Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
EasyState whether each of the following statements is true or false:
$(1)$ Area of a parallelogram $= \text{base} \times \text{corresponding altitude}$.
$(2)$ Area of a rhombus $= \frac{1}{2} \times \text{Product of its diagonals}$.
$(3)$ Area of a square $= (\text{Side})^2$.
$(1)$ Area of a parallelogram $= \text{base} \times \text{corresponding altitude}$.
$(2)$ Area of a rhombus $= \frac{1}{2} \times \text{Product of its diagonals}$.
$(3)$ Area of a square $= (\text{Side})^2$.
(A) $(1)$ False. The area of a parallelogram is given by the product of its base and the corresponding altitude,not half of it.
$(2)$ False. The area of a rhombus is half the product of its diagonals,i.e.,$\frac{1}{2} \times d_1 \times d_2$.
$(3)$ True. The area of a square is indeed the square of its side length.
$(2)$ False. The area of a rhombus is half the product of its diagonals,i.e.,$\frac{1}{2} \times d_1 \times d_2$.
$(3)$ True. The area of a square is indeed the square of its side length.
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Similar Questions
In $\Delta ABC$,$AD$ is a median. If $ar(\Delta ABC) = 50 \, cm^2$,then $ar(\Delta ADC) = \dots \dots \dots cm^2$.
Easy
View SolutionIn $\Delta PQR$,$PM$ is a median and $N$ is the midpoint of $PM$. If $\text{ar}(PQN) = 36 \text{ cm}^2$,then $\text{ar}(PQR) = \dots \text{ cm}^2$.
Medium
View SolutionWrite True or False and justify your answer:
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$
Medium
View SolutionIn $\Delta ABC$,point $D$ lies on side $BC$. $E$ is the midpoint of $AD$. Prove that,$ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Medium
View SolutionState 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
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