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.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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 $\frac{1}{2} \times \text{Product of the sides forming the right angle}$.

Explore More

Similar Questions

$ABCD$ is a trapezium in which $AB \parallel DC$,$DC = 30 \, cm$ and $AB = 50 \, cm$. If $X$ and $Y$ are,respectively,the mid-points of $AD$ and $BC$,prove that $\operatorname{ar}(DCYX) = \frac{7}{9} \operatorname{ar}(XYBA)$.

Difficult
View Solution

If a triangle and a parallelogram are on the same base and between the same parallels,then the ratio of the area of the triangle to the area of the parallelogram is:

In $\Delta ABC$,point $D$ lies on side $BC$. $E$ is the midpoint of $AD$. Prove that,$ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.

In the given figure,$PQM$ is a line and $SQ || RM$. Prove that $ar(PQR) = ar(PMS)$.

$X$ and $Y$ are points on the side $LN$ of the triangle $LMN$ such that $LX = XY = YN$. Through $X$,a line is drawn parallel to $LM$ to meet $MN$ at $Z$ (See figure). Prove that $\operatorname{ar}(LZY) = \operatorname{ar}(MZYX)$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo