**Exercise 6.4 Page: 143**

**1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm ^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.**

**Solution: **Given, ΔABC ~ ΔDEF,

Area of ΔABC = 64 cm^{2}

Area of ΔDEF = 121 cm^{2}

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,

= AC^{2}/DF^{2} = BC^{2}/EF^{2}

∴ 64/121 = BC^{2}/EF^{2}

⇒ (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

**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) = AB^{2}/CD^{2}

= (2CD)^{2}/CD^{2} [∴ AB = 2CD]

∴ Area of (ΔAOB)/Area of (ΔCOD)

= 4CD^{2}/CD^{2} = 4/1

Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

**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.

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

**Solution:**

SayΔABC 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) = BC^{2}/QR^{2}

⇒ BC^{2}/QR^{2} =1 [Since, Area(ΔABC) = (ΔPQR)

⇒ BC^{2}/QR^{2}

⇒ BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

**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:**

Given, D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC.

In ΔABC,

F is the mid-point of AB (Already given)

E is the mid-point of AC (Already given)

So, by the mid-point theorem, we have,

FE || BC and FE = 1/2BC

⇒ FE || BC and FE || BD [BD = 1/2BC]

Since, opposite sides of parallelogram are equal and parallel

∴ BDEF is parallelogram.

Similarly, in ΔFBD and ΔDEF, we have

FB = DE (Opposite sides of parallelogram BDEF)

FD = FD (Common sides)

BD = FE (Opposite sides of parallelogram BDEF)

∴ ΔFBD ≅ ΔDEF

Similarly, we can prove that

ΔAFE ≅ ΔDEF

ΔEDC ≅ ΔDEF

As we know, if triangles are congruent, then they are equal in area.

So,

Area(ΔFBD) = Area(ΔDEF) ……………………………**(i)**

Area(ΔAFE) = Area(ΔDEF) ……………………………….**(ii)**

and,

Area(ΔEDC) = Area(ΔDEF) ………………………….**(iii)**

Now,

Area(ΔABC) = Area(ΔFBD) + Area(ΔDEF) + Area(ΔAFE) + Area(ΔEDC) ………**(iv)**

Area(ΔABC) = Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF)

From equation **(i)**, **(ii)** and **(iii)**,

⇒ Area(ΔDEF) = (1/4)Area(ΔABC)

⇒ Area(ΔDEF)/Area(ΔABC) = 1/4

Hence, Area(ΔDEF): Area(ΔABC) = 1:4

**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.

We have to prove: Area(ΔABC)/Area(ΔDEF) = AM^{2}/DN^{2}

Since, ΔABC ~ ΔDEF (Given)

∴ Area(ΔABC)/Area(ΔDEF) = (AB^{2}/DE^{2}) ……………………………**(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) = AB^{2}/DE^{2} = AM^{2}/DN^{2}

Hence, proved.

**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 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) = (AC^{2}/BC^{2}) = AC^{2}/BC^{2}

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:**

**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.

∴ BD = DC = 1/2BC

Let each side of triangle is 2*a*.

As, ΔABC ~ ΔBDE

∴ Area(ΔABC)/Area(ΔBDE) = AB^{2}/BD^{2} = (2*a*)^{2}/(*a*)^{2} = 4*a*^{2}/*a*^{2} = 4/1 = 4:1

Hence, the correct answer is (C).

**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.

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) = AB^{2}/DE^{2 }

∴ Area(ΔABC)/Area(ΔDEF) = (4/9)^{2 }= 16/81 = 16:81

Hence, the correct answer is (D).