Question

if base of a parallelogram is 18 cm and its height is 11 cm find its area​

Answer 1

$\red{\huge{\underline{\overline{\mid{\bold{\purple{\:\: SOLUTION \: \: \red{\mid}}}}}}}}$

$\bold{Let,} \\ \\ \bold{ABCD \: \: is \: \: a \: \: parallelogram \:, \: where} \\ \\ \bold{Base =AD = BC = 18 \: \: cm} \\ \\ \bold{Height = AM = CN = 11 \: \: cm}$

$\bold{Now,} \\ \\ \bold{Area \: \: of \: \: the \: \: \triangle \: ABC = \frac{1}{2} \times AM \times BC} \\ \\ \bold{ =( \frac{1}{2} \times 11 \times 18) \: \: sq.cm} \\ \\ \bold{ = 99 \: cm}$

$\bold{ \therefore \: Area \: \: of \: \: the \: \: parallelogram } \\ \\ \bold{= 2 \times Area \: \: of \: \: \triangle \: ABC } \\ \\ \bold{ =( 2 \times 99) \: \: sq.cm} \\ \\ \bold{ = \underline{\red { \: \:198 \: \: sq.cm\: \: }}}$

$\bold{\huge{\underline{\red{\: \: OR \: \:}}}}$

$\bold{\underline{\: \: We\: \: know \: \: that \: \:}}$

$\boxed{\bold{Area \: \: of \: \: a \: \: parallelogram \: = \: Base \times Height}}$

$\bold{Here, } \\ \\ \bold{\: \: Base \: \: of \: \: the \: \: parallelogram , AD=BC= 18 \: \: cm} \\ \\ \bold{\: \: Height \: \: of \: \: the \: \: parallelogram , AM = CN= 11 \: \: cm} \\ \\ \bold{ \therefore \: Area \: \: of \: \: the \: \: parallelogram }\\ \\ \bold{= Base \times Height} \\ \\ \bold{ AD \times AM}\\ \\ \bold{ = ( \: 18 \: \times \: 11 \: ) \: \: sq.cm }\\ \\ \bold{ = \underline{\red{\: \: 198 \: \: sq.cm\: \: }}}$

Answer 2

Answer:

above one Answer

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