Telangana SCERT Class 9 Math Solution Chapter 11 Areas Exercise 11.3

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:


Updated: October 2, 2021 — 4:32 pm

1 Comment

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  1. Super but some images not clear ok reaset it in thos all problems are there super, excellent, marbles

    E.charan teja

    Thank you

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