Ex 7.2, 10
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. [Hint : Area of a rhombus = 1/2 (product of its diagonals)]
Let the vertices be
A (3, 0) , B (4, 5)
C (−1, 4) , D (−2, −1)
We know that
Area of Rhombus = 1/2 (Product of diagonals)
= 1/2 × AC × BD
We need to find AC & BD using distance formula
Finding AC
AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
Here, x1 = 3 , y1 = 0
x2 = −1, y2 = 4
Putting values
AC = √((−1−3 )2+(4 −0)2)
= √((−4 )2+(4)2)
= √((4 )2+(4)2)
= √(2(4)2)
= √2 × 4
= 4√2
Finding BD
BD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2)
Here, x1 = 4
y1 = 5
x2 = −2
y2 = −1
Putting values
AC = √((−2 −4 )2+(−1−5)2)
= √((−6 )2+(−6)2)
= √((6)2+(6)2)
= √(2(6)2)
= √2 × 6
= 6√2
Now,
Area of Rhombus = 1/2 × AC × BD
= 1/2 × 4√2 × 6√2
= 2√2 × 6√2
= (2 × 6) × (√2 × √2)
= (12) × (2)
= 24 square units
Hence, Area of rhombus = 24 square units

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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