167-NCERT-10-6-Triangles - Ex- 6.4

NCERT

10th Mathematics

Exercise 6.4

Topic: 6 Triangles

Click here for ⇨ NCERT-10-6-Triangles - Ex- 6.3

EXERCISE 6.4

Q1. Let ∆ ABC ~ ∆ DEF and their areas be 64 cm2 and 121 cm2 respectively. 

If EF = 15.4 cm, find BC.

Solution:

1) As ∆ ABC ~ ∆ DEF, we know that ratios of the areas of similar triangles are equal

to the ratios of the square of corresponding sides of the tringles, so, we have,

A(∆ ABC)/A(∆ DEF) = [(BC)²/(EF)²]

64/121 = [(BC)/(EF)]²

[(BC)/(EF)]² = 64/121

[(BC)/(15.4)] = 8/11

(BC) = (15.4 x 8)/11

(BC) = (1.4 x 8)

(BC) = 11.2 cm.

2) So, BC = 11.2 cm.

Q2. 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:

1) In trapezium ABCD,  AB || DC, and in ∆ OAB and ∆ OCD, we have,

a) < OAB = < OCD --------- (alternate angles)
b) < OBA = < ODC --------- (alternate angles)
c) < AOB = < COD --------- (vertically opposite angles)

2) So by AAA similarity test, we have,

∆ OAB ~ ∆ OCD

3) We know that ratios of the areas of similar triangles are equal to the ratios of

the square of corresponding sides of the triangles, so, we have,

A(∆ OAB)/A(∆ OCD) = [(AB)²/(CD)²]

A(∆ OAB)/A(∆ OCD) = [(AB)/(CD)]²

A(∆ OAB)/A(∆ OCD) = [2(CD)/(CD)]²

A(∆ OAB)/A(∆ OCD) = [2/1]²

A(∆ OAB)/A(∆ OCD) = 4/1

4) So, the ratio of the areas of triangles AOB and COD is 4:1.

Q3. In the following fig., ABC and DBC are two triangles on the same base BC.

If AD intersects BC at O, show that

[area (∆ ABC)]/[area (∆ DBC)] = AO/DO.

 

Solution:

1) We know that the area of a triangle = 1/2 x base x height.

a) A(∆ ABC)/A(∆ DCB) = [1/2 x (BC) x (AM)]/[1/2 x (BC) x (DN)]
b) A(∆ ABC)/A(∆ DCB) = (AM)/(DN) -------- equation 1

2) In ∆ AMO and ∆ DNO,

a) < AMO = < DNO ------ (AM ⊥ BC and AM ⊥ BC)

b) < AOM = < DON ------- (vertically opposite angles)

3) So by AA similarity test, we have,

∆ AMO ~ ∆ DNO

4) Here we have, 

(AM)/(DN) = (AO)/(DO) -------- equation 2

5) From equations 1 and 2 we have, 

A(∆ ABC)/A(∆ DCB) = (AO)/(DO), hence proved.

Q4. If the areas of two similar triangles are equal, prove they are congruent. 

Solution:

1) As ∆ ABC ~ ∆ PQR, we know that ratios of the areas of similar triangles are equal

to the ratios of the square of corresponding sides of the tringles, so, we have,

A(∆ ABC)/A(∆ PQR) = [(AB)²/(PQ)²]

1 = [(AB)²/(PQ)²]

[(AB)²/(PQ)²] = 1

(AB)² = (PQ)²

(AB) = (PQ)

2) Similarly we can prove that, 

(AC) = (PR)

(BC) = (QR)

3) So by the SSS test of congruence, we have,

∆ ABC ≅ ∆ PQR, hence proved.

Q5. 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:

1) As D, E, and F are the mid-points of DE || AC, and by mid-point

theorem, we have,

DF = (1/2) CB ----------- equation 1

DF = BE       (As E is mid-point of BC) ----------- equation 2

2) So, ▱ DFEB is a parallelogram,

a) < DFE = < DBE --- (opposite angles of a parallelogram) ---- equation 3

3) So, ▱ FDEC is a parallelogram,

a) < FDE = < FCE --- (opposite angles of a parallelogram) ---- equation 4

4) In ∆ DEF and ∆ ABC, and from equations 3 and 4, we have,

a) < DFE = < CBA --- from equation 3

b) < FDE = < BCA --- from equation 4

5) By AA similarity test, we have,

a) ∆ DEF ~ ∆ CAB

6) We know that ratios of the areas of similar triangles are equal to the ratios of

the square of corresponding sides of the triangles, so, we have,

A(∆ DEF)/A(∆ CAB) = [(DF)²/(CB)²]

A(∆ DEF)/A(∆ CAB) = [(DF)/(CB)]²
A(∆ DEF)/A(∆ CAB) = [(1/2)(CB)/(CB)]² ---- from equation 1
A(∆ DEF)/A(∆ CAB) = [(1/2)]²
A(∆ DEF)/A(∆ CAB) = 1/4

7) So, the ratio of the areas of ∆ DEF and ∆ ABC = 1/4.

