Triangles Test

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Triangles Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, $$QA$$ and $$PB$$ are perpendiculars to $$AB.$$ If $$AO=10$$ cm, $$BO=6$$ cm and $$PB=9$$ cm, find $$AQ.$$

A
$$5$$ cm

B
$$10$$ cm

C
$$15$$ cm

D
$$20$$ cm

Solution

In $$\triangle $$ $$AOQ$$ and $$\triangle BOP$$, we have

$$\angle OAQ=\angle OBP$$ (Each equal to $$90^{\circ}$$)

$$\angle AOQ=\angle POB$$ (Vertically opp. $$\angle $$s)

$$\therefore $$ By AA-similarity,

$$\triangle AOQ\sim \triangle BOP$$

$$\Rightarrow $$ $$\displaystyle \frac{AO}{BO}=\frac{OQ}{OP}=\frac{AQ}{BP}$$

$$\displaystyle \frac{AO}{BO}=\frac{AQ}{9}= \frac{10}{6}\\\Rightarrow AQ=\dfrac{10\times 9}{6}\\=15\ cm$$
Question 2
1 / -0

The areas of two similar triangles are $$121$$ cm$$^{2}$$ and $$64$$ cm$$^{2}$$, respectively. If the median of the first triangle is $$12.1$$ cm, then the corresponding median of the other is:

A
$$6.4$$ cm

B
$$10$$ cm

C
$$8.8$$ cm

D
$$3.2$$ cm

Solution

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding medians. Therefore,
$$\displaystyle \frac{121}{64}=\frac{\left ( 12.1 \right )^{2}}{x^{2}},$$ where $$x$$ is the median of the other $$\triangle .$$
$$\Rightarrow $$ $$\displaystyle x^{2}=\frac{\left ( 12.1 \right )^{2}\times 64}{121}\Rightarrow x=\sqrt{\frac{121}{100}\times 64}$$
$$\displaystyle =\frac{11}{10}\times 8=8.8$$ cm.
Question 3
1 / -0

Triangles $$ABC$$ and $$DEF$$ are similar. If the length of the perpendicular $$AP$$ from $$A$$ on the opposite side $$BC$$ is $$2$$ cm and the length of the perpendicular $$DQ$$ from $$D$$ on the opposite side $$EF$$ is $$1$$ cm, then what is the area of $$\triangle ABC\: ?$$

A
One and half times the area of the triangle $$DEF.$$

B
Four times the area of triangle $$DEF.$$

C
Twice the area of the triangle $$DEF.$$

D
Three times the area of triangle $$DEF.$$

Solution

Areas of similar triangles are in the ratio of square of altitues.
$$\dfrac{Ar(\triangle ABC)}{Ar(\triangle DEF)} = \dfrac{AP^2}{DQ^2}$$
$$\Rightarrow Ar(\triangle ABC) = 4 Ar(\triangle DEF)$$
Question 4
1 / -0

If $$\triangle ABC$$ is similar to $$\triangle DEF$$ such that $$BC=3$$ cm, $$EF=4$$ cm and area of $$\triangle ABC=54\: \text{cm}^{2}.$$ Find the area of $$\triangle DEF.$$ (in cm$$^2$$)

A
$$54$$

B
$$36$$

C
$$72$$

D
$$96$$

Solution

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides,

Therefore, $$\displaystyle \frac{ar\left ( \triangle ABC \right )}{ar\left ( \triangle DEF \right )}=\dfrac{BC^{2}}{EF^{2}}$$

$$\Rightarrow $$ $$\displaystyle \dfrac{54}{ar\left ( \triangle DEF \right )}=\dfrac{3^{2}}{4^{2}}$$

Thus $$\displaystyle ar\left ( \triangle DEF \right )=\dfrac{54\times 16}{9}=96\: \text{cm}^{2}$$
Question 5
1 / -0

$$D$$ and $$E$$ are respectively the points on the sides $$AB$$ and $$AC$$ of a $$\triangle ABC$$ such that $$AB = 12 \text{ cm},$$ $$AD = 8 \text{ cm},$$ $$AE = 12 \text{ cm}$$ and $$AC = 18 \text{ cm}$$, then

