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Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.4
  • Last Updated : 12 Mar, 2021

Question 1. Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Solution:

According to the theorem 1, we get

\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2} = \frac{BC^2}{EF^2}

\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{BC^2}{EF^2}



\frac{64}{121} = \frac{BC^2}{15.4^2}

\frac{8^2}{11^2} = \frac{BC^2}{15.4^2}

\frac{8}{11} = \frac{BC}{15.4}

BC = \frac{8}{11}  × 15.4

BC = 11.2 cm

Question 2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD. 

Solution:

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.



In △AOB and △COD,

∠ AOB = ∠ COD (Opposite angles)

∠ 1 = ∠ 2 (Alternate angles of parallel lines)

△AOB ~ △COD by AA property.

According to the theorem 1, we get

\frac{ar(ΔAOB)}{ar(ΔCOD)} = \frac{AB^2}{CD^2}

As, AB = 2CD

\frac{(2CD)^2}{CD^2}

\frac{4CD^2}{CD^2}

\frac{4}{1}

ar(AOB) : ar(COD) = 4 : 1

Question 3. In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \frac{ar(ΔABC)}{ar(ΔDBC)} = \frac{AO}{DO}.

Solution:

Let’s draw two perpendiculars AP and DM on line BC.

Area of triangle = ½ × Base × Height

\frac{ar(ΔABC)}{ar(ΔDBC)} = \frac{½ × BC × AP}{½ × BC × DM}

\frac{ar(ΔABC)}{ar(ΔDBC)} = \frac{AP}{DM}  ……………………………(1)

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each 90°)

∠AOP = ∠DOM (Vertically opposite angles)

ΔAPO ~ ΔDMO by AA similarity

\frac{AP}{DM} = \frac{AO}{DO}  ……………………………(2)

From (1) and (2), we can conclude that

\mathbf{\frac{ar(ΔABC)}{ar(ΔDBC)} = \frac{AO}{DO}}

Question 4. If the areas of two similar triangles are equal, prove that they are congruent.

Solution:

As it is given, ΔABC ~ ΔDEF

According to the theorem 1, we have

\frac{Area of (ΔABC)}{Area of (ΔDEF)} = \frac{BC^2}{EF^2}

\frac{BC^2}{EF^2}  =1 [Since, Area(ΔABC) = Area(ΔDEF)

BC2 = EF2

BC = EF

Similarly, we can prove that

AB = DE and AC = DF

Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

Question 5. D, E and F are respectively the mid-points of sides AB, BC, and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC.

Solution:

As, it is given here

DF = ½ BC

DE = ½ AC

EF = ½ AB

So, \frac{DF}{BC} = \frac{DE}{AC} = \frac{EF}{AB} = \frac{1}{2}

Hence, ΔABC ~ ΔDEF

According to theorem 1,

\frac{ar(ΔDEF)}{ar(ΔABC)} = \frac{DE^2}{AC^2} = \frac{EF^2}{AB^2} = \frac{DF^2}{AC^2} = \frac{1^2}{2^2}

\frac{ar(ΔDEF)}{ar(ΔABC)} = \frac{1}{4}

ar(ΔDEF) : ar(ΔABC) = 1 : 4

Question 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:

Given: AM and DN are the medians of triangles ABC and DEF respectively.

ΔABC ~ ΔDEF

According to theorem 1,

\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}

So, \frac{AB}{DE} = \frac{BC}{EF} = \frac{2BP}{2EQ} = \frac{BP}{EQ}

\frac{AB}{DE} = \frac{BP}{EQ}  ……………………….(1)

∠B = ∠E (because ΔABC ~ ΔDEF)

Hence, ΔABP ~ ΔDEQ [SAS similarity criterion]

\frac{BP}{EQ} = \frac{AP}{DQ}  ……………………….(2)

From (1) and (2), we conclude that

\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{AP^2}{DQ^2}

Hence, proved!

Question 7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:

Let’s take side of square = a

Diagonal of square AC = a√2

As, ΔBCF and ΔACE are equilateral, so they are similar

ΔBCF ~ ΔACE

According to theorem 1,

\frac{ar(ΔACE)}{ar(ΔBCF)} = \frac{AC^2}{BC^2}

= \frac{(a√2)^2}{a^2}

\frac{ar(ΔACE)}{ar(ΔBCF)}  = 2

Hence, Area of (ΔBCF) = ½ Area of (ΔACE)

Tick the correct answer and justify:

Question 8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is

(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

Solution:

Here,

AB = BC = AC = a

and, BE = BD = ED = ½a

ΔABC ~ ΔEBD (Equilateral triangle)

According to theorem 1,

\frac{ar(ΔABC)}{ar(ΔEBD)} = \frac{AB^2}{EB^2} = \frac{a^2}{(½a)^2}

\frac{ar(ΔABC)}{ar(ΔEBD)} = \frac{4}{1}

Area of (ΔABC) : Area of (ΔEBD) = 4 : 1

Hence, OPTION (C) is correct.

Question 9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

(A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81 

Solution:

ΔABC ~ ΔDEF

\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{4}{9}

According to theorem 1,

\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{AB^2}{DE^2}

\frac{ar(ΔABC)}{ar(ΔDEF)} = \frac{4^2}{9^2} = \frac{16}{81}

Area of (ΔABC) : Area of (ΔDEF) = 16 : 81

Hence, OPTION (D) is correct.

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