Find the area of rombhus in which each side is 15 cm and one of its diognal is 24 cm
Answers
Answer:
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Step-by-step explanation:
1/2*15*24
1*15*12
=180
here is the explanation :-
1st method
let ABCD is a rhombus with diagonal AC and BD which intersect each other at O.
AC = 24 => AO = 12
let BO = x and AB = 15 cm (given)
by Pythagorean theorem
c²= a² + b²
225 = 144 + x²
x² = 225 - 144
x² = 81
x = 9 cm
BO = 9cm
diagonal BD = 2 × 9 = 18cm.
Area = ½ × ( product of diagonal )
½ × 24 × 18
Area = 216sq.cm
2nd method
let's call it ABCD and name the mid point E .
the diagonal AC has length 24 cm
AE is half of AC , so the length is 12 cm
ABE is a right triangle with sides AB = 15cm and AE = 12 cm
using Pythagoras theorem, BE = 9cm
diagonal BD is twice the length of BE there fore it is 18 cm
using the area formula A = d¹ d²/2 ,we have 18×24/2
= 216 ²
3rd method
(AB)² = (OA)² + (OB)²
(15)² = (12)² + (OB)²
(OB)²= (15)² - (12)²
(OB)² = 225-144
= 81
(OB)² = (9)²
(OB) = 9 cm
BD= 2 × OB
= 2 × 9
= 18
hence area of Rhombus ABCD = ½ × AC × BD
= ( 1/2 × 24 × 18 )
= 216 cm ²