^{2} = \frac{225-81}{25}$

$\Rightarrow h^{2} = \frac{144}{25}$

$\Rightarrow h = \sqrt{\frac{144}{25}}$

$\Rightarrow h = \frac{12}{5} = 2.40\;\text{cm}$

$\therefore$  The shortest distance between $\text{PQ}$ and $\text{SR (in cm)}$ is $2.40\;\text{cm}.$

Correct Answer $:\text{B}$

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