Question

1) The lengths of the diagonals of a rhombus are 20 and 48 meters. Find the perimeter of the rhombus.

2) The perimeter of a rhombus is 200 cm and one of its diagonal has a length of 60 cm. Find the area of the rhombus.

Solve Both Questions. Need Urgently​

Answer 1

Q1

given

d¹=20m

d²=48m

perimeter of rhombus=2√(d¹)²+(d²)²

=2√20²+48²

=2√400+2304

=2√2704

=2*52=>104m

hence:-

the perimeter of rhombus is 104m

Q2

perimeter=200

diagonal=60

so,

4a=200

a=50

here we have to divide by 2

60/2=30

so,

other diagonal=2√50²-30²

=2√2500-900

=2√1600=40*2=80cm

hence

the area of rhombus=1/2*d¹*d²

=1/2*60*80

=2400cm²

Answer 2

1) The lengths of the diagonals of a rhombus are 20 and 48 meters. Find the perimeter of the rhombus.

AnswEr :

• Diagonal₁ = 48 metre
• Diagonal₂ = 20 metre
• Find Perimeter of Rhombous?

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(9,1.3){\mathsf{10 m}}\put(10,1.3){\mathsf{24 m}}\put(7.7,0.9){\large{B}}\put(11.1,0.9){\large{C}}\put(9.9,2.1){\large{O}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(11,1){\line(-1,1){2}}\put(8,1){\line(2,1){4}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\end{picture}$

Rhombous ABCD with Diagonals AC and BD. we know that Diagonals are Perpendicular to Each Other, therefore they'll bisect each other, where OB = 10 m, and OC = 24 m.

• By Considering Right Triangle BOC and applying Pythagoras Theorem :

$\longrightarrow\tt BC^{2}= OB^{2} + OC^{2} \\ \\\longrightarrow\tt BC^{2}= (10)^{2} +(24)^{2} \\ \\\longrightarrow\tt BC^{2}= 100+576 \\ \\\longrightarrow\tt BC^{2}= 676 \\ \\\longrightarrow\tt BC = \sqrt{676} \\ \\\longrightarrow\tt BC = \sqrt{26 \times 26} \\ \\\longrightarrow \blue{\tt BC =26\:m}$

Now, we Got Side of Rhombous i.e. 26 m

• P E R I M E T E R :

⇒ Perimeter = 4 × Side

⇒ Perimeter = 4 × 26 m

⇒ Perimeter = 104 metre

⠀

∴ Perimeter of Rhombous ABCD is 104 m.

$\rule{300}{2}$

2) The perimeter of a rhombus is 200 cm and one of its diagonal has a length of 60 cm. Find the area of the rhombus.

• Perimeter = 200 cm
• Diagonal₁ = 60 cm
• Find Area of Rhombous?

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(8.6,3){\large{A}}\put(9,1.3){\mathsf{30 cm}}\put(7.7,0.9){\large{B}}\put(11.1,0.9){\large{C}}\put(9.9,2.1){\large{O}}\put(8,1){\line(1,0){3}}\put(11,1){\line(1,2){1}}\put(9,3){\line(3,0){3}}\put(11,1){\line(-1,1){2}}\put(8,1){\line(2,1){4}}\put(8,1){\line(1,2){1}}\put(12.1,3){\large{D}}\end{picture}$

Rhombous ABCD with Diagonals AC and BD. we know that Diagonals Bisect Each Other, therefore OB = BC/2 = 60/2 = 30 cm.

• Calculation of Sides of Rhombous :

⇒ Perimeter = 4 × Side

⇒ 200 cm = 4 × Side

• Dividing both term by 4

⇒ Side = 50 cm

• In Right ∆BOC, By Pythagoras theorem :

$\longrightarrow\tt BC^{2}= OB^{2} + OC^{2} \\ \\\longrightarrow\tt (50)^{2}=(30)^{2} +OC^{2} \\ \\\longrightarrow\tt 2500=900+OC^{2} \\ \\\longrightarrow\tt OC^{2}=2500 - 900 \\ \\\longrightarrow\tt OC^{2} =1600 \\ \\\longrightarrow\tt OC = \sqrt{40 \times 40} \\ \\\longrightarrow \blue{\tt OC =40\:cm}$

◗ AC = 2(OC) = 2(40 cm) = 80 cm

$\rule{300}{1}$

• Area of Rhombous ABCD :

$\implies \tt Area_{\tiny ABCD} = \dfrac{D_1 \times D_2}{2}\\ \\\implies \tt Area_{\tiny ABCD} = \dfrac{AC\times BD}{2} \\ \\\implies \tt Area_{\tiny ABCD} = \dfrac{80 \times 60}{2} \\ \\\implies \tt Area_{\tiny ABCD} = \cancel\dfrac{4800}{2} \\ \\\implies \boxed{\orange{\tt Area_{\tiny ABCD} =2400 \:{cm}^{2}}}$

⠀

∴ Area of Rhombous ABCD is 2400 cm².

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