Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)
In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).
D and E are points on sides AB and AC respectively of ΔABC such that
ar (DBC) = ar (EBC). Prove that DE || BC.
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)
In the given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.
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 the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-
(i) ΔMBC ≅ ΔABD
(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
(iv) ΔFCB ≅ ΔACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar (CYXE) = ar(ACFG)
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)
Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
In the following figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
