मराठी

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

Advertisements
Advertisements

प्रश्न

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

उत्तर

Since ΔABQ and ΔPBQ lie on the same base BQ and are between the same parallels AP and BQ,

∴ Area (ΔABQ) = Area (ΔPBQ) ... (1)

Again, ΔBCQ and ΔBRQ lie on the same base BQ and are between the same parallels BQ and CR.

∴ Area (ΔBCQ) = Area (ΔBRQ) ... (2)

On adding equations (1) and (2), we obtain

Area (ΔABQ) + Area (ΔBCQ) = Area (ΔPBQ) + Area (ΔBRQ)

⇒ Area (ΔAQC) = Area (ΔPBR)

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Areas of Parallelograms and Triangles - Exercise 9.3 [पृष्ठ १६४]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 9
पाठ 9 Areas of Parallelograms and Triangles
Exercise 9.3 | Q 14 | पृष्ठ १६४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).


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 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.]


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 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 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).


PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.


The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×