Question

The diagonals of a quadrilateral are of length 12 cm and 16 cm. If the diagonals bisect each other and are perpendicular to each other.

1) What special name can be given to the quadrilateral?
2) What is the length of each side of the quadrilateral?
3) Find its perimeter also.​

Answer 1

$\boxed{\pink{\sf{Given:-}}}$

• The diagonals of a quadrilateral are of length 12 cm and 16 cm.
• The diagonals bisect each other and are perpendicular to each other.

$\boxed{\pink{\sf{To\:\:be\:\:found :-}}}$

1. What special name can be given to the quadrilateral?
2. What is the length of each side of the quadrilateral?
3. Find its perimeter also.​

$\red{\underline{\underline{\tt{Solution:-}}}}$

[ See the attached image ]

1. $\green{\huge{\underline{ \sf Rhombus.}}}$

As the conditions satisfy the property of rhombus-

• the diagonals bisect each other and are perpendicular to each other.

2. $\green{\huge{\underline{ \sf 10cm}}}$

As the diagonals bisect each other and are perpendicular to each other, so OB = OD = 6cm [as BD = 12] and OA = OC = 8cm [AC = 16cm]

Taking center as O,

We get right-angle triangles i.e ΔAOB, ΔBOC, ΔOCD, ΔAOD

Taking any one triangle ΔAOB

Using Pythagoras theorem,

$\boxed{\bf (Hypotenuse)^{2} = (base)^{2} + (perpendicular)^{2}}$

→ (AD)² = (6)² + (8)²

→ (AD)² = 36 + 64

→ (AD)² = 100

→ AD = $\sqrt{100}$

→ AD = 10 cm

And as the quadrilateral is rhombus so all sides are equal.

AD = AB = BC = DC = 10 cm each

3. $\green{\huge{\underline{ \sf 40cm}}}$

Perimeter = sum of all sides

= 10+10+10+10 = 40 cm

Answer 2

1. Rhombus

as the diagonals bisect each other at 90 degree (given)

2. using Pythagoras theorem

we get 10cm

x^{2} = 36+64

=> x = √100

=> x = 10

3.

ans.4× side = 4 × 10 = 40 cm