NCERT solutions for class 10 Maths
Chapter 6
Triangles
Exercise 6.4
Ex 6.4 Question 1.
Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Solution:
Given that, ΔABC ~ ΔDEF,
ar.(ΔABC)/ar.(ΔDEF) = AB²/DE² = BC²/EF² = AC²/DF²
Area of ΔABC = 64 cm2
Area of ΔDEF = 121 cm2
EF = 15.4 cm
∴ ar.(ΔABC)/ar.(ΔDEF) = BC²/EF²
∴ 64/121 = BC2/EF2
⇒ (8/11)2 = (BC/15.4)2
⇒ 8/11 = BC/15.4
⇒ BC = 8×15.4/11
⇒ BC = 8 × 1.4
⇒ BC = 11.2 cm
Ex 6.4 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.
Solution:
Given, AB || DC.
∠OAB = ∠OCD & ∠OBA = ∠OCD [Alternate angles]
In ΔAOB and ΔCOD,
∠AOB = ∠COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate angles]
∠OBA = ∠ODC [Alternate angles]
∴ ΔAOB ~ ΔCOD [AAA similarity criterion]
∴ Area of (ΔAOB)/Area of (ΔCOD) = AB2/CD2
= (2CD)2/CD2 [∴ AB = 2CD]
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD2/CD2 = 4/1 = 4:1
Ex 6.4 Question 3.
In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.
Solution:
Let AP and DM are the perpendiculars on line BC.
Area of a triangle = 1/2 × Base × Height
∴ ar(ΔABC)/ar(ΔDBC) = ½BC x AP/ ½BC x DM = AP/DM ……(i)
In ΔAPO and ΔDMO,
∠APO = ∠DMO [Each 90°]
∠AOP = ∠DOM [Vertically opposite angles]
∴ ΔAPO ~ ΔDMO [AA similarity criterion]
∴ AP/DM = AO/DO
⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO. [From equation (i)]
Ex 6.4 Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:
Let, ΔABC ~ ΔPQR
∴ ar(ΔABC)/ar(ΔPQR) = BC2/QR2
⇒ BC2/QR2 =1 [Since, Area(ΔABC) = (ΔPQR)]
⇒ BC2/QR2
⇒ BC = QR
Similarly,
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [SSS congruencecy]
Ex 6.4 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.
Solution:
In ΔABC,
F is the mid-point of AB [Given]
E is the mid-point of AC [Given]
By mid-point theorem,
FE || BC and FE = 1/2BC
⇒ FE || BC and FE || BD [BD = 1/2BC]
[Opposite sides of the parallelogram are equal and parallel]
∴ BDEF is parallelogram.
Similarly, in ΔFBD and ΔDEF,
FB = DE [Opposite sides of parallelogram BDEF]
FD = FD [Common]
BD = FE [Opposite sides of parallelogram BDEF]
∴ ΔFBD ≅ ΔDEF
Similarly,
ΔAFE ≅ ΔDEF
ΔEDC ≅ ΔDEF
[when triangles are congruent, they are equal in the area.]
ar(ΔFBD) = ar(ΔDEF) ……………(i)
ar(ΔAFE) = ar(ΔDEF) ………….(ii)
and,
ar(ΔEDC) = ar(ΔDEF) ………….(iii)
Now,
ar(ΔABC) = ar(ΔFBD) + ar(ΔDEF) + ar(ΔAFE) + ar(ΔEDC) ………(iv)
ar(ΔABC) = ar(ΔDEF) + ar(ΔDEF) + ar(ΔDEF) + ar(ΔDEF)
From equation (i), (ii) and (iii),
⇒ ar(ΔDEF) = (1/4)ar(ΔABC)
⇒ ar(ΔDEF)/ar(ΔABC) = 1/4
Hence, ar(ΔDEF): ar(ΔABC) = 1:4
Ex 6.4 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: ΔABC ~ ΔDEF.
To prove: Area(ΔABC)/Area(ΔDEF) = AM2/DN2
Since, ΔABC ~ ΔDEF (Given)
∴ Area(ΔABC)/Area(ΔDEF) = (AB2/DE2) …………(i)
and, AB/DE = BC/EF = CA/FD ………………(ii)
⇒AB/DE = ½BC/½EF = CD/FD
In ΔABM and ΔDEN,
Since ΔABC ~ ΔDEF
∴ ∠B = ∠E
AB/DE = BM/EN [Proved above]
∴ ΔABC ~ ΔDEF [SAS similarity]
⇒ AB/DE = AM/DN …………..(iii)
∴ ΔABM ~ ΔDEN
[As the areas of two similar triangles are proportional to the squares of the corresponding sides.]
∴ area(ΔABC)/area(ΔDEF) = AB2/DE2 = AM2/DN2
Ex 6.4 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:
Area(ΔBQC) = ½ Area(ΔAPC)
ΔAPC and ΔBQC are both equilateral triangles,
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC2/BC2) = AC2/BC2
Diagonal = √2 side = √2 BC = AC
(√2 BC/BC)2 =2
⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)
Ex 6.4 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
Solution:
Both the triangles are equilateral and each angle of both the triangles are 60°.
By AAA similarity, ΔABC ~ ΔBDE
The side of ΔBDE = x/2
∴ ar(ΔABC)/ar(ΔBDE) = AB2/BD2 = (x/x/2)2= 4/1 = 4:1
Hence, (C) is the correct answer.
Ex 6.4 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:-
The ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.
The ratio of the area of two triangles = (4/9)2 = 16/81 = 16:81
Hence, (D) is the correct answer.
