हिंदी

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

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • True

  • False

MCQ
सत्य या असत्य

उत्तर

This statement is False.

Explanation:

Given in the question, ABCD is a parallelogram and X is the mid-point of AB.

So, area(ABCD) = area(AXCD) + area(ΔXBC)  ...(i)

Now, diagonal AC of a parallelogram divides it into two triangles of equal area.

area(ABCD) = 2area(ΔABC)  ...(ii)

Similarly, X is the mid-point of AB,

So, area(ΔCXB) = `1/2`area(ΔABC)   ...(iii) [Median divides the triangle in two triangles of equal area]

2area(ΔABC) = `24 + 1/2` area(ΔABC)  ...[By using equation (i), (ii) and (iii)]

Now, 2area(ΔABC) – `1/2`area(ΔABC) = 24

`3/2`area(ΔABC) = 24

Therefore, area(ΔABC) = `(2 xx 24)/3` = 16 cm2.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Areas of Parallelograms & Triangles - Exercise 9.2 [पृष्ठ ८८]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
अध्याय 9 Areas of Parallelograms & Triangles
Exercise 9.2 | Q 1. | पृष्ठ ८८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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)


A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.


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 is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-

(i) ΔMBC ≅ ΔABD

(ii) ar (BYXD) = 2 ar(MBC)

(iii) ar (BYXD) = ar(ABMN)

(iv) ΔFCB ≅ ΔACE

(v) ar(CYXE) = 2 ar(FCB)

(vi) ar (CYXE) = ar(ACFG)

(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.


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×