Question

length of diagonal of rhombus are in the ratio 3 ratio 4 if the perimeter is 40 cm find the length of its side and diagonal ​

Answer 1

Solution :

Let ABCD be a rhombus.

$\therefore \sf \: AB = BC = CD = AD \:$

$\therefore \sf \: Side \: of \: rhombus = \dfrac{1}{4} \times 40$

$\sf \: Side = \large{\boxed{ \sf \: 10 \: cm}}$

Let BD = 3x and AC = 4x

$\therefore \sf \: OD = \dfrac{3}{2}x \: and \: OC = \dfrac{4}{2} x$

Now,Triangle DOC is a right-angled triangle.

$\therefore \sf \: {OD}^{2} + \: {OC}^{2} = {CD}^{2} \:$

$\hookrightarrow \sf \: ( { \dfrac{3}{2}x) }^{2} + { (\dfrac{4}{2} x})^{2} = {10}^{2}$

$\hookrightarrow \sf \: (\dfrac{9 {x}^{2} }{4} ) + (\dfrac{ {16x}^{2} }{4} ) = 100$

$\hookrightarrow \sf \: 25 {x}^{2} = 100 \times 4$

$\hookrightarrow \sf \: 25 {x}^{2} =400$

$\hookrightarrow \sf \: {x}^{2} = \cancel\dfrac{400}{25}$

$\hookrightarrow \sf \: {x}^{2} = 16$

$\hookrightarrow \sf \: {x} = \sqrt{16}$

$\hookrightarrow \sf \: {x} = \sqrt{4 \times 4}$

$\sf \large \boxed{ \sf \: x = 4}$

Diagonals of BD and AD of rhombus are :

$\star \sf \: 3x = 3 \times 4 = \boxed{ \sf \: 12 \: cm}$

$\star \sf \: 4x = 4 \times 4 = \boxed{ \sf \: 16 \: cm}$

Length of side of Diagonal = 10 cm

Answer 2

QUESTION:-

length of diagonal of rhombus are in the ratio 3 ratio 4 if the perimeter is 40 cm find the length of its side and diagonal

SOLUTION:-

GIVEN

{Length of diagonal are in 3:4

{Perimeter of rhombus=40cm

FIND:-length of its side and diagonal

Let

length the of diagonal be 3x and 4x

As per the question

side of rhombus=1/4*40

=10cm

by pathogorus theorem

$⇒ {side}^{2} = {d1}^{2} + {d2}^{2} \\ \\⇒ {10}^{2} = (3x) {}^{2} + (4x) {}^{2} \\ \\ ⇒100 = 9 {x}^{2} + 16 {x}^{2} \\ \\ ⇒100 = 25 {x}^{2} \\ \\ ⇒ {x}^{2} = \frac{100}{25} \\ \\ ⇒ {x}^{2} = 4 \\ \\ ⇒x = \sqrt{4} = 2$

HENCE:-

length of d¹=3*2*2=12cm

length of d²=4*2*2=16cm