Question

Find the perimeter of the rhombus, if lengths of its diagonals are 12cm and 16cm. ​

Answer 1

Answer:

$\huge{\pink{\boxed{\green{\sf{PERMETER=40cm}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: { \orange{given}} \\ {\pink {\boxed{ \green{diagonal \: 1st(d1) = 12cm}}}} \\ {\pink {\boxed{ \green{diagonal \: 2nd(d2) = 16}}}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: { \blue{to \: find}}\\ {\purple {\boxed{ \red{perimeter \: of \: rhombus = }}}}$

☆ According to given question :

☆ We know the properties of rhombus :

$\to \: diagonals \: bisects \: each \: other \: at \: 90 \degree \\ \to \: all \: sides \: are \: equal \: in \: rhombus \\ \to \: perimeter \: of \: rhombus = 4 \times side$

☆ So first we find side of rhombus :

☆ we take a right angled traingle whose base is 6cm and perpendicular is 8cm.

$\to {h}^{2} = {p}^{2} + {b}^{2} \\ \to {h}^{2} = {8}^{2} + {6}^{2} \\ \to {h}^{2} = 64 + 36 \\ \to {h} = \sqrt{100} \\ { \pink{ \boxed{ \green{\to h = 10cm}} }}\\ \\ \to perimeter \: of \: rhombus = 4 \times side \\ \to perimeter = 4 \times 10 \\{ \pink{ \boxed{ \green{ \to perimeter = 40cm}}}}$

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