Q1 of 77 Page 16

If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find

are equilateral triangles

We know that,


Area of equilateral triangle = a2


D is the mid-point of BC then,


Area () = * ()2


= *


Now,


Area (: Area (


* a2: *


1:


4: 1


Hence,


Area (: Area ( is 4: 1


More from this chapter

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29

In Fig. 15.90, D and E are two points on BC such that BD=DE=EC. Show that ar(Δ ABD) = ar(Δ ADE) =ar(Δ AEC)

 30

In Fig. 15.91, 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) Δ MBC ≅ ABD (ii) ar(BYXD) =2ar(Δ MBC)


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


(iv) Δ FCB ≅ Δ ACE


(v) ar(CYXE) = 2ar(Δ FCB)


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


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

 2

In Fig. 15.104, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.

 3

In Fig. 15.104, find the area of ΔGEF.