Advertisements
Advertisements
Question
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
Advertisements
Solution
Given: In a parallelogram PQRS, O is any point on the diagonal PR.
To prove: ar (ΔPSO) = ar (ΔPQO)
Construction: Join SQ which intersect PR at B.
Proof: We know that, diagonals of a parallelogram bisect each other, so B is the mid-point of SQ.
Here, PB is a median of ΔQPS and we know that, a median of a triangle divides it into two triangles of equal area.
∴ ar (ΔBPQ) = ar (ΔBPS) ...(i)
Also, OB is the median of ΔOSQ.
∴ ar (ΔOBQ) = ar (ΔOBS) ...(ii)
On adding equations (i) and (ii), we get
ar (ΔBPQ) + ar (ΔOBQ) = ar (ΔBPS) + ar (ΔOBS)
⇒ ar (ΔPQO) = ar (ΔPSO)
Hence proved.
APPEARS IN
RELATED QUESTIONS
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)
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.
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 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.]
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).
[Hint : From A and C, draw perpendiculars to BD.]
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
In the following figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).
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).
