Question

6. DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
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Answer 1

Answer:

area of IIgm=base × atitude

1470=AB×DL

1470=35×DL

DL=42cm

similarly

1470=AD×BM

BM=30cm

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

• ABCD is a parallelogram such that AB = 35 cm and AD = 49 cm.

• Area of parallelogram ABCD = 1470 $\rm \: cm^2$

• DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD.

Now,

ABCD is a parallelogram with base AB and corresponding height is DL.

So,

$\rm \: Area_{(parallelogram \: ABCD)} = AB \times DL$

On substituting the values, we get

$\rm \: 1470 = 35 \: \times \: DL$

$\bf\implies \:DL \: = \: 42 \: cm$

Again,

ABCD is a parallelogram with base AD and corresponding height is BM.

So,

$\rm \: Area_{(parallelogram \: ABCD)} \: = \: AD \times BM$

$\rm \: 1470 = 49 \times BM$

$\bf\implies \:BM \: = \: 30 \: cm$

Hence,

$\begin{gathered}\begin{gathered}\bf\: \bf\implies \:Length \: of \: \begin{cases} &\tt{DL \: = \: 42 \: cm} \\ \\ &\tt{BM \: = \: 30 \: cm} \end{cases}\end{gathered}\end{gathered}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$