Question 38

If the diagonal d of a quadrilateral is doubled and the heights $h_{1}$ and $h_{2}$ falling on d are halved, then the area of quadrilateral is ___.

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Solution

$\frac{1}{2} (h_{1} + h_{2}) d$
Let ABCD be a quadrilateral, where $h_{1}$ and $h_{2}$ are altitudes on the diagonal BD = d
Then, area of quadrilateral ABCD $= \frac{1}{2} (h_{1} + h_{2}) \times B D$

If altitudes are halved and the diagonal is doubled, then the new heights are $\frac{h_{1}}{2}$ , $\frac{h_{2}}{2}$ respectively.
Also the new diagonal is $2 d$ .

Area of quadrilateral ABCD $= \frac{1}{2} (\frac{h_{1}}{2} + \frac{h_{2}}{2}) \times 2 d = \frac{1}{2} (\frac{h_{1} + h_{2}}{2}) \times 2 d$
$= \frac{1}{2} (h_{1} + h_{2}) \times d$

Hence, area remains unchanged.

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Similar questions

What will be the area A of the quadrilateral in each case with their diagonal $d c m$ & the sum of their heights $(h_{1} + h_{2}) c m$ .

i) $d = 20, (h_{1} + h_{2}) = 10$

ii) $d = 15, (h_{1} + h_{2}) = 20$

iii) $d = 20, (h_{1} + h_{2}) = 25$

iv) $d = 40, (h_{1} + h_{2}) = 30$

10

h_{1} = 5

h_{2}

7

\frac{1}{2} \times d \times (h_{1} + h_{2})

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