मराठी

The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.

Advertisements
Advertisements

प्रश्न

The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.

बेरीज
Advertisements

उत्तर

Given: The medians BE and CF of a triangle ABC intersect at G

To prove: ar (ΔGBC) = ar (AFCE)

Proof: As median CF divides a triangle into triangle of equal area. So, ar (ΔBCF) = ar (ΔACF)

ar (ΔGBF) + ar (ΔGBC) = ar (AFGE) + ar (ΔGCE)  ...(I)

Now, median BE divides a triangle into two triangle of equal area.

So, ar (ΔGBF) + ar (AFGE) = ar (ΔGCE) + ar (ΔGBC)  ...(II)

Now, subtracting (II) from (I), we get

ar (ΔGBC) – ar (AFGE) = ar (ΔAFGE) – ar (ΔGBC)

ar (ΔGBC) + ar (ΔGBC) = ar (ΔAFGE) + ar (ΔAFGE)

2ar (ΔGBC) = 2ar (AFGE)

Hence, ar (ΔGBC) = ar (AFGE).

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Areas of Parallelograms & Triangles - Exercise 9.4 [पृष्ठ ९४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
पाठ 9 Areas of Parallelograms & Triangles
Exercise 9.4 | Q 3. | पृष्ठ ९४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE)


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)


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 given figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.


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.]


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 below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).


In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:

(1) ar (ΔLCM ) = ar (ΔLBM )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)


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 ______.


ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar (BDE) = `1/4` ar (ABC).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×