NCERT Solutions Class 10th Maths Chapter – 6 Triangles Exercise – 6.4

June 4, 2022

NCERT Solutions Class 10th Maths Chapter – 6 Triangles

Textbook	NCERT
Class	10^th
Subject	Mathematics
Chapter	6^th
Chapter Name	Triangles
Grade	Class 10th Mathematics
Medium	English
Source	last doubt

NCERT Solutions Class 10th Maths Chapter – 6 Triangles Exercise – 6.4 were prepared by Experienced Lastdoubt.com Teachers. Detailed answers of all the questions in Chapter 6 Maths Class 10 Triangles Exercise 6.4 provided in NCERT TextBook.

NCERT Solutions Class 10th Maths Chapter – 6 Triangles

Chapter – 6

Triangles

Exercise – 6.4

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 1

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

Solution: Given, ΔABC ~ ΔDEF,
Area of ΔABC = 64 cm²
Area of ΔDEF = 121 cm²
EF = 15.4 cm

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,
= AC²/DF² = 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

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 2

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

In ΔAOB and ΔCOD, we have
∠1 = ∠2 (Alternate angles)
∠3 = ∠4 (Alternate angles)
∠5 = ∠6 (Vertically opposite angle)
∴ ΔAOB ~ ΔCOD [AAA similarity criterion]
As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,
Area of (ΔAOB)/Area of (ΔCOD) = AB²/CD²
= 4CD²/CD² = 4/1
∴ Area of (ΔAOB)/Area of (ΔCOD)
= 4CD2/CD2 = 4/1
Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 3

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: Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.
We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO
Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = 1/2 × Base × Height

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.

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 4

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

Solution: Say ΔABC and ΔPQR are two similar triangles and equal in area

Now let us prove ΔABC ≅ ΔPQR.
Since, ΔABC ~ ΔPQR
∴ Area of (ΔABC)/Area of (ΔPQR) = BC²/QR²
⇒ BC²/QR² =1 [Since, Area(ΔABC) = (ΔPQR)
⇒ BC²/QR²
⇒ BC = QR
Similarly, we can prove that
AB = PQ and AC = PR
Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 5

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:

D, E, and F are the mid-points of

ΔABC

∴

DE || AC and

DE = (1/2) AC (Midpoint theorem) \dots. (1)

ΔBED

and

ΔBCA

∠BED = ∠BCA

(Corresponding angles)

∠BDE = ∠BAC

(Corresponding angles)

∠EBD = ∠CBA (Common angles)

∴ ΔBED \sim ΔBCA

(AAA similarity criterion)

\frac{ar (ΔBED) / ar (}{ΔBCA)} =({\frac{D E/}{A C)}}^{2}

\Rightarrowar (\frac{ΔBED) / ar (}{ΔBCA)} = (\frac{1/}{4)} [From (1)]

\Rightarrowar (ΔBED) = (\frac{1/}{4) ar (} ΔBCA)

Similarly,

ar (ΔCFE) = (\frac{1/}{4) ar (} C B A)

and ar (

ΔADF) = (\frac{1/}{4) ar (} ΔADF) = (\frac{1/}{4) ar (} ΔABC)

Also,

ar (ΔDEF) = ar (ΔABC) - [ar (ΔBED) + ar (ΔCFE) + ar (ΔADF)]

\Rightarrowar (ΔDEF) = ar (ΔABC) - (\frac{3/}{4) ar} (ΔABC) = (\frac{1/}{4)} a r (ΔABC)

\Rightarrowar (\frac{ΔDEF) /}{a r (ΔABC)} = (\frac{1/}{4)}

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 6

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 and ΔABC ~ ΔDEF.
Chapter - 6 Exercise 6.4

We have 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)

In ΔABM and ΔDEN,
Since ΔABC ~ ΔDEF
∴ ∠B = ∠E
AB/DE = BM/EN [Already Proved in equation (i)]
∴ ΔABC ~ ΔDEF [SAS similarity criterion]
⇒ 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
Hence, proved.

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 7

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:
Chapter - 6 Exercise 6.4

Given, ABCD is a square whose one diagonal is AC. ΔAPC and ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.
Area(ΔBQC) = ½ Area(ΔAPC)
Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,
∴ ΔAPC ~ ΔBQC [AAA similarity criterion]
∴ area(ΔAPC)/area(ΔBQC) = (AC²/BC²) = AC²/BC²
Since, Diagonal = √2 side = √2 BC = AC

⇒ area(ΔAPC) = 2 × area(ΔBQC)
⇒ area(ΔBQC) = 1/2area(ΔAPC)
Hence, proved.
Tick the correct answer and justify:

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 8

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: Given, ΔABC and ΔBDE are two equilateral triangle. D is the midpoint of BC.

Chapter - 6 Exercise 6.4
∴ BD = DC = 1/2BC
Let each side of triangle is 2a.
As, ΔABC ~ ΔBDE
∴ Area(ΔABC)/Area(ΔBDE) = AB²/BD² = (2a)²/(a)² = 4a²/a² = 4/1 = 4:1
Hence, the correct answer is (C).

Ncert Solution Class 10th (Chapter – 6) Exercise – 6.4 Question No. 9

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: Given, Sides of two similar triangles are in the ratio 4 : 9.

Chapter - 6 Exercise 6.4
Let ABC and DEF are two similar triangles, such that,
ΔABC ~ ΔDEF
And AB/DE = AC/DF = BC/EF = 4/9
As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,
∴ Area(ΔABC)/Area(ΔDEF) = AB²/DE²
∴ Area(ΔABC)/Area(ΔDEF) = (4/9)²= 16/81 = 16:81
Hence, the correct answer is (D).

NCERT Solutions Class 10th Maths All Chapter

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