Math, asked by rituravidev, 10 months ago

The diagonals AC and BD of a quadrilateral ABCD are of length 10 cm and 24 cm respectively. If the diagonals bisect each other at 900, find the perimeter of the quadrilateral.

Answers

Answered by burhaanikIQ
49

Step-by-step explanation:

let the diagonals bisect each other at point O

They are perpendicular to each other.

90°

This means that ABCD is a rhombus

AO=1/2AC

AO=1/2×10

AO=5cm

BO=1/2BD

BO=1/2×24

BO=12cm

In right angled ∆AOB,

angle AOB=90°

By Pythagoras' theorem,

AB^2=AO^2+OB^2

AB^2=5^2+12^2

AB^2=25+144

AB^2=169

AB=√169 ..... {taking square roots }

AB=13cm

Area of rhombus=(side)^2

=13^2

169cm^2

Answered by SlicerOG
1

Answer:

Step-by-step explanation:

let the diagonals bisect each other at point O

They are perpendicular to each other.

90°

This means that ABCD is a rhombus

AO=1/2AC

AO=1/2×10

AO=5cm

BO=1/2BD

BO=1/2×24

BO=12cm

In right angled ∆AOB,

angle AOB=90°

By Pythagoras' theorem,

AB^2=AO^2+OB^2

AB^2=5^2+12^2

AB^2=25+144

AB^2=169

AB=√169 ..... {taking square roots }

AB=13cm

Area of rhombus=(side)^2

=13^2

169cm^2,

All sides of rhombus are equal so AB=13,BC=13,CD=13,DA-13

Perimeter of rhombus = 4S

                                    = 4*13

                                    = 62 cm

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