Question

The side of a rhombus is 5 cm. If the length of one diagonal of the rhombus is 8 cm, then find the length of the other diagonal.​

Answer 1

$\star$ Reference of Image is in diagram :

$\setlength{ \unitlength}{20} \begin{picture}(6,6) \linethickness{1} \put(1,4){\line(1,0){6}} \put(4,4){\line(0,1){3}} \put(4,4){\line(0, - 1){3}}\qbezier(1,4)(4,1)(4,1)\qbezier(4,1)(7,4)(7,4)\qbezier(7,4)(7,4)(4,7)\qbezier(4,7)(4,7)(1,4)\put(0.8,3.5){ \bf{ A } }\put(4,0.5){ \bf{ B} }\put(7,4){ \bf{ C} }\put(4,7){ \bf{ D } }\put(4.2,3.5){ \bf{ O } }\end{picture}$

━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

Given: Side of a rhombus is 5 cm and Length of one diagonal of the rhombus is 8 cm respectively.

To find: The length of the other Diagonal.

◗ Let's Head to the Question now:

Half of the length of Diagonal AC, 8/2 = OC = 4 cm.

$\underline{\textsf{By Pythagoras Theorem in \Delta OBC :}}$

• Pythagoras Theorem (Sum of squares of two sides of a right angled triangle is equal to the third side).

$\bf{Here}\begin{cases}\sf{\:\;\;OC = 4\;cm}\\\sf{\;\;\;BC = 5\;cm}\end{cases}$

Now,

$:\implies\sf (OB)^2 + (OC)^2 = (BC)^2 \\\\\\:\implies\sf (OB)^2 + 4^2 = 5^2 \\\\\\:\implies\sf OB^2 + 16 = 25\\\\\\\implies\sf OB^2 = 25 - 16\\\\\\:\implies\sf OB^2 = 9\\\\\\\implies\sf OB = \sqrt{9}\\\\\\:\implies{\underline{\boxed{\frak{\pink{OB = 3\;cm}}}}}\;\bigstar$

⠀⠀⠀⠀

$\therefore{\underline{\sf{Hence,\; required\;value\;of\;OB\;is\; \bf{ 3\;cm}.}}}$

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

◗Length of the other Diagonal is 2 OB.

Therefore,

$:\implies\sf 2 \times OB\\\\\\:\implies\sf 2 \times 3 \\\\\\:\implies{\underline{\boxed{\frak{\purple{6\;cm}}}}}\;\bigstar$

$\therefore{\underline{\sf{Hence,\;Length \; of \; the \; other\; Diagonal\;is\; {\textbf{6\;cm }.}}}}$

Answer 2

Answer:

Given :-

• Side of rhombus = 5 cm
• 1st Diagonal = 8 cm

To Find :-

Second Diagonal

Solution :-

Half of Diagonal 1 = 8/2 = 4 cm

Now,

By Using Pythagoras theorem

AB² + AC² = BC²

AB² = 5² - 4²

AB² = 25 - 16

AB² = 9

AB = √9

AB = 3

Now,

AB = 2 × 3 = 6 cm