Question

plz solve this problem​

Answer 1

$\large\underline{\sf{sollution: }}$

$\large\underline{\sf{the \: given \: figure \: can \: be \: }} \\ \large\underline{\sf{broken \: it \: 3 \: triangles \: and}} \\ \large\underline{\sf{a \: quadrilateral}}$

$\large{\sf{the \: line \: it \: the \: middle \: has \:}} \\ \large{\sf{total \:length = 34cm }} \\ \large{\sf{ + 6cm + 10cm = 50cm}}$

$\large{\sf{the \: area \: of \: lower \: triangle }} \\ \large{\sf{ = \frac{1}{2} \times 50cm \times 16cm }}$

$\large{\sf{ {400cm}^{2} }}$

$\large\underline{\sf{similarly = the \: area \: of}} \\ \large\underline{\sf{other \: two \: can \: be \: found}}$

$\large{\sf{area \: of \: top \: left \: triangle = }} \\ \large{\sf{ \frac{1}{2} \times 12cm \times 6cm = {36cm}^{2} }}$

$\large{\sf{area \: of \: top \: right \: triangle}} \\ \large{\sf{ = \frac{1}{2} \times 10cm \times 4cm = {20cm}^{2} }}$

$\large\underline{\sf{so \: the \: area \: of \: figure \: is \: = }} \\ \large\underline{\sf{ {400cm}^{2} + {36cm}^{2} + {20cm}^{2} }} \\ \large\underline{\sf{ + area \: of \: trapezium}}$

$\large\underline{\sf{henceforth \: the \: area \: of \: the}} \\ \large\underline{\sf{figure \: is \: {456cm}^{2} }} \\ \large\underline{\sf{area \: of \: trapezium}}$

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$\large\underline{\sf{hope \: \: helpful}}$