Similarity of Triangles NCERT Exercise 6.3 part three Class Ten Mathematics

Triangle

Exercise 6.3 Part 3

Question 10: CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ∼ Δ FEG, show that:

similar triangles exercise solution

(a) `(CD)/(GH)=(AC)/(FG)`

Solution: Δ ABC ∼ Δ FEG (given)

Hence; ∠ABC = ∠FEG

∠ACB = ∠FGE

∠BAC = ∠GFE -------------(1)


Hence; ∠ACD = ∠FGH (Halves of ∠BCA and ∠FGE) -------------(2)

From equations (1) and (2);

ΔACD ∼ Δ FGH

Hence; in these triangles,

`(CD)/(AC)=(GH)/(FG)`

Or, `(CD)/(GH)=(AC)/(FG)`

Proved

(b) Δ DCB ∼ Δ HGE

Solution: In Δ DCB and Δ HGE

∠DCB = ∠HGE (Halves of ∠BCA and ∠FGE)

∠DBC = ∠HEG (Because they are common to ∠ABC and ∠FEG)

Hence; Δ DCB ∼ Δ HE proved

(c) Δ DCA ∼ Δ HGF

Solution: In Δ DCA and Δ HGF

ΔACD ∼ Δ FGH [proved in question (a)]

Hence; ΔACD ∼ Δ FGH proved

(Note: points C and G are in middle and sequence of other points has been changed in naming these triangles. Hence; triangles in question (a) are similar in this question.)


Question 11: In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ∼ Δ ECF.

similar triangles exercise solution

Solution: In Δ ABD and Δ ECF

∠ADB = ∠EFC (Right angle)

∠ABC = ∠ECF (Angles opposite to equal sides)

Hence; Δ ABD ∼ Δ ECF (AAA criterion)

Question 12: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR. Show that Δ ABC ∼ Δ PQR

similar triangles exercise solution

Solution: In triangle Δ ABC and Δ PQR

`(AB)/(PQ)=(AC)/(PR)=(AD)/(PM)` (given) ---------(1)

Hence; ∠BAD = ∠QPM

∠DAC = ∠MPR

(These are angles made by a side and median of one triangle and corresponding side and median of another triangle)

Hence; ∠BAD + ∠DAC = ∠QPM + ∠MPR

Or, ∠BAC = ∠QPR -----------(2)

From equation (1) and (2);

Δ ABC ∼ Δ PQR proved (SAS criterion)


Question 13: D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.

similar triangles exercise solution

Solution: In ΔBAC and ΔADC;

∠BAC = ∠ADC (given)

∠ACB = ∠DCA (Common angle)

Hence; ΔBAC ∼ ΔADC

Hence; `(CA)/(CB)=(CD)/(CA)`

(corresponding sides are in same ratio)

Or, `CA xx CA = CB xx CD`

Or, `CA^2 = CB xx CD` proved

Question 14: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

similar triangles exercise solution

Solution: Height of pole = AB = 6 m and its shadow = BC = 4 m

Height of tower = PQ = ? and its shadow = QR = 28 m

The angle of elevation of the sun will be same at a given time for both the triangles.

Hence; ΔABC ∼ ΔPQR

This means;

`(AB)/(AC)=(PQ)/(QR)`

Or, `6/4=(PQ)/(28)`

Or, `PQ=(6xx28)/(4)=42 cm`

Height of tower = 42 m

Question 15: If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ∼ Δ PQR, prove that `(AB)/(PQ)=(AD)/(PM)`

similar triangles exercise solution

Solution: Δ ABC ∼ Δ PQR (Given)

Hence; `(AB)/(AD)=(PQ)/(PM)`

(A side and the median of one triangle are in same ratio as a corresponding side and median of another triangle)

`(AB)/(PQ)=(AD)/(PM)`

Proved



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