Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumState 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.
(A) $(1)$ True. $A$ diagonal of a parallelogram divides it into two triangles of equal area. Therefore,$ar(ABCD) = 2 \times ar(ABC) = 2 \times 96 \, cm^2 = 192 \, cm^2$.
$(2)$ False. The area of a right triangle is given by $\frac{1}{2} \times \text{Product of the sides forming the right angle}$.
$(2)$ False. The area of a right triangle is given by $\frac{1}{2} \times \text{Product of the sides forming the right angle}$.
Explore More
Similar Questions
$O$ is any point on the diagonal $PR$ of a parallelogram $PQRS$. Prove that $\operatorname{ar}(\triangle PSO) = \operatorname{ar}(\triangle PQO)$.
Difficult
View SolutionState 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 $= (\text{Side})^2$.
Easy
View SolutionWrite True or False and justify your answer:
$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \, cm$. $A$ is any point on $PQ$. If $PS = 5 \, cm$,then $\text{ar}(PAS) = 30 \, cm^2$.
$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \, cm$. $A$ is any point on $PQ$. If $PS = 5 \, cm$,then $\text{ar}(PAS) = 30 \, cm^2$.
Difficult
View SolutionThe medians $BE$ and $CF$ of a triangle $ABC$ intersect at $G$. Prove that the area of $\triangle GBC = \text{area of the quadrilateral } AFGE.$
Difficult
View SolutionWhich of the following figures lie on the same base and between the same parallels? Write the common base and the two parallels for the figure for which the answer is affirmative.
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo