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NCERT solution class 10 chapter 6 Triangles exercise 6.4 mathematics

EXERCISE 6.4


Page No 143:

Question 1:

Let  and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Answer:

Video Solution

Question 2:

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

Answer:

Since AB || CD,

∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∴ ΔAOB ∼ ΔCOD (By AAA similarity criterion)


Page No 144:

Question 3:

In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that 

Answer:

Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = 

.

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ∼ ΔDMO (By AA similarity criterion)

Video Solution

Question 4:

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

Answer:

Let us assume two similar triangles as ΔABC ∼ ΔPQR.

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Question 5:

D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

Answer:

D and E are the mid-points of ΔABC.

Video Solution

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.

Answer:

Let us assume two similar triangles as ΔABC ∼ ΔPQR. Let AD and PS be the medians of these triangles.

 ΔABC ∼ ΔPQR

…(1)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PS are medians,

∴ BD = DC = 

And, QS = SR = 

Equation (1) becomes

 … (3)

In ΔABD and ΔPQS,

∠B = ∠Q [Using equation (2)]

And,  [Using equation (3)]

∴ ΔABD ∼ ΔPQS (SAS similarity criterion)

Therefore, it can be said that

 … (4)

From equations (1) and (4), we may find that

And hence,

Video Solution

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.

Answer:

Let ABCD be a square of side a.

Therefore, its diagonal 

Two desired equilateral triangles are formed as ΔABE and ΔDBF.

Side of an equilateral triangle, ΔABE, described on one of its sides = a

Side of an equilateral triangle, ΔDBF, described on one of its diagonals 

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Video Solution

Question 8:

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

(A) 2 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4

Answer:

We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Let side of ΔABC = x

Therefore, side of 

Hence, the correct answer is (C).

Video Solution

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

Answer:

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9.

Therefore, ratio between areas of these triangles = 

Hence, the correct answer is (D).


 

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