Advertisements
Advertisements
प्रश्न
In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
Advertisements
उत्तर
Draw a line through A parallel to BC
Given that, BD= DE = EC
We observe that the triangles ABD and AEC are on the equal bases and between the same
parallels C and BC. Therefore, Their areas are equal.
Hence, ar ( ABD) = ar (ΔADE) = ar ( ΔACDE)
APPEARS IN
संबंधित प्रश्न
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that
ar (ABE) = ar (ACF)
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that
(i) ar(PRQ) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
In ∆ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Figure), then prove that ar (BPQ) = `1/2` ar (∆ABC).
In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)
In the following figure, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ABP) = ar (ACQ).
