Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 12 | Page 45
Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal
solution
It is given that,
AB2 + AD2 = 130 sq. cm
BD = 14 cm
Diagonals of a parallelogram bisect each other.
i.e. O is the midpoint of AC and BD.
In ∆ABD, point O is the midpoint of side BD.
[BO = OD = frac{1}{2}BD = 7 cm]
[{AB}^2 + {AD}^2 = 2 {AO}^2 + 2 {BO}^2 left( text{by Apollonius theorem} right)]
[ Rightarrow 130 = 2 {AO}^2 + 2 left( 7 right)^2 ]
[ Rightarrow 130 = 2 {AO}^2 + 2 times 49]
[ Rightarrow 130 = 2 {AO}^2 + 98]
[ Rightarrow 2 {AO}^2 = 130 – 98]
[ Rightarrow 2 {AO}^2 = 32]
[ Rightarrow {AO}^2 = 16]
[ Rightarrow AO = 4 cm]
Since point O is the midpoint of side AC.
∴ AC = 2AO = 8 cm
Hence, the length of the other diagonal is 8 cm.
Explanation:-
Let ABCD be a parallelogram with diagonals AC and BD intersecting at O. Let AB = a, BC = b, AC = c and BD = d. We know that:
a^2 + b^2 = 130 (given)
We also know that the diagonals of a parallelogram bisect each other. Therefore, we have:
AO = CO = 1/2 * AC = 1/2 * c
BO = DO = 1/2 * BD = 1/2 * d
We can apply the Pythagorean theorem in triangles AOB and COD to get:
OA^2 + AB^2 = OB^2 and OC^2 + CD^2 = OD^2
Substituting the values of OA, OB, OC and OD, we get:
(1/2c)^2 + a^2 = (1/2d)^2 and (1/2c)^2 + b^2 = (1/2d)^2
Simplifying these equations, we get:
c^2/4 + a^2 = d^2/4 and c^2/4 + b^2 = d^2/4
Adding these equations, we get:
c^2/2 + a^2 + b^2 = d^2/2
But we know that a^2 + b^2 = 130. Therefore, we get:
c^2 + 130 = d^2
Given that one diagonal of the parallelogram is 14 cm, we can assume without loss of generality that c is the shorter diagonal. Therefore, we have:
c^2 + 130 = 196
Simplifying, we get:
c^2 = 66
Hence, the length of the shorter diagonal is:
c = sqrt(66) cm
To find the length of the longer diagonal, we can use the equation:
c^2 + 130 = d^2
Substituting the value of c, we get:
66 + 130 = d^2
Simplifying, we get:
d^2 = 196
Therefore, the length of the longer diagonal is:
d = 14 cm
Hence, we have found that the length of the longer diagonal of the parallelogram is 14 cm.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Problem Set 2 | Q 12 | Page 45