Question

2. The parallel sides of a trapezium are 15 cm and 13 cm and its height is 8 cm. Find its
area .
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Answer 1

Step-by-step explanation:

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Answer 2

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt The\: parallel\: sides\: of \:a \:trapezium \:are\: 15\: cm \:and\\\tt 13\: cm \:and\: its\: height\: is \:8\: cm. \:Find\: its\: area.$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\sf \bf Base \:of \:trapezium (b_1) = 15 \:cm$
• $\sf \bf Base\:of \:trapezium (b_2) = 13 \:cm$
• $\sf \bf Height \:of \:trapezium (h) = 8 \:cm$

$\sf \bf {\boxed {\mathbb {TO\:SOLVE }}}$

• $\sf \bf Area\:of\:the\:trapezium$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

${\boxed {\boxed {\color {blue} {\sf \bf A=\dfrac{b_1+b_2}{2}h}}}}$

• $\sf A=area\:of\:trapezium$
• $\sf b_1=1st \:base\:of \:trapezium$
• $\sf b_2=2nd \:base\:of \:trapezium$
• $\sf h=height \:of\:trapezium$

${\underbrace {\overbrace {\color {orange} {\sf \bf Substitute \:the \:values}}}}$

$\sf \bf \implies A=\dfrac{15+13}{2}\times 8$

$\sf \bf \implies A=\dfrac{28}{2}\times 8$

$\sf \bf \implies A=14 \times 8$

$\implies{\boxed {\color {green} {\sf \bf A=112\:{cm}^{2}}}}$

${\color {purple} \underline {\tt \therefore Area\:of \:trapezium \:is\:112\:{cm}^{2}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\sf \bf Area\:of\:trapezium =\dfrac{b_1+b_2}{2}h$

$\sf \bf Perimeter\:of\:trapezium =a+b+c+d$

$\sf \bf Height\:of\:trapezium =2\dfrac{A}{b_1+b_2}$