हिंदी

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).

Advertisements
Advertisements

प्रश्न

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).

Advertisements

उत्तर

Draw a line through A parallel to BC

Given that, BD= DE = EC
We observe that the triangles ABD and AEC are on the equal bases and between the same
parallels C and BC. Therefore, Their areas are equal.
Hence, ar ( ABD) = ar (ΔADE) = ar ( ΔACDE)

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Areas of Parallelograms and Triangles - Exercise 14.3 [पृष्ठ ४६]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 9
अध्याय 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 15 | पृष्ठ ४६

वीडियो ट्यूटोरियलVIEW ALL [1]

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.


In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)


In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).


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.]


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 ______.


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.


In ∆ABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (LOB) = ar (MOC)


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).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×