मराठी

In a δAbc, If L and M Are Points on Ab and Ac Respectively Such that Lm || Bc. Prove That: (1) Ar (δLcm ) = Ar (δLbm ) (2) Ar (δLbc) = Ar (δMbc) (3) Ar (δAbm) Ar (δAcl) (4) Ar (δLob) Ar (δMoc) - Mathematics

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प्रश्न

In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:

(1) ar (ΔLCM ) = ar (ΔLBM )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)

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उत्तर

(1)   Clearly Triangles LMB and LMC are on the same base LM and between the same
parallels LM and BC.

∴ ar (ΔLMB) = ar (ΔLMC)      ......(1)

(2) We observe that triangles LBC and MBC area on the same base BC and between the
same parallels LM and BC
 ∴ arc  ΔLBC = ar (MBC)                  ..........(2)

 (3)  We have
ar (ΔLMB) = ar  (ΔLMC)                       [from (1)]
⇒  ar  ( ΔALM) + ar (ΔLMB) = ar (ΔALM) +  ar (LMC) 
⇒  ar (ΔABM)  = ar (ΔACL)

(4)  We have
ar(ΔCBC) = ar (ΔMBC)              ∴ [from (1)]

⇒ ar  (ΔLBC) =  ar (ΔBOC) =  a (ΔMBC) - ar (BOC)

⇒  ar (ΔLOB) = ar (ΔMOC)

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पाठ 14: Areas of Parallelograms and Triangles - Exercise 14.3 [पृष्ठ ४८]

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आरडी शर्मा Mathematics [English] Class 9
पाठ 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 28 | पृष्ठ ४८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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


XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that

ar (ABE) = ar (ACF)


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)]


Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).

[Hint : From A and C, draw perpendiculars to BD.]


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)


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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)


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