Question

The Diagonals of a rhombus are 8cm and 6cm. Find the sides of the Rhombus.​

Answer 1

Answer:

in rhombus opposite sides are parallel

so

the answer is 64 plus 36

hundred

hundred is the correct answer

Answer 2

Question :-

• The Diagonals of a rhombus are 8cm and 6cm. Find the sides of the Rhombus.

Diagram :-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,3){1.5}}\put(0,0){\line(1,0){5}}\put(5,0){\line(1,3){1.5}}\put(1.5,4.5){\line(1,0){5}}\qbezier(1.56,4.5)(1.56,4.5)(5,0)\qbezier(6.45,4.5)(6.45,4.5)(0,0)\put(-0.5,-0.5){\sf B}\put(1,4.8){\sf A}\put(5.2,-0.5){\sf C}\put(6.7,4.75){\sf D}\put(3,1.6){\sf O}\end{picture}$

Given :-

• First Diagonal ( AC ) = 6cm
• Second Diagonal ( BD ) = 8cm

Solution :-

$\qquad \quad {:} \longrightarrow \sf{\bf{OA \: = \: \frac{AC}{2} \: = \: \frac{6}{2} \: = \: { \tt \boxed{3cm} } }}$

$\qquad \quad {:} \longrightarrow \sf{\bf{OB \: = \: \frac{BD}{2} \: = \: \frac{8}{2} \: = \: { \tt \boxed{4cm} } }}$

$\qquad \quad {:} \longrightarrow \sf{\bf{ \triangle \: AOB \: = \: \angle \: AOB \: = \: \boxed{{90}^{ \circ} } }}$

Applying Pythagoras Theorem :-

$\qquad \quad {:} \longrightarrow \sf \boxed{\tt{ {(AB)}^{2} \: = \: {(OA)}^{2} \: + \: {(OB)}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf {\sf{ {(AB)}^{2} \: = \: {(3)}^{2} \: + \: {(4)}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf {\sf{ {(AB)}^{2} \: = \: 9 \: + \: 16 }}$

$\qquad \quad {:} \longrightarrow \sf {\sf{ {(AB)}^{2} \: = \: 25 }}$

$\qquad \quad {:} \longrightarrow \sf {\sf{ AB \: = \: \sqrt{25} }}$

$\qquad \quad {:} \longrightarrow \sf \boxed{\bf{ AB \: = \: 5cm }}$

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All the Sides of a Rhombus are equal

$\bf \therefore \; All\;sides = 5cm$

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