हिंदी

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

Advertisements
Advertisements

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)

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)

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 14: Areas of Parallelograms and Triangles - Exercise 14.3 [पृष्ठ ४८]

APPEARS IN

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

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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)]


In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).

Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]


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.


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 ______.


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


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).


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)


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×