English

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

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

Given: In the following figure, CD || AE and CY || BA

To prove: ar (ΔCBX) = ar (ΔAXY) .

Proof: We know that, triangles on the same base and between the same parallels are equal in areas.

Here, ΔABY and ΔABC both lie on the same base AB and between the same parallels CY and BA.

ar (ΔABY) = ar (ΔABC)

⇒ ar (ABX) + ar (AXY) = ar (ABX) + ar (CBX)

⇒ ar (AXY) = ar (CBX)  ...[Eliminating ar (ABX) from both sides]

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Areas of Parallelograms & Triangles - Exercise 9.4 [Page 95]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 9
Chapter 9 Areas of Parallelograms & Triangles
Exercise 9.4 | Q 4. | Page 95

RELATED QUESTIONS

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


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)]


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.]


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


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.]


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 ______.


ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.


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


X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Prove that ar (LZY) = ar (MZYX)


If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×