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
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
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