Question

in a rhombus each side measures 13 cm and one of the diagonal is 10 cm find the length of the Other diagonal also find the area of the Rhombus​

Answer 1

AnswEr :

Reference of Image is in the attachment :

• ABCD is a Rhombus.
• AB = BC = CD = DA = 13 cm
• AC = 10 cm
• OA = OC = 5 cm

we know that Diagonal of Rhombus Bisect Each Other. Therefore BD will Bisect AC at O and that Angle will be 90°.

• In Triangle BOC ; By Heron's Formula :

⇒ BC² = OC² + OB²

⇒ ( 13 )² = ( 5 )² + OB²

⇒ OB² = ( 13 )² - ( 5 )²

• (a² - b²) = (a + b)(a - b)

⇒ OB² = (13 + 5)(13 - 5)

⇒ OB² = 18 × 8

⇒ OB² = 144

⇒ OB = √144

⇒ OB = √(12 × 12)

⇒ OB = 12 cm

Now : BD = 2 × OB = (2 × 12 cm) = 24 cm

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• Area of Rhombus ABCD :

⇒ Area = 1 /2 × D₁ × D₂

⇒ Area = 1 /2 × AC × BD

⇒ Area = 1 /2 × 10 cm × 24 cm

⇒ Area = 10 cm × 12 cm

⇒ Area = 120 cm²

∴ Area of the Rhombus will be 120 cm² and Other Diagonal will be 24 cm.

Answer 2

$\bf{\Huge{\underline{\boxed{\sf{\green{ANSWER\::}}}}}}$

Given:

In a rhombus each side measures 13cm and one of the diagonal is 10cm.

To find:

The length of the other diagonal & area of the rhombus.

$\bf{\Large{\underline{Explanation\::}}}}$

We have,

• Each rhombus side= 13cm
• Diagonal of one rhombus,[d1]= 10cm

From attachment figure of rhombus:

• AB=BC=CD=DA= 13cm
• AC,d1= 10cm
• AC ⊥ BD      [Property of rhombus]

• OA = OC
• OB = OD

Now,

In ΔOAB, [Using Pythagoras Theorem]

→ [Hypotenuse]² = [Base]² + [Perpendicular]²

→ (AB)² = (OA)² + (OB)²

→ (13cm)² = (5cm)² + (OB)²

→ 169cm² = 25cm² + (OB)²

→ OB² = 169cm² - 25cm²

→ OB² = 144cm²

→ OB = √144cm²

→ OB = 12cm

• Other diagonal of the rhombus:

→ BD = OB + OD

→ BD = 2 × OB

→ BD = (2 × 12)cm

→ BD = 24cm

Now,

• Area of rhombus ABCD

Diagonal,d1 [AC]= 10cm

Diagonal,d2 [BD]= 24cm

→ Area of rhombus= $\frac{1}{2} *d1*d2$

→ Area of rhombus= $(\frac{1}{2} *10*24)cm^{2}$

→ Area of rhombus= $(\frac{1}{\cancel{2}} *10*\cancel{24})cm^{2}$

→ Area of rhombus= (10 × 12)cm²

→ Area of rhombus = 120cm²