हिंदी

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

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.

योग
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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Areas of Parallelograms & Triangles - Exercise 9.4 [पृष्ठ ९४]

APPEARS IN

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

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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.]


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)


The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure). Show that

ar (ABCD) = ar (PBQR).

[Hint: Join AC and PQ. Now compare area (ACQ) and area (APQ)]


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 following figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar (DPC) = `1/2` ar (EFGD).


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)


In the following figure, CD || AE and CY || BA. Prove that ar (CBX) = ar (AXY).


In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×