RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.4

Question 1.
Let △ABC ~ △DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
△ABC ~ △DEF, area of △ABC = 64 cm² and area of △DEF = 121 cm² and EF = 15.4 cm.

△ABC ~ △DEF
∴ \(\frac{a r(\Delta \mathrm{ABC})}{a r(\Delta \mathrm{DEF})}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}\) = \(\frac{\mathrm{AC}^{2}}{\mathrm{DE}^{2}}\) = \(\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)
[∵ We know that if two triangles are similar then the ratio of their area is equal to square of the ratio of their corresponding sides.]

∴ BC = \(\frac{8 \times 15.4}{11}\) = 8 × 1.4
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:
ABCD is a trapezium in which AB || DC. Its diagonals AC and BD intersect each other at point O. AB = 2 CD.

Now, in △AOB and △COD,
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angles)
∴ △AOB ~ △COD
(By AAA similarity criterion)

∴ Ratio of required ar△AOB and ar△COD
= 4:1.

Question 3.
In Fig., ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \(\frac{a r(\mathrm{ABC})}{a r(\mathrm{DBC})}\) = \(\frac{\mathbf{A O}}{\text { DO }}\).

Solution:
Given : △ABC and △DBC aretwo triangles on the same base BC. AD intersects BC at O.

To Prove : \(\frac{a r(\mathrm{ABC})}{a r(\mathrm{DBC})}\) = \(\frac{\mathrm{AO}}{\mathrm{DO}}\)
Construction : Draw AL ⊥ BC, DM ⊥ BC
Proof: In △ALO and △DMO,
∠1 = ∠2 (Vertically opposite angles)
∠L = ∠M (Each 90°)
∴ △ALO ~ △DMO (AA similarity criterion)
∴ \(\frac{\mathrm{AL}}{\mathrm{DM}}\) = \(\frac{\mathrm{AO}}{\mathrm{DO}}\)
(∵ If two triangles are similar, then their corresponding sides are proportional)

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Given : Two triangles ABC and DEF are similar and are equal in area.
To prove : △ABC ~ △DEF

Proof: Since △ABC ~ △DEF,
∴ \(\frac{a r(\triangle \mathrm{ABC})}{a r(\triangle \mathrm{DEF})}\) = \(\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)
⇒ \(\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\) = 1
[∵ ar (△ABC) = ar(△DEF)]
⇒ BC² = EF²
⇒ BC = EF
Also, since △ABC ~ △DEF, therefore they are equiangular.
and ∠B = ∠E
and ∠C = ∠F
Now in △ABC and △DEF,
∠B = ∠E, ∠C = ∠F
and BC = EF
△ABC ≅ △DEF
(ASA congmence theorem) (Hence Proved)

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:
Given : D, E and F are respectively the mid-points of sides AB, BC and CA of △ABC.
Required : To find ar(△DEF): ar(△ABC)
Proof : It is given that in △ABC, D and E are respectively the mid-points of the sides AB and BC.
∴ DE || AC
⇒ DE || FC ……(i)

Similarly in △ABC, D and F are the mid-points of sides AB and AC respectively.
∴ DF || BC
⇒ DF || EC ……..(ii)
From (i) and (ii) it is clear that DECF is a parallelogram.
Similarly △DEF is a parallelogram.
Now in △DEF and △ABC
∠DEF = ∠A
(From parallelogram ADEF)
and ∠EDF = ∠C
(From parallelogram DECF)
∴ By AA similarity
△DEF ~ △CAB

∴ ar(△DEF) : ar(△ABC) = 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 : ABC and DEF are two similar triangles.
AP and DM are respectively the medians of sides BC and EF.

To prove: \(\frac{a r(\triangle \mathrm{ABC})}{a r(\Delta \mathrm{DEF})}\) = \(\frac{A P^{2}}{D M^{2}}\)
Proof: △ABC ~ △DEF (Given)
∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{BC}}{\mathrm{EF}}\) = \(\frac{\mathrm{2BP}}{\mathrm{2EM}}\)
(∵ AP and DM are medians)
∴ BC = 2BP and EF = 2EM AB = BP
⇒ \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{BP}}{\mathrm{EM}}\) ….(1)
In △ABP and △DEM,
∠B = ∠E (∵ △ABC ~ △DEF)
\(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{BP}}{\mathrm{EM}}\) [Proved in (i)]
∴ △ABP ~ △DEM (By SAS similarity criterion)
∴ \(\frac{\mathrm{AB}}{\mathrm{DE}}\) = \(\frac{\mathrm{AP}}{\mathrm{DM}}\) ….(ii)
Since the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
∴ \(\frac{a r(\triangle \mathrm{ABC})}{a r(\triangle \mathrm{DEF})}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}\) = \(\frac{\mathrm{AP}^{2}}{\mathrm{DM}^{2}}\)
(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:
Given : ABCD is a square. Equilateral triangle has been described on side AB of the square and equilateral triangle ACF has been described on diagonals AC.

To prove : \(\frac{a r(\triangle \mathrm{ABE})}{a r(\triangle \mathrm{ACF})}\) = \(\frac{1}{2}\)
Proof: In right △ABC,
AB² + BC² = AC²
(By Pythagoras Theorem)
or AB² + AB² = AC²
(∵ AB = BC, sides of the same square)
2AB² = AC² ………(i)
Now, each of △ABE and △ACF is equilateral, therefore is equiangular and therefore similar,
or △ABE ~ △ACF
Here any side of first triangle is parallel to any side of the other triangle.
∴ \(\frac{a r(\triangle \mathrm{ABE})}{a r(\triangle \mathrm{ACF})}\) = \(\frac{\mathrm{AB}^{2}}{\mathrm{AC}^{2}}\)
[∵ The ratio of the areas of non-similar triangles is equal to the square of the ratio of their corresponding sides]
⇒ \(\frac{\mathrm{AB}^{2}}{2 \mathrm{AB}^{2}}\) = \(\frac{1}{2}\) [Using (i)]
Hence area of △ABE = \(\frac{1}{2}\)area of △ACF
(Hence Proved)
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:
△ABC and △BDE are two equilateral triangles such that D is the mid-point of the side BC

∴ BD = DC = \(\frac{1}{2}\) BC
or BC : BD = 2 : 1
Let the side of the equilateral triangle be 2a. △ABC ~ △BDE
Since in an equilateral triangle angles are the same.

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:
△ABC ~ △DEF (Given)

∴ Correct choice = (D)