TOPICS

Exercise - 6.4

Triangles

**Question-1 :-** Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.

Given that Δ ABC ~ Δ DEF Therefore, ar(Δ ABC)/ar(Δ DEF) = (AB/DE)^{2}= (BC/EF)^{2}= (AC/DF)^{2}Also, given that ar(Δ ABC) = 64 cm^{2}& ar(Δ DEF) = 121 cm^{2}, EF = 15.4 cm Now, ar(Δ ABC)/ar(Δ DEF) = (BC/EF)^{2}64/121 = BC^{2}/(15.4)^{2}8/11 = BC/15.4 BC = (15.4 x 8)/11 BC = 1.4 x 8 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.

Given that AB || CD, ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles) In Δ AOB and Δ COD, ∠AOB = ∠COD (Vertically opposite angles) ∠OAB = ∠OCD (Alternate interior angles) ∠OBA = ∠ODC (Alternate interior angles) Therefore, Δ AOB ∼ Δ COD (By AAA similarity criterion) Hence, ar(Δ AOB)/ar(Δ COD) = (AB/CD)^{2}Also given, AB = 2.CD ar(Δ AOB)/ar(Δ COD) = (2.CD/CD)^{2}ar(Δ AOB)/ar(Δ COD) = 4/1 ar(Δ AOB)/ar(Δ COD) = 4 : 1

**Question-3 :-** In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ar(Δ ABC)/ar(Δ DBC) = AO/DO.

Given that : ABC and DBC are two triangles on the same base BC. AD intersects BC at O. Prove that : ar(Δ ABC)/ar(Δ DBC) = AO/DO Construction : Let us draw two perpendiculars AP and DM on line BC. Proof : We know that area of triangle = 1/2 x Base x Height Therefore, ar(Δ ABC)/ar(Δ DBC) = (1/2 x BC x AP)/(1/2 x BC x DM) = AP/DM In Δ APO and Δ DMO, ∠APO = ∠DMO (Each = 90^{o}) ∠AOP = ∠DOM (Vertically opposite angles) Δ APO ∼ Δ DMO (By AA similarity criterion) Hence, AP/DM = AO/DO ar(Δ ABC)/ar(Δ DBC) = AO/DO

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

Given that : Δ ABC ∼ Δ PQR and ar(Δ ABC) = ar(Δ PQR) Prove that : Δ ABC ≅ Δ PQR Proof : Let us assume two similar triangles as Δ ABC ∼ Δ PQR. ar(Δ ABC)/ar(Δ PQR) = (AB/PQ)^{2}= (BC/QR)^{2}= (AC/PR)^{2}......(i) Given that ar(Δ ABC) = ar(Δ PQR) So, ar(Δ ABC)/ar(Δ PQR) = 1 Putting the value in equation (i), we get 1 = (AB/PQ)^{2}= (BC/QR)^{2}= (AC/PR)^{2}AB = PQ, BC = QR and AC = PR Therefore, Δ ABC ≅ Δ PQR (By using SSS congruency rule)

**Question-5 :-** D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find the ratio of the areas of Δ DEF and Δ ABC.

Here, D and E are the mid-points of ΔABC. Therefore, DE || AC and DE = AC/2 In Δ BED and Δ ABC ∠BED = ∠BCA (Corresponding angles) ∠BDE = ∠BAC (Corresponding angles) ∠EBD = ∠CBA (Common angle) Δ BED ∼ Δ ABC (By AAA similarity rule) ar(Δ BED)/ar(Δ ABC) = (DE/AC)^{2}ar(Δ BED)/ar(Δ ABC) = 1/4 ar(Δ BED) = 1/4 x ar(Δ ABC) Similarly, ar(Δ CFE) = 1/4 x ar(Δ ABC) and ar(Δ ADF) = 1/4 x ar(Δ ABC) Also, ar(Δ DEF) = ar(Δ ABC) - [ar(Δ BED) + ar(Δ CFE) + ar(Δ ADF)] ar(Δ DEF) = ar(Δ ABC) - 3/4 x ar(Δ ABC) = 1/4 x ar(Δ ABC) ar(Δ DEF)/ar(Δ ABC) = 1/4 ar(Δ DEF) : ar(Δ 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.

Let us assume two similar triangles as Δ ABC ∼ Δ PQR. Let AD and PS be the medians of these triangles. Therefore, Δ ABC ∼ Δ PQR AB/PQ = BC/QR = AC/PR ......(i) ∠BAC = ∠QPR, ∠ABC = ∠PQR, ∠ACB = ∠PRQ ......(ii) Since, AD and PS are medians, Hence, BD = DC = BC/2 and QS = SR = QR/2 By equation (i) becomes AB/PQ = BD/QS = AC/PR .......(iii) In ΔABD and ΔPQS, ∠ABD = ∠PQS (Using equation (ii)) AB/PQ = BD/QS (Using equation (iii)) Δ ABD ∼ Δ PQS (SAS similarity criterion) Therefore, we can be said that AB/PQ = BD/QS = AD/PS .........(iv) ar(Δ ABC)/ar(Δ PQR) = (AB/PQ)^{2}= (BC/QR)^{2}= (AC/PR)^{2}From equations (i) and (iv), we get AB/PQ = BC/QR = AC/PR = AD/PS and Hence, ar(Δ ABC)/ar(Δ PQR) = (AD/PS)^{2}

**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.

Let ABCD be a square of side a. Therefore, its diagonal = √2a Two desired equilateral triangles are formed as Δ ABE and Δ DBF. Side of an equilateral triangle, Δ ABE, described on one of its sides = a Side of an equilateral triangle, Δ DBF, described on one of its diagonals = √2a We know that equilateral triangles have all its angles as 60^{o}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. ar(Δ ABE)/ar(Δ DBF) = (a/√2a)^{2}= 1/2

**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

We know that equilateral triangles have all its angles as 60^{o}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 Δ ABC = x Therefore, side of Δ BDE = x/2 ar(Δ ABC)/ar(Δ BDE) = [x/(x/2)]^{2}= 4/1 ar(Δ ABC) : ar(Δ BDE) = 4 : 1 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

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 = (4/9)^{2}= 16/81 Hence, the correct answer is (D).

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