Math, asked by mmlagrawal, 4 months ago

Find the area of rombhus in which each side is 15 cm and one of its diognal is 24 cm​

Answers

Answered by harishyadavmuthyala2
1

Answer:

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Step-by-step explanation:

1/2*15*24

1*15*12

=180

Answered by Anonymous
6

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

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