Question

Two adjacent sides of a parallelogram are 24 cm and 18 cm. If the distance between
longer sides is 12 cm, find the distance between shorter sides.

**steps required**​

Answer 1

Answer:

hope this helps you sis................✌❤

Answer 2

Given:

• Two adjacent sides of a parallelogram are 24 cm and 18 cm.
• The distance between longer sides is 12 cm

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To find:

• Distance between shorter sides?

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Solution:

$\setlength{\unitlength}{1.6cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(8.6,3){\large\tt{D}}\put(7.7,0.9){\large\tt{A}}\put(9.5,0.7){\sf{\large{24 cm}}}\put(11.1,0.9){\large\tt{B}}\put(8,1){\line(1,0){3}}\qbezier(11,1)(11.5,2)(12,3)\put(9,3){\line(3,0){3}}\put(9.1,1.8){\sf{\large{12 cm}}}\put(9,1){\line(0,1){2}}\qbezier(8,1)(8.5,2)(9,3)\qbezier(11.5,2)(11.5,2)(9,3)\put(12.1,3){\large\tt{C}}\put(11.5,1.6){\sf{\large{18 cm}}}\put(11.65,2){\large\tt{F}}\put(8.9,0.7){\large\tt{E}}\end{picture}$

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Here,

• AB = 24 cm
• BC = 18 cm
• DE = 12 cm

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☯ Let Distance between shorter sides, DF be x cm.

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We know that,

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$\star\;{\boxed{\sf{\purple{Area_{\;(parallelogram)} = Base \times Height}}}}$

Now, Finding area of parallelogram, considering AB as base :]

$:\implies\sf AB \times DE$

$:\implies\sf 24 \times 12$

$:\implies{\boxed{\sf{\pink{288\;cm^2}}}}\;\bigstar$

Now, again finding area of parallelogram, considering BC as base :]

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$:\implies\sf BC \times DF$

$:\implies\sf 18 \times x$

$:\implies{\boxed{\sf{\pink{18x\;cm^2}}}}\;\bigstar$

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We know that,

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Area of parallelogram will be the same, if we consider AB as base or BC as base.

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Therefore,

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$:\implies\sf AB \times DE = BC \times DF$

$:\implies\sf 288 = 18x$

$:\implies\sf x = \cancel{ \dfrac{288}{18}}$

$:\implies{\boxed{\sf{\purple{16\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Distance\;between\;the\;shorter\;sides\;is\; \bf{16\;cm}.}}}$