Question

In Fig. 15.83, CD||AE and CY||BA.
(i) Name a triangle equal in area of Δ CBX
(ii) Prove that ar(Δ ZDE) = ar(Δ CZA)
(iii) Prove that ar(Δ BCYZ) = ar(Δ EDZ)

Answer 1

Given : CD || AE and CY || BA.

 

To prove :

(i) Name a triangle equal in area of Δ CBX

(ii) Prove that ar(Δ ZDE) = ar(Δ CZA)

(iii) Prove that ar(Δ BCYZ) = ar(Δ EDZ)

 

Solution :  

(i) ΔAYC and Δ BCY are on the same base CY and between the same parallels CY || AB.

We know that triangles on the same base and between the same parallels are equal in areas.

∴ ar (ΔAYC) = ar (ΔBCY)

On Subtracting ΔCXY from both sides we obtain :  

ar (ΔAYC) – ar (ΔCXY) = ar (ΔBCY) – ar (ΔCXY)  

ar (ΔCBX) = at (ΔAXY)

 

(ii) Since, ΔACC and ΔADE are on the same base AF and between the same parallels CD || AF.

We know that triangles on the same base and between the same parallels are equal in areas.

∴ ar (∆ACE) = ar (∆ADE)

On Subtracting ΔAZE from both sides we obtain :  

ar (∆ACE) -  ar (ΔAZE) = ar (ΔADE) - ar (∆AZE)

ar (∆ACZ) = ar (∆ZDE)

 

(iii) Since, ΔBCY and ΔACY are on the same base CY and between the same parallels CY || BA.

∴ ar (ΔACY) = Area (ΔBCY) ……..(1)

Now ar (∆ACZ) = ar (∆EDZ)

ar (∆ACY) + ar (∆CYZ) = ar (∆EDZ)

ar (∆BCY) + ar (∆CYZ) = ar (∆EDZ)

[From eq 1]

ar (quad. BCZY) = ar (∆EDZ)

 

HOPE THIS ANSWER WILL HELP YOU…..

 

Some questions of this chapter :

A point D is taken on the side BC of a Δ ABC such that BD = 2DC. Prove that

ar(Δ ABD) = 2ar(Δ ADC)

https://brainly.in/question/15909774

In Fig. 15.82, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, Show that ar(Δ ABC) = ar(Δ ABD).

https://brainly.in/question/15909772

Answer 2

$\bf \fbox{ \huge\red{ SOLUTION}}$

since, ΔBCA and ΔBYA are on the same base BA and between asme parallels BA and CY Then area (ΔBCA)

$\bf \red{→}ar(BYA)$

$\bf \pink{→}ar(Δ CBX) + ar(Δ BXA) =$

$\bf = ar(Δ BXA) + ar(Δ AXY)$

$\bf \red{→}ar(Δ CBX) = ar(ΔBXA) =$

$\bf = ar(Δ BXA) + ar(Δ AXY)$

$\bf \pink→ ar (Δ CBX )= ar(Δ AXY) \: \: ...(i)$

Since, ΔACE are on the same base AE and between same parallels CD and AE

Then, ar (ΔACE) ar (ΔADE)

= (ΔCLA) + ar (ΔAZE) = ar (ΔAZE) + ar (ΔDZE)

= ar (ΔCZA) = (ΔDZE) ...... (2)

Since, ΔCBY and ΔCAY ar on the same base CY and between same parallels

BA and CY

Then ar (ΔCBY) ar (ΔCAY)

Adding ar (CYG) on both sides, we get

= ar (ΔCBX) + ar (ΔCYZ) = (ΔCAY) + ar (ΔCYZ)

= ar (BCZX) = ar (ΔCZA) .......... (3)

Compare equation (2) and (3) are (BCZY) = ar (ΔDZE)