Exercise 6.4
1.Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Sol. ∆ ABC ~ ∆ DEF |Given
$$\therefore \frac{ar(∆ ABC)}{ar(∆ DEF)} = (\frac{BC}{EF})^{2}$$
|$$\because$$ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
Putting given values,
$$\frac{64}{121} = (\frac{BC}{15.4})^{2} \Rightarrow (\frac{8}{11})^{2} = (\frac{BC}{15.4})^{2}$$
$$\Rightarrow \frac{8}{11} = \frac{BC}{15.4}$$ |Taking square root on both sides
Cross – multiplying
11 X BC = 8 X 15.4
$$\Rightarrow BC = \frac{8 X 15.4}{11}$$ = 8 X 1.4
$$\Rightarrow$$ BC = 11.2 cm.
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.
Sol.In ∆ AOB and ∆ COD,
$$\angle AOB = \angle COD$$ |Vert.opp.$$\angle S$$
$$\angle OAB = \angle OCD$$ |Alt. Int. $$\angle S$$
[$$\because$$ AB || DC (given)]
$$\therefore$$ ∆ AOB ~ ∆ COD
|AA similarity criterion
$$\therefore \frac{ar(∆ AOB}{ar(∆ COD)} = (\frac{AB}{CD})^{2}$$
|$$\because$$ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
= $$(\frac{2CD}{CD})^{2}$$ |$$\because$$ AB = 2CD (given)
= $$(\frac{2}{1})^{2} = \frac{4}{1}$$ = 4 : 1.
3.In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that $$\frac{ar(ABC)}{ar(DBC)} = \frac{AO}{DO}$$.
Sol.Given : In figure, ABC and DBC are two triangles on the same base BC and AD intersects BC at O.
To Prove : $$\frac{ar(∆ ABC)}{ar(∆ DBC)} = \frac{AO}{DO}$$
Construction : Draw $$AP \perp BC$$ and $$DQ \perp BC$$.
Proof : In ∆ AOP and ∆ DOQ
$$\angle APO = \angle DQO$$
|Each equal to $$90^{0}$$ (By construction)
$$\angle AOP = \angle DOQ$$ |vert.opp.$$\angle S$$
$$\therefore$$ ∆ AOP ~ ∆ DOQ
|AA similarity criterion
$$\therefore \frac{AP}{DQ} = \frac{AO}{DO}$$ … (1)
|$$\because$$ Corresponding sides of two similar triangles are proportional
Now, $$\frac{ar(∆ ABC}{ar(∆ DBC} = \frac{\frac{1}{2}(BC)(AP)}{\frac{1}{2}(BC)(DQ)}$$
= $$\frac{AP}{DQ}$$
= $$\frac{AO}{DO}$$ |From .... (1)
4.If the areas of two similar triangles are equal, prove that they are congruent.
Sol.Given : ∆ ABC and ∆ DEF are two similar triangle such that
ar(∆ ABC) = ar(∆ DEF)
To Prove : $$∆ ABC \cong ∆ DEF$$
Proof : ∆ ABC ~ ∆ DEF |Given
$$\therefore \frac{ar(∆ ABC)}{ar(∆ DEF)} = (\frac{BC}{EF})^{2}$$
|$$\therefore$$ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
But ar (∆ ABC) = ar (∆ DEF) (given)
$$\Rightarrow 1 = (\frac{BC}{EF})^{2}$$
Taking square roots,
1 = $$\frac{BC}{EF}$$
$$\Rightarrow$$ BC = EF … (1)
Also, $$\angle B = \angle E$$ …. (2)
and $$\angle C = \angle F$$ … (3)
|$$\because$$ ∆ ABC ~ ∆ DEF (given)
In view of (1), (2) and (3),
$$∆ ABC \cong ∆ DEF$$
|ASA congruence criterion
5.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.
Sol.Given :D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC.
To Determine. Ratio of the areas of ∆ DEF and ∆ ABC.
Determination.
Because D and E are mid – points of sides AB and BC respectively of ∆ ABC (given),
$$\therefore$$ DE || AC i.e., DE || AF
Similarly EF || AD
$$\therefore$$ Quadrilateral ADEF is a parallelogram
$$\angle DEF = \angle A$$
[$$\because$$ Opposite angles of a parallelogram are equal]
Arguing as above, BEFD is also a parallelogram.
$$\therefore \angle DFE = \angle B$$ … (2)
[$$\because$$ Opposite angles of a parallelogram are equal]
In view of (1) and (2),
∆ DEF ~ ∆ ABC
|AA similarity criterion
$$\therefore \frac{ar(∆ DEF)}{ar(∆ ABC)} = (\frac{DF}{BC})^{2}$$
|$$\because$$ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
= $$\left ( \frac{\frac{1}{2}BC}{BC} \right )^{2}$$
|$$\because$$ D and F are the mid-points of AB and AC respectively $$\therefore$$ DF || BC and DF = $$\frac{1}{2} BC$$
= $$(\frac{1}{2})^{2} = \frac{1}{4}$$
$$\therefore$$ ar (∆ DEF): ar (∆ ABC) = 1 : 4.
Remark. $$\therefore$$ ar (∆ ABC) : 4 ar (∆ DEF)
6.Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Sol.Given : Two triangles ABC and DEF such that
∆ ABC ~ ∆ DEF
Let AM and DN be their corresponding medians.
To Prove : $$\frac{ar(∆ ABC)}{ar(∆ DEF)} = (\frac{AM}{DN})^{2}$$
Proof : ∆ ABC ~ ∆ DEF |Given
$$\therefore \frac{ar(∆ ABC)}{ar(∆ DEF)} = (\frac{AB}{DE})^{2}$$ ….. (1)
|$$\because$$ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides and $$\frac{AB}{DE} = \frac{BC}{EF}$$
|$$\because$$ The corresponding sides of two similar triangles are proportional
$$\Rightarrow \frac{AB}{DE} = \frac{2BM}{2EN}$$
|$$\because$$ AM and DN are the medians $$\Rightarrow$$ M is mid-point of BC and N is mid-point of EF.
= $$\frac{BM}{EN}$$ …. (2)
Also, $$\angle ABM = \angle DEN$$ …. (3)
|$$\because$$ ∆ ABC ~ ∆ DEF
$$\therefore \angle ABC = \angle DEF$$
(Corresponding angles of two similar triangles are equal)
In view of (2) and (3),
∆ ABM ~ ∆ DEN
|SAS similarity criterion
$$\therefore \frac{AB}{DE} = \frac{BM}{EN} = \frac{AM}{DN}$$ …. (4)
|$$\because$$ Corresponding sides of two similar triangles are proportional
Putting $$\frac{AB}{DE} = \frac{AM}{DN}$$ from (4) in (1), we have
$$\frac{ar(∆ ABC)}{ar(∆ DEF)} = (\frac{AM}{DN})^{2}$$.
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.
Sol.Given : ABCD is a square whose one diagonal is AC.
∆ BQC and ∆ APC are two equilateral triangles described on the side BC and diagonal AC of the square ABCD.
To Prove : ar (∆ BQC) = $$\frac{1}{2}$$ ar(∆ APC)
Proof : $$\because$$ ∆ APC and ∆ BQC are both equilateral triangles
$$\therefore$$ ∆ APC ~ ∆ BQC
|AAA similarity criterion
$$\therefore \frac{ar(∆ APC)}{ar(∆ BQC)} = (\frac{AC}{BC})^{2}$$
|$$\therefore$$ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
= $$(\frac{\sqrt{2}BC}{BC})^{2}$$
|$$\because$$ Diagonal of square = $$\sqrt{2}$$ side
= $$(\sqrt{2})^{2}$$ = 2
$$\Rightarrow \frac{ar(∆ APC)}{ar(∆ BQC)}$$ = 2
Cross – multiplying
2 ar (∆ BQC) = ar (∆ APC)
$$\Rightarrow ar (∆ BQC) = \frac{1}{2} ar(∆ APC)$$.
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 areas of triangles ABC and BDE is
(A)2 : 1 (B)1 : 2 (C)4 : 1 (D)1 : 4.
Sol.$$\because$$ ∆ ABC and ∆ BDE are both equilateral triangles
$$\therefore$$ ∆ ABC ~ ∆ BDE
|AAA similarity criterion
$$\therefore \frac{ar(∆ ABC)}{ar(∆ BDE)} = (\frac{AB}{BD})^{2}$$
|$$\because$$ The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
= $$(\frac{BC}{BD})^{2}$$ |$$\because$$ AB = BC = CA for equilateral ∆ ABC
= $$(\frac{2BD}{BD})^{2}$$
|$$\because$$ D is the mid – point of BC
= $$\frac{4}{1}$$
$$\Rightarrow$$ ar (∆ ABC) : ar (∆ BDE) = 4 : 1
Hence, (C) 4 : 1 is the correct answer.
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
Sol.Ratio of the areas of these triangles = $$4^{2} : 9^{2}$$
|$$\because$$ The ratio of the similar triangles is equal to the square of the ratio of their corresponding sides
Hence, (D) 16 : 81 is the correct answer.
0 Doubts's