In Delta ABC,point D lies on side BC. E is th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(N/A) Given: In $\Delta ABC$,$D$ is a point on $BC$ and $E$ is the midpoint of $AD$.To prove: $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.Proof:$1$.

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) Given: In $\Delta ABC$,$D$ is a point on $BC$ and $E$ is the midpoint of $AD$.
To prove: $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Proof:
$1$. In $\Delta ABD$,$E$ is the midpoint of $AD$. Since $BE$ is a median of $\Delta ABD$,it divides the triangle into two triangles of equal area. Therefore,$ar(\Delta EBD) = \frac{1}{2} ar(\Delta ABD)$.
$2$. Similarly,in $\Delta ADC$,$E$ is the midpoint of $AD$. Since $CE$ is a median of $\Delta ADC$,it divides the triangle into two triangles of equal area. Therefore,$ar(\Delta ECD) = \frac{1}{2} ar(\Delta ADC)$.
$3$. Adding the two equations: $ar(\Delta EBD) + ar(\Delta ECD) = \frac{1}{2} ar(\Delta ABD) + \frac{1}{2} ar(\Delta ADC)$.
$4$. $ar(\Delta EBC) = \frac{1}{2} [ar(\Delta ABD) + ar(\Delta ADC)]$.
$5$. Since $ar(\Delta ABD) + ar(\Delta ADC) = ar(\Delta ABC)$,we get $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Hence proved.

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

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.

In the figure,$PSDA$ is a parallelogram. Points $Q$ and $R$ are taken on $PS$ such that $PQ = QR = RS$ and $PA \parallel QB \parallel RC$. Prove that $\operatorname{ar}(PQE) = \operatorname{ar}(CFD)$.

$(1)$ If a planar region formed by a figure $T$ is made up of two non-overlapping planar regions formed by figures $P$ and $Q$,then $\operatorname{ar}(T) = \dots$
$(2)$ Area of a parallelogram $= \dots$

$ABCD$ 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 $\operatorname{ar}(DCYX) = \frac{7}{9} \operatorname{ar}(XYBA)$.

$(1)$ The part of the plane enclosed by a simple closed figure is called a $\ldots \ldots \ldots$
$(2)$ $\ldots \ldots \ldots$ of the planar region corresponding to a closed figure is called its area.

Vedclass Test Series

Exam Paper Generator

Online Exam Module