Telangana SCERT Solution Class IX (9) Math Chapter 11 Areas Exercise 11.3
(1) In a triangle ABC (see figure), E is the midpoint of median AD, show that
(i) ar ΔABE = ar ΔACE
(ii) ar( ABC) = ¼ arΔABE
Solution:
Given,
In △ABC
E is the mid point of medium AD
BE is the medium of △ABD
CE is the medium of △AOC
Join AO altitude of △ABC
In △ABD & △ADC
Ar, △ABD = ½ x b x h
= ½ x BD x AO – (i)
Ar, △AOC = ½ x b x h
= ½ x DC x AO – (ii)
Ar △ABC = 1/22 x b x h
= ½ x BC x AO
Adding equation (i) & (ii) we get
Ar △ABD + ar △ADC = ½ x BD x AO + ½ x DC x AO
= ½ AO (BD + DC)
= ½ x AO x BC
= ar △ABC
Now, BD = DC [∴AD median]
ar △ABD = ar △ADC = ½ x AD x BD
∴ We can conclude that area of two triangles foretned with median of the larger triangle is ½ of the area of the larger triangles.
∴ Similarly In △ABD
ar △ABE = ½ ar △ABD [BE is median of △ABD]
Similarly in △ADC
Ar △AEC = ½ ar △ADC [CE is median of △ADC]
Now, ar △ABD = ar △ADC
(i) ar △ABE = ar △AEC = ½ ar △ABD proved
(ii) ar △ABE = ½ ar △ABD
= ½ x ½ ar △ABC [∵median divides the triangle into two equal triangles]
= ¼ ar △ABC proved
(2.) Show that the diagonals of a parallelogram divide it into four triangles of equal area
Given,
ABCD is a ∥ gm
AC & BD diagonals of ∥ gm ABCD
We know,
diagonals of ∥ gm divided the ∥ gm into two equal △s
∴ Ar △ADC, ar △ABC
In △ADC
O is the mid point of AC [∵ diagonals of bisect each other]
∴ DO is the median of △ADC
We know, median of a triangle divided the triangle into two equal angles
∴ ar △ADC = ar △ABD + ar △DOC
= 2 ar △AOD
Ar ∥ gm ABCD = 2 ar △AOB + 2 ar △AOD
= or △AOB + ar △DOC + ar △AOD + ar △DOC
Now, ar △AOB = ar △DOC & ar △AOD = ar △DOC
Now, In △BOC & △ADB
Ar of △DOC = ar of △BOC [∵OC is median of △BDC]
Ar of △ABD = ar of △ADB [∵AO is median △ABD]
△AOB = ar △DOC = ar △AOD = ar △DOC
∴ Diagonals of divided it into 4 equal triangles
(3) In the figure, ΔABC and ΔABD are two triangles on the same base /AB. If line segment CD is bisected by AB at O, show that.
ar (ΔABC) = ar (ΔABD).
Solution:
From,
The figure
CD bisects AB
And AB bisects CD
In △BOC & △AOD
(i) OA = OB [∵AB & DC bisects each other]
(ii) OC = OD [∵AB & DC bisects each other]
(iii) ∠BOC = ∠AOD [vertically opposite angles]
∴ △BOC ≅ △AOD by SAS
∴ AD = BC [sides of congruent triangle]
∠OAD = ∠OBC [angles of congruent triangle]
∠ODA = ∠OCB [angles of congruent triangle]
∴We can say that
∠OAD and ∠ODA are alternate angles to ∠OBC & ∠OCB
∴ AD ∥ BC
∴ Quad ABCD is a ∥ gm AB & CD are diagonals of ∥ gm ABCD
∴ △ABC ≅ (△ABD) [∵AB is a diagonals of ∥ gm ABCD and diagonals divides the ||gm into two congruent triangles]
∴ Ar △ABC = ar (△ABD)
Since congruent as has same area.
(4) In the figure, ΔABC, D, E, F are the midpoints of sides BC, CA and AB respectively. Show that
(i) BDEF is a parallelogram
(ii) ar(ΔDEF) = 1/4 ar(ΔABC)
(iii) ar(BDEF) = 1/2ar(ΔABC)
Solution:
Given,
D, E, F are mid point of side BC, AC of △ABC
∴ In △ABC
EF ∥ BC & EF = ½ BC
[∵by mid point theorem, line joining two midpoint of each side of △ is parallel and half of the third side]
BD = ½ BC [∵D is mid point of BC]
BD ∥ EF [∵BD is part of BC & parts of parallel lines area also parallel]
∴ BD = EF & BO ∥ EF
∴ In quad BDEF opposite sides BD & EF are parallel to each other and equal in length
(i) BDEF is a ∥ gm
(ii) In △BFD & △AFE
(i) EF = BD [opposite side of ∥ gm]
(ii) EBD = AFE [corresponding angles EP ∥ BD]
(iii) BF = AF [∵F is mid point of AB]
∴ △BFD ≅ △AFE by SAS
Similarly △AFE ≅ △DEC by SAS and △BFD ≅ △DEF [∵ diagonal FD divides the ∥ gm BDEF into two congruent triangles]
∴ △AFE ≅ △BFD ≅ △DEF ≅ △DEC
∴ Ar △AFE = ar △BFD = ar △DEF = ar △DEC
[∵congruent triangles has same area]
∴ ar △AFE + ar △BFD + ar △DEF + ar △DEC = ar △ABC
Or, ar △DEF + ar △DEF + ar △DEF + ar △DEF = ar △ABC
Or, 4 ar △DEF = ar △ABC
Or, ar △DEF = ¼ ar △ABC (proved) – (i)
Ar of BPEF = ar △BED + ar △DEF
Or, BDEF = ar △DEF + ar △DEP [∵△BED ≅ △DCF]
Or, ar BDEF = 2 ar △DEF
Or, ar BDEF = 2 x ¼ or △ABC [from equation (ii)]
Or, ar BDEF = ½ ar △ABC (proved)
(5) In the figure D, E are points on the sides AB and AC respectively of ΔABC such that ar(ΔDBC) = ar(ΔEBC). Prove that DE || BC
Solution:
Super but some images not clear ok reaset it in thos all problems are there super, excellent, marbles
E.charan teja
Thank you
Just looking like a wow….