Q6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. 

Solution:

1) It is given that ∆ ABC ~ ∆ PQR,

a) < ABC = < PQR --------- equation 1.

b) < BCA = < QRP --------- equation 2.
a) < CAB = < RPQ --------- equation 3.

2) Similarity, we have,

a) (AB)/(PQ) = (BC)/(QR) = (CA)/(RP) --------- equation 4.

3) We know that ratios of the areas of similar triangles are equal to the ratios of

the square of corresponding sides of the triangles, so, we have,

A(∆ ABC)/A(∆ PQR) = (AB)²/(PQ)² = (BC)²/(QR)² = (AC)²/(PR)² --- equation 5.

4) In ∆ ABD and ∆ PQM, we have,

a) < ABD = < PQM ---- from equation 1.

b) (AB)/(PQ) = (BD)/(QM) ----- (D and M are mid-points of BC and QR).

5) So, by the SAS similarity test,

∆ ABD ~ ∆ PQM --------- equation 6.

6) We know that ratios of the areas of similar triangles are equal to the ratios of

the square of corresponding sides of the triangles, so, we have,

A(∆ ABD)/A(∆ PQM) = (AB)²/(PQ)² = (BD)²/(QM)² = (AD)²/(PM)² --- equation 7.

7) So from equation 7, we have,

A(∆ ABD)/A(∆ PQM) = (AD)²/(PM)²

8) Hence, it is proved that,

the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Q7. 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:

1) ∆ PAB and ∆ QAC are equilateral triangles, so we have,

∆ PAB ~ ∆ QAC

2) We know that ratios of the areas of similar triangles are equal to the ratios of

the square of corresponding sides of the triangles, so, we have,

A(∆ PAB)/A(∆ QAC) = (AB)²/(AC)² --- equation 1

3) In ▱ ABCD is a square, so we have,

(AC) = √2 (AB) --- equation 2

4) From equations 1 and 2,

 A(∆ PAB)/A(∆ QAC) = (AB)²/(AC)²

A(∆ PAB)/A(∆ QAC) = (AB)²/(√2 (AB))²

A(∆ PAB)/A(∆ QAC) = 1/(√2)²

A(∆ PAB)/A(∆ QAC) = 1/2
A(∆ PAB) = (1/2) x [A(∆ QAC)] hence proved.

Q8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is 

(A) 2: 1 (B) 1: 2 (C) 4: 1 (D) 1: 4 

Solution:

1) As D is the mid-point of BC,

(BC) = 2 (BD) -------- equation 1

2) ∆ ABC and ∆ EBD are equilateral triangles, so we have,

∆ ABC ~ ∆ EBD.

3) We know that ratios of the areas of similar triangles are equal to the ratios of

the square of corresponding sides of the triangles, so, we have,

A(∆ ABC)/A(∆ EBD) = (BC)²/(BD)² --- equation 2.

4) From equations 1 and 2 we have,

A(∆ ABC)/A(∆ EBD) = (BC)²/(BD)² 
A(∆ ABC)/A(∆ EBD) = [2(BD)]²/(BD)²
A(∆ ABC)/A(∆ EBD) = [2]²
A(∆ ABC)/A(∆ EBD) = 4/1

5) The ratio of the areas of triangles ABC and EBD is 4:1

6) Answer: (C) 4:1.

Q9. 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:

1) ∆ ABC and ∆ PQR are similar triangles, so we have,

∆ ABC ~ ∆ PQR.

2) It is given that (AB)/(PQ) = (BC)/(QR) = (AC)/(PR) = 4/9 --- equation 1. 

3) We know that ratios of the areas of similar triangles are equal to the ratios of

the square of corresponding sides of the triangles, so, we have,

A(∆ ABC)/A(∆ PQR) = (BC)²/(QR)² --- equation 2.

4) From equations 1 and 2 we have,

A(∆ ABC)/A(∆ PQR) = (BC)²/(QR)² 
A(∆ ABC)/A(∆ PQR) = [4/9]²
A(∆ ABC)/A(∆ PQR) = 16/81

5) The ratio of the areas of triangles ABC and PQR is 16:81

6) Answer: (D) 16:81.

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Click here for ⇨ NCERT-10-6-Triangles - Ex- 6.5

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