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Exercise 6.4 - Chapter 6 Triangles class 10 ncert solutions Maths - SaraNextGen [2024]


Question 1:

Let https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2780/Chapter%206_html_m23857e30.gif and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2780/Chapter%206_html_m589c3eb0.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2780/Chapter%206_html_m5166952f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2780/Chapter%206_html_m360d02f1.gif

 

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2783/Chapter%206_html_m4e1f03ea.jpg

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)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2783/Chapter%206_html_m572b99b3.gif

 

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 https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2785/Chapter%206_html_m6ddc4d1c.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2785/Chapter%206_html_m4da082d6.jpg

Answer:

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2785/Chapter%206_html_m47ae31c1.jpg

We know that area of a triangle = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2785/Chapter%206_html_3ae18f7.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2785/Chapter%206_html_m6c62023c.gif.

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ∼ ΔDMO (By AA similarity criterion)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2785/Chapter%206_html_55b8aa0c.gif

 

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.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2787/Chapter%206_html_m640a527d.gif

 

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2789/Chapter%206_html_m1d51deca.jpg

D and E are the mid-points of ΔABC.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2789/Chapter%206_html_m426f8d14.gif

 

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_5683849c.jpg

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_6290838.gif ΔABC ∼ ΔPQR

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_m76f24b48.gif…(1)

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PS are medians,

∴ BD = DC = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_644d0f24.gif

And, QS = SR = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_3ecffbc8.gif

Equation (1) becomes

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_m1bb554bb.gif … (3)

In ΔABD and ΔPQS,

∠B = ∠Q [Using equation (2)]

And, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_m555fb1e5.gif [Using equation (3)]

∴ ΔABD ∼ ΔPQS (SAS similarity criterion)

Therefore, it can be said that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_b58b8be.gif … (4)

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_6077fde.gif

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

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_127263af.gif

And hence,

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2792/Chapter%206_html_313701d7.gif

 

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2793/Chapter%206_html_m165bf345.jpg

Let ABCD be a square of side a.

Therefore, its diagonal https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2793/Chapter%206_html_620cd942.gif

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 https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2793/Chapter%206_html_620cd942.gif

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.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2793/Chapter%206_html_m3571e658.gif

 

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:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2794/Chapter%206_html_41f6f08c.jpg

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 https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2794/Chapter%206_html_m3a1e27d.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2794/Chapter%206_html_m5d6d01e6.gif

Hence, the correct Answer is (C).

 

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 = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/133/2796/Chapter%206_html_42a795b.gif

Hence, the correct Answer is (D).

Also Read : Exercise-6.5-Chapter-6-Triangles-class-10-ncert-solutions-Maths

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