Question

Find the area of a rhombus having each side equal to 13 cm and one of whose diagonals is 24 cm.​

Answer 1

$\small\fbox{★ MATHS QUESTION !! ★}</p><p>$

$\small\fbox\purple{answer - }</p><p>$

$\small\bold{\: ☯︎ \:given - }$

$\pink{\mapsto \:each \: side \: of \: rhombus = 13 \: cm } \\ \pink{\mapsto \:one \: diagonal \: = 24 \: cm}$

$\small\bold{☯︎ \: find - }</p><p>$

$\color{gold}\star \: area \: of \: the \: rhombus$

$\huge\mathfrak{solution}</p><p>$

$\longmapsto \: one \: side \: of \: rhombus = 13 \: cm$

$\longmapsto \: length \: of \: one \: diagonal = 24 \: cm$

$\red{ \star\small\fbox{ property \: of \: rhombus }}</p><p>$

$\color{plum} \leadsto\: diagonals \: bisect \: each \: other\: at \: right \: angles$

$\orange{ \to \: now \: taking \: \frac{1}{4}th \: potion \: of \: rhombus \: (triangle) }$

$\blue{ \hookrightarrow \: base = 12 \: cm }\\ \blue{\hookrightarrow \: hypotenuse = 13 \: cm}$

$\circ \underline{\: applying \: pythagoras \: theorem \: - }$

$\mapsto \: 2\sqrt{ {13}^{2} - {12}^{2} } \\ \mapsto \: 2\sqrt{169 - 144} \\ \hookrightarrow \: 2\sqrt{25} = 2 \times 5 = 10$

$\pink{ \star \: other \: diagonal \: = 10}$

$now \: - \\ \green{area = \frac{1}{2} \times (product \: of \: diagonals) } \\ \color{lightgreen}\frac{1}{2} \times 24 \times 10 = {120 \: cm}^{2}$

$\small\color{teal} { = 120 \: cm}^{2} {\checkmark }$

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