Question

1. The perimeter of a rhombus is 60cm. If the length of its longer diagonal
measures 24cm, the length of the shorter diagonal is .......... cm.
18
d) 9
15
a) 20
c) 15
​

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

The perimeter of a rhombus is 60 cm. If the length of its longer diagonal measures 24 cm.

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

As we know that formula of the perimeter of rhombus;

$\boxed{\bf{Perimeter = 4 \times side}}$

A/q

$\mapsto\tt{Perimeter\:of\:rhombus = 4 \times side}$

$\mapsto\tt{60 = 4 \times side}$

$\mapsto\tt{Side = \cancel{60/4}}$

$\mapsto\tt{Side = 15\:cm}$

Therefore,the all side of rhombus will be 15 cm .

Now, attachment a figure, a/c question:

In ΔOCB :

AC = 24 cm

OC = 1/2 AC

OC = 1/2 × 24

OC = 12 cm

Using by Pythagoras Theorem :

→ (Hypotenuse)² = (Base)² + (perpendicular)²

→ (BC)² = (OC)² + (OB)²

→ (15)² = (12)² + (OB)²

→ 225 = 144 + 0B²

→ OB² = 225 - 144

→ OB² = 81

→ OB = √81

→ OB = 9 cm

&

BD = 2 × OB

BD = 2 × 9

BD = 18 cm

Thus,

The shorter diagonal will be 18 cm .

Answer 2

Given -

• The perimeter of a rhombus = 60 cm.
• The length of its longer diagonal = 24 cm.

To find -

• Length of its shorter diagonal.

Solution -

As we know,

Perimeter=4 × side

= 60 = 4 × side

Side= 60/4

Side = 15cm

•°• All the sides of rhombus measures = 15 cm.

In ΔOBC :

AC = 24 cm

OC = 1/2 AC

OC = 1/2 × 24

OC = 12 cm

Using by Pythagoras Theorem :

= (Hypotenuse)² = (Base)² + (perpendicular)²

= (BC)² = (OC)² + (OB)²

= (15)² = (12)² + (OB)²

= 225 = 144 + 0B²

= OB² = 225 - 144

= OB² = 81

= OB = √81

= OB = 9 cm

BD = 2 × OB

BD = 2 × 9

BD = 18 cm

The shorter diagonal = 18 cm .

Figure -

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A}\put(6,0.5){\sf B}\put(1.4,4.3){\sf D}\put(6.6,4.3){\sf C}\end{picture}$