each side of rhombus is 3root5 and its one diagonal is twice the other ,then find the area of the quadrilateral . this question is from Mensuration of class8 question no 29
Answers
Solution :-
Given : Each side of rhombus is 3√5 cm.
Let the length of one diagonal be x
Length of other diagonal = 2x
We know that : Diagonals of a rhombus bisect each other at 90°.
Suppose that : ABCD is a rhombus and AC and BD are the two diagonals of the rhombus which bisect each other at 90°.
By applying Pythagoras theorem,
AB² = AO² + BO²
=> (3√5)² = (2x/2)² + (x/2)²
=> 45 = x² + x²/4
=> 45 = (4x² + x²)/4
=> 45 × 4 = 5x²
=> x² = 180/5
=> x = √36 = 6
Length of one diagonal = x = 6 cm
Length of other diagonal = 2x = 2 × 6 = 12 cm
Area of Rhombus = ½(product of its diagonals) sq. units
= ½(6 × 12) cm²
= 72/2 = 36 cm²
Hence,
Area of rhombus = 36 cm²
Answer
Explanation
Given :-
Each side of rhombus = 3√5 cm
Let the length of the diagonal be β
So,
length of other diagonal is 2β
_________________________
We know that diagonals bisects each other at 90°
_________________________
Now, see the attached picture
PQ² = OP² + OQ²
_________[By Pythagoras theorem]
(3√5)² = (2β/2)² + (x/2)²
45 = β² + β²/4
45 = 4β² + β² / 4
45 * 4 = 5β²
5β² = 180
β² = 180/5
β² = 36
β = √36
β = 6
β = 6cm
____________________
One diagonal = β
» (6)
» 6cm
_______________
Other diagonal = 2β
» 2(6)
» 12 cm
____________________
So, the diagonals are 6cm and 12cm
