मराठी

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.

पर्याय

  • True

  • False

MCQ
चूक किंवा बरोबर

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Areas of Parallelograms & Triangles - Exercise 9.2 [पृष्ठ ८८]

APPEARS IN

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

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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)


D and E are points on sides AB and AC respectively of ΔABC such that

ar (DBC) = ar (EBC). Prove that DE || BC.


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, 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 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 a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :

(1) ar (Δ PBQ) = ar (Δ ARC)

(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)

(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .


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)


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)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×