Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumProve that the area of a rhombus is half the product of its diagonals.
(N/A) Let $ABCD$ be a rhombus with diagonals $AC$ and $BD$ intersecting at point $O$.
By the properties of a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Thus,$AC \perp BD$ and $AO = OC = \frac{1}{2} AC$,$BO = OD = \frac{1}{2} BD$.
The area of the rhombus $ABCD$ is the sum of the areas of $\triangle ABC$ and $\triangle ADC$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BO$.
Area of $\triangle ADC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times OD$.
Total Area $= \frac{1}{2} \times AC \times BO + \frac{1}{2} \times AC \times OD = \frac{1}{2} \times AC \times (BO + OD)$.
Since $BO + OD = BD$,the Area $= \frac{1}{2} \times AC \times BD$.
Therefore,the area of a rhombus is half the product of its diagonals.
By the properties of a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.
Thus,$AC \perp BD$ and $AO = OC = \frac{1}{2} AC$,$BO = OD = \frac{1}{2} BD$.
The area of the rhombus $ABCD$ is the sum of the areas of $\triangle ABC$ and $\triangle ADC$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times BO$.
Area of $\triangle ADC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AC \times OD$.
Total Area $= \frac{1}{2} \times AC \times BO + \frac{1}{2} \times AC \times OD = \frac{1}{2} \times AC \times (BO + OD)$.
Since $BO + OD = BD$,the Area $= \frac{1}{2} \times AC \times BD$.
Therefore,the area of a rhombus is half the product of its diagonals.
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Similar Questions
State whether each of the following statements is true or false:
$(1)$ If $ar(ABC) = 96 \, cm^2$ for the parallelogram $ABCD$,then $ar(ABCD) = 192 \, cm^2$.
$(2)$ Area of a right triangle = Product of the sides forming the right angle.
$(1)$ If $ar(ABC) = 96 \, cm^2$ for the parallelogram $ABCD$,then $ar(ABCD) = 192 \, cm^2$.
$(2)$ Area of a right triangle = Product of the sides forming the right angle.
Medium
View SolutionIn the figure,$ABCD$ is a parallelogram. Points $P$ and $Q$ on $BC$ trisect $BC$ into three equal parts. Prove that $\operatorname{ar}(APQ) = \operatorname{ar}(DPQ) = \frac{1}{6} \operatorname{ar}(ABCD)$.
Difficult
View SolutionWrite True or False and justify your answer:
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.
Medium
View Solution$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:
Easy
View SolutionIn $\Delta ABC$,$AD$ is a median and $AM$ is an altitude. The side $BA$ of $\Delta ABC$ is produced to any point $E$,so that $AB = AE$. If $BC = 16\, cm$ and $AM = 8\, cm$,then find the area of $\Delta EBD$ in $cm^2$.
Medium
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