A
$$DE\parallel BD$$

B
$$DE \parallel BC$$

C
$$AD \parallel BD$$

D
$$AD \parallel CD$$

Solution

If $$ \dfrac { AD }{ BD } =\dfrac { AE }{ CE } $$ then by converse of basic proportionality theorem we can say that $$DE\;|\;|BC.$$

$$\begin{aligned}{}\frac{{AD}}{{BD}} &= \frac{{AE}}{{CE}}\quad\quad\quad\dots(i)\\\frac{{BD}}{{AD}} &= \frac{{CE}}{{AE}}\\\frac{{BD}}{{AD}} + 1 &= \frac{{CE}}{{AE}} + 1\\\frac{{BD + AD}}{{AD}}& = \frac{{CE + AE}}{{AE}}\\\frac{{AB}}{{AD}}& = \frac{{AC}}{{AE}}\\\frac{{AD}}{{AB}} &= \frac{{AE}}{{AC}}\quad\quad\quad\dots(ii)\end{aligned}$$

So, the above calculation implies that instead of an equation $$(i)$$ if we prove that equation $$(ii)$$ is true then also we can say that $$DE\;|\;|BC.$$
$$\begin{aligned}{}\frac{{AD}}{{AB}} &= \frac{{AE}}{{AC}}\\\frac{8}{{12}} &= \frac{{12}}{{18}}\\\frac{2}{3} &= \frac{2}{3}\quad\quad\quad[\text{Which is true}]\end{aligned}$$

Hence, proved $$DE\;|\;|BC.$$
Question 6
1 / -0

In a $$\Delta ABC,\;D\;and\;E$$ are points on the sides $$AB$$ and $$AC$$ respectively such that $$DE\;\parallel\;BC$$. If $$AD=4x-3,\;AE=8x-7,\;BD=3x-1\;and\;CE=5x-3$$, find the value of $$x$$.

A
$$1$$

B
$$2$$

C
$$\dfrac{1}{2}$$

D
$$3$$

Solution

In $$\Delta ABC$$, we have
$$DE\;\parallel\;BC$$
$$\therefore\;\displaystyle\frac{AD}{DB}=\displaystyle\frac{AE}{EC}\;\;\;\;\;$$[By Basic Proportionality Theorem]
$$\Rightarrow\;\displaystyle\frac{4x-3}{3x-1}=\displaystyle\frac{8x-7}{5x-3}$$
$$\Rightarrow\;20x^2-15x-12x+9=24x^2-21x-8x+7$$
$$\Rightarrow\;20x^2-27x+9=24x^2-29x+7$$
$$\Rightarrow\;4x^2-2x-2=0$$
$$\Rightarrow\;2x^2-x-1=0$$
$$\Rightarrow\;(2x+1)(x-1)=0$$
$$\Rightarrow\;x=1\;or\;x=-\displaystyle\frac{1}{2}$$
So, the required value of $$x$$ is $$1$$.
[$$x=-\displaystyle\frac{1}{2}$$ is neglected as length can not be negative].
Question 7
1 / -0

In a $$\displaystyle \bigtriangleup ABC $$, AB = AC = 2.5 cm, BC = 4 cm. Find its height from A to the opposite base.

A
1.5 cm

B
1 cm

C
2 cm

D
3 cm

Solution

$$ \textrm{In }\triangle ABC $$ In order to find height we need to consider that $$ AD \perp BC $$
Hence in right angled $$ \triangle ADC $$
$$ AC^2 = AD^2 + DC^2 $$
$$ AD^2 = AC^2 - DC^2 $$
$$ AD^2 = (2.5)^2 - (2)^2 $$
$$ AD^2 = 2.25 $$
$$ AD = 1.5 \, cm $$
Option A is the correct answer.
Question 8
1 / -0

If $$\Delta ABC\sim \Delta DEF$$ such that area of $$\Delta ABC$$ is $$9 cm^2$$ and area of $$\Delta DEF$$ is $$16 cm^2$$ and $$BC=1.8 cm$$, then EF is

A
2.4 cm

B
1.35 cm

C
2.1 cm

D
3.2 cm

Solution

$$ar(\triangle ABC)=9cm^2,\,ar(\triangle DEF)=16cm^2$$ and $$BC=1.8cm$$

$$\triangle ABC\sim\triangle DEF$$ [ Given ]

$$\Rightarrow$$ $$\dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=\dfrac{(BC)^2}{(EF)^2}$$ [ Area of similar triangle theorem ]

$$\Rightarrow$$ $$\dfrac{9}{16}=\dfrac{(1.8)^2}{(EF)^2}$$
Taking square root on both sides,

$$\Rightarrow$$ $$\dfrac{3}{4}=\dfrac{1.8}{EF}$$

$$\Rightarrow$$ $$EF=\dfrac{1.8\times 4}{3}$$

$$\Rightarrow$$ $$EF=\dfrac{7.2}{3}$$

$$\Rightarrow$$ $$EF=2.4\,cm$$
Question 9
1 / -0

$$\displaystyle \angle BAC={ 90 }^{ o }$$, AD is its bisector. If $$\displaystyle DE\bot AC$$, prove that $$\displaystyle DE\times \left( AB+AC \right) =AB\times AC$$

A
$$\displaystyle DE\times \left( AB+AC \right) =AB\times AC$$ is proved

B
$$\displaystyle DE\times \left( AB+AC \right) =AB\times AC$$is not proved

C
$$\displaystyle DE\times \left( AB+AC \right) > AB\times AC$$

D
$$\displaystyle DE\times \left( AB+AC \right) < AB\times AC$$

E
Blank

Solution

It is given that AD is the bisector of $$\displaystyle \angle A$$ of $$\displaystyle \Delta ABC$$.
$$\displaystyle \therefore \frac { AB }{ AC } =\frac { BD }{ DC } $$
$$\displaystyle \Rightarrow \frac { AB }{ AC } +1=\frac { BD }{ DC } +1$$ [Adding 1 on both sides]
$$\displaystyle \Rightarrow \frac { AB+AC }{ AC } =\frac { BD+DC }{ DC } $$
$$\displaystyle \Rightarrow \frac { AB+AC }{ AC } =\frac { BC }{ DC } $$........(i)
In $$\displaystyle \Delta 's\quad CDEandCBA$$, we have
$$\displaystyle \angle DCE=\angle BCA$$ [Common]
$$\displaystyle \angle DEC=\angle BAC$$ [Each equal to $$\displaystyle { 90 }^{ o }$$]
So, by AA-criterion of similarity
$$\displaystyle \Delta CDE\sim \Delta CBA$$
$$\displaystyle \Rightarrow \frac { CD }{ CB } =\frac { DE }{ BA } $$
$$\displaystyle \Rightarrow \frac { AB }{ DE } =\frac { BC }{ DC } $$.......(ii)
From (i) and (ii), we have
$$\displaystyle \Rightarrow \frac { AB+AC }{ AC } =\frac { AB }{ DE } $$
$$\displaystyle \Rightarrow DE\times \left( AB+AC \right) =AB\times AC$$.
Question 10
1 / -0

In the figure, $$\displaystyle PQ\parallel BC$$ and AP : PB = 1 : 2. Find $$\displaystyle \frac { ar\left( \Delta APQ \right) }{ ar\left( \Delta ABC \right) } $$.

A
1 : 4

B
4 : 1

C
1 : 9

D
2 : 9

Solution

If $$AP : PB = 1 : 2$$ then $$AP : AB$$ will be $$1 : 3$$.

So, in similar triangles, the ratio of the area of $$\dfrac { \Delta \quad APQ }{ \Delta \quad ABC } