Triangles Test

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Triangles Test - 37

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Weekly Quiz Competition

Question 1
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In Figure , $$AB = BC$$. If the area of $$\triangle$$ $$ABE$$ is $$x$$, calculate the area of $$\triangle$$ $$ACD$$.

A
$$x\sqrt{2}$$

B
$$x\sqrt{3}$$

C
$$2x$$

D
$$3x$$

E
$$4x$$

Solution

$$AB=BC$$         ....Given
So, $$AB: AC = 1:2$$          ...(I)
$$\angle BED + \angle BEA = 180^o$$         ...Angles in linear pair
$$\therefore 118^o + \angle BEA = 180^o$$
$$\therefore \angle BEA = 62^o = \angle CDA$$    ...(II)
In $$\triangle ABE$$ and $$\triangle ACD$$
$$\angle A$$ is the common angle.
$$\therefore \angle BEA = \angle CDA$$    ... from (II)
$$\triangle ABE \sim \triangle ACD$$         ....AAA test of similarity
So, $$\dfrac {A(\triangle ABE)}{A(\triangle ACD)} = \left(\dfrac {AB}{AC}\right)^2$$           ....Theorem on ratio of areas of similar triangles
$$\Rightarrow \dfrac {A(\triangle ABE)}{A(\triangle ACD)} = \left(\dfrac 12\right)^2$$
$$\Rightarrow \dfrac {x}{A(\triangle ACD)} = \dfrac 14$$
$$A(\triangle ACD) = 4x$$ sq. units
Question 2
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In the figure below, $$\angle ABC=\angle DFE,\angle BAC=\angle FDE$$, $$D$$ and $$F$$ are on $${ AB } ,\ { AD } ={ FB } $$ and the distances are shown in the figure itself in centimeters. Calculate the length of $${ AD } $$, in centimeters.

A
$$5$$

B
$$4$$

C
$$3$$

D
$$2$$

E
$$1$$

Solution

In $$\triangle ABC$$ and $$\triangle DFE,$$
$$\angle ABC =\angle DFE$$ [Given]
$$\angle BAC =\angle FDE$$ [Given]

Hence, by $$AAA$$ rule of similarity,
$$\triangle ABC \sim \triangle DFE$$

As, the two triangles are similar their sides will be proportional as well,
$$\therefore \dfrac {AB}{DF} = \dfrac {BC}{FE} = \dfrac {AC}{DE}\quad\quad\quad\quad\dots(i)$$

Let $$AD=BF=x$$
$$AB = AD+DF+BF$$
$$AB = 6+2x$$

From equation $$(i),$$
$$\begin{aligned}{}\frac{{AB}}{{DF}} &= \frac{{AC}}{{DE}}\\\frac{{6 + 2x}}{6} &= \frac{{20}}{{12}}\\72 + 24x& = 120\\24x& = 48\\x& = 2\end{aligned}$$
Hence,
$$AD=BF=2\ cm.$$
Question 3
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In the following figure, what is the value of $$x$$ in terms of $$y$$?

A
$$10y$$

B
$$10y^{2}$$

C
$$y\sqrt {10}$$

D
$$\sqrt {10y}$$

E
$$10\sqrt {y}$$

Solution

As per Pythagoras theorem,
hypotenuse$$^{ 2 }=$$ base $$^{ 2 }+$$ perpendicular $$^{ 2 }$$
In the given figure, $$h=x, b=3y, p=y$$
As per theorem, we have
$$x^{ 2 }={ \left( 3y \right) }^{ 2 }+{ y }^{ 2 }$$
$$x=\sqrt { 9{ y }^{ 2 }+{ y }^{ 2 } } $$
$$x=\sqrt { { 10y }^{ 2 } } $$$$=y\sqrt { 10 } $$
Question 4
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In the given figure, find the value of $$AE$$ if $$AC = 5.6$$ cm and $$DE || BC$$.

A
2.4 cm

B
2.1 cm

C
2.2 cm

D
2.5 cm

Solution

In given figure $$\angle ADE=\angle ACB$$ and $$AC=5.6$$ cm

$$DE\parallel BC$$

$$\angle BAC=\angle DAE$$ ...... (Common angle)

Then $$\angle ABC=\angle AED$$

$$\therefore \Delta ADE\sim \Delta ABC$$ ...... [By AAA postulates of similarity]

$$\therefore \dfrac{AD}{AB}=\dfrac{AE}{AC}$$

$$\Rightarrow \dfrac{3}{3+5}=\dfrac{AE}{5.6}$$

$$\Rightarrow AE=\dfrac{3}{8}\times 5.6=2.1$$ cm
Question 5
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In similar triangles $$\triangle ABC$$ and $$\triangle FDE, DE = 4 cm, BC = 8 cm$$ and area of $$\triangle FDE = 25 cm^2$$. What is the area of $$\Delta ABC$$?

A
144 cm$$^2$$

B
121 cm$$^2$$

C
100 cm$$^2$$

D
81 cm$$^2$$

Solution

$$DE=4\,cm,\,BC=8\,cm$$ and $$ar(\triangle FDE)=25\,cm^2$$ [ Given ]

$$\Rightarrow$$ $$\triangle ABC\sim\triangle FDE$$ [ Given ]
$$\Rightarrow$$ $$\dfrac{ar(\triangle ABC)}{ar(\triangle FDE)}=\dfrac{(BC)^2}{(DE)^2}$$ [ By area of similar triangle theorem ]

$$\Rightarrow$$ $$\dfrac{ar(\triangle ABC)}{25}=\dfrac{(8)^2}{(4)^2}$$

$$\Rightarrow$$ $$ar(\triangle ABC)=\dfrac{64}{16}\times 25$$

$$\Rightarrow$$ $$ar(\triangle ABC)=4\times 25$$
$$\therefore$$ $$ar(\triangle ABC)=100\,cm^2$$
Question 6
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What is the ratio of the areas of two similar triangles whose corresponding sides are in the ratio 15:19?

A
$$\sqrt{15} : \sqrt{19}$$

B
$$15 : 19$$

C
$$225 : 361$$

D
$$125 : 144$$

Solution

We know ratio of areas of two similar triangles is equal to the ratio of the square of their corresponding sides.
Ratio of sides $$=\dfrac{15}{19}$$
Ratio of areas $$={ \left( \dfrac { 15 }{ 19 } \right) }^{ 2 }=\dfrac { 225 }{ 361 } $$
So, option $$C$$ is correct.
Question 7
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In the figure shown above, $$DE$$ is parallel to $$BC$$ and the ratio of the areas of $$\triangle ADE$$ and trapezium $$BDEC$$ is $$4 : 5$$. What is $$DE : BC$$?

A
$$1 : 2$$

B
$$2 : 3$$

C
$$4 : 5$$

D
None of the above
Question 8
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The areas of two similar triangles are 100 $$\mathrm{cm^2}$$ and 64 $$\mathrm{cm^2}$$ respectively. If the median of greater side of the first triangle is 13 cm, find the corresponding median of the other triangle.

A
20 cm

B
15 cm

C
10 cm

D
10.4 cm

Solution

Given that the areas of two similar triangles are $$100 \ \mathrm{cm^2}$$ and $$64 \ \mathrm{cm^2}$$.
and median of greater triangle is $$13 \ \mathrm{cm}$$.
Let the median of the other triangle be $$x \ \mathrm{cm}$$

We know that, the areas of two similar triangles are in the ratio of the squares of their corresponding sides or medians.

The ratio of the areas of triangles $$=\dfrac{100}{64}=\dfrac{25}{16}$$

$$\therefore \dfrac{(13)^{2}}{(x)^{2}}=\dfrac{25}{16}$$

$$\Rightarrow \dfrac{169}{x^{2}}=\dfrac{25}{16}$$

$$\Rightarrow 25x^{2}=169\times 16$$

$$\Rightarrow x^{2}=\dfrac{2704}{25}=108.16$$

$$\Rightarrow x=10.4\ \mathrm{cm}$$

Hence, the median of the other triangle is $$10.4 \ \mathrm{cm}$$
Question 9
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In Figure , $$\overline{QR} = \overline{RS}$$. If the area of $$\triangle RST$$ is $$\dfrac{c}{2}$$, what is the area of $$\triangle QSP$$?

A
$$c\sqrt{2}$$

B
$$c\sqrt{3}$$

C
$$c$$

D
$$2c$$

E
$$3c$$

Solution

$$QR=RS$$         ....Given
So, $$SR: SQ = 1:2$$          ...(I)
$$\angle RTP + \angle RTS = 180^o$$         ...Angles in linear pair
$$\therefore 126^o + \angle RTS = 180^o$$
$$\therefore \angle RTS = 54^o = \angle SPQ$$    ...(II)
In $$\triangle RST$$ and $$\triangle QSP$$
$$\angle S$$ is the common angle.
$$\therefore \angle RTS = \angle SPQ$$    ... from (II)
$$\triangle RST \sim \triangle SPQ$$         ....AAA test of similarity
So, $$\dfrac {A(\triangle RST)}{A(\triangle SPQ)} = \left(\dfrac {RS}{SQ}\right)^2$$           ....Theorem on ratio of areas of similar triangles
$$\Rightarrow \dfrac {A(\triangle RST)}{A(\triangle SPQ)} = \left(\dfrac 12\right)^2$$
$$\Rightarrow \dfrac {\frac c2}{A(\triangle SPQ)} = \dfrac 14$$
$$A(\triangle SPQ) = 2c$$ sq. units
Question 10
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In Figure , $$\overline{EG} || \overline{DH}$$, and the lengths of segments $$\overline{DE}$$ and $$\overline{EF}$$ are as marked. If the area of $$\triangle EFG$$ is a, calculate the area of $$\triangle DFH$$ in terms of $$a$$.

A
$$\dfrac{4a}{5}$$

B
$$\dfrac{16a}{25}$$

C
$$\dfrac{16a}{20}$$

D
$$\dfrac{25a}{16}$$

Solution

Given: $$EG \parallel DH$$
By basic proportionality theorem,
$$\dfrac {ED}{EF} = \dfrac {GH}{GF}$$
So, $$\dfrac {EF+ED}{EF} = \dfrac {GF+GH}{GF}$$       ....By componendo
$$\therefore \dfrac {FD}{EF} = \dfrac {FH}{GF} = \dfrac 54$$         ....(I)
In $$\triangle FEG$$ and $$\triangle FDH$$
$$\angle F$$ is the common angle
And, $$\dfrac {EF}{ED} = \dfrac {GF}{GH}$$     ....From (I)
$$\therefore \triangle FEG \sim \triangle FDH$$         ....S.A.S test of similarity
By theorem on area of similar triangles, we get
$$\dfrac {A(\triangle FDH)}{A(\triangle FEG)} = \left(\dfrac {FD}{EF} \right)^2$$
$$\Rightarrow \dfrac {A(\triangle FDH)}{a} = \left(\dfrac {5}{4} \right)^2$$
$$\Rightarrow A(\triangle FDH) = \dfrac {25}{16} a$$