Advertisements
Advertisements
प्रश्न
Prove that the area of a rhombus is equal to half the rectangle contained by its diagonals.
उत्तर
Since the diagonals of a rhombus intersect at right angles,
Therefore, OB ⊥ AC and OD ⊥AC
Now, ar(rhombus ABCD)
= ar(ΔABC) + ar(ΔADC)
= `(1)/(2)("AC" xx "BO") + (1)/(2)("AC" xx "DO")`
= `(1)/(2){"AC" xx ("BO" + "DO")}`
= `(1)/(2)("AC" xx "BD")`
Therefore, the area of a rhombus is equal to half the rectangle contained by its diagonals.
APPEARS IN
संबंधित प्रश्न
In the following figure, ABCD is a rhombus and DCFE is a square.
If ∠ABC =56°, find:
(i) ∠DAE
(ii) ∠FEA
(iii) ∠EAC
(iv) ∠AEC
Prove that the quadrilateral formed by joining the mid-points of consecutive sides of a rectangle is a rhombus.
Prove that the quadrilateral formed by joining the mid-points of consecutive sides of a rhombus is a rectangle.
Prove that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus.
Diagonals of a rhombus are equal and perpendicular to each other.
A rhombus can be constructed uniquely if both diagonals are given.
The diagonals of a rhombus are 8 cm and 15 cm. Find its side.
In rhombus BEAM, find ∠AME and ∠AEM.
