Advertisements
Advertisements
प्रश्न
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).

Advertisements
उत्तर
Given: X and Y are the mid-points of AC and AB respectively. Also, QP || BC and CYQ, BXP are straight lines.
To prove: ar (ΔABP) = ar (ΔACQ)
Proof: Since, X and Y are the mid-points of AC and AB respectively.
So, XY || BC
We know that, triangles on the same base and between the same parallels are equal in area.
Here, ΔBYC and ΔBXC lie on same base BC and between the same parallels BC and XY.
So, ar (ΔBYC) = ar (ΔBXC)
On subtracting ar (ΔBOC) from both sides, we get
ar (ΔBYC) – ar (ΔBOC) = ar (ΔBXC) – ar (ΔBOC)
=» ar (ΔBOY) = ar (ΔCOX)
On adding ar (ΔXOY) both sides, we get
ar (ΔSOY) + ar (ΔXOY) = ar (ΔCOX) + ar (ΔXOY)
⇒ ar (ΔBYX) = ar (ΔCXY) ...(i)
Hence, we observe that quadrilaterals XYAP and YXAQ are on the same base XY and between the same parallels XY and PQ.
ar (XYAP) = ar (YXAQ) ...(ii)
On adding equations (i) and (ii), we get
ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ)
⇒ ar (ΔABP) = ar (ΔACQ)
Hence proved.
APPEARS IN
संबंधित प्रश्न
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
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)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(i) ar (BDE) = 1/4 ar (ABC)
(ii) ar (BDE) = 1/2 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of
BC. AE intersects BC in F. Prove that

(1) ar (Δ BDE) = `1/2` ar (ΔABC)
(2) Area ( ΔBDE) `= 1/2 ` ar (ΔBAE)
(3) ar (BEF) = ar (ΔAFD)
(4) area (Δ ABC) = 2 area (ΔBEC)
(5) ar (ΔFED) `= 1/8` ar (ΔAFC)
(6) ar (Δ BFE) = 2 ar (ΔEFD)
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).

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

A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
