Advertisements
Advertisements
प्रश्न
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)
Advertisements
उत्तर
It is given that
XY || BC ⇒ EY || BC
BE || AC ⇒ BE || CY
Therefore, EBCY is a parallelogram.
It is given that
XY || BC ⇒ XF || BC
FC || AB ⇒ FC || XB
Therefore, BCFX is a parallelogram.
Parallelograms EBCY and BCFX are on the same base BC and between the same parallels BC and EF.
∴ Area (EBCY) = Area (BCFX) ... (1)
Consider parallelogram EBCY and ΔAEB
These lie on the same base BE and are between the same parallels BE and AC.
∴ Area (ΔABE) = 1/2Area (EBCY) ... (2)
Also, parallelogram BCFX and ΔACF are on the same base CF and between the same parallels CF and AB.
∴ Area (ΔACF) = 1/2Area (BCFX) ... (3)
From equations (1), (2), and (3), we obtain
Area (ΔABE) = Area (ΔACF)
संबंधित प्रश्न
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?
[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]
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.
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔABD)
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
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).
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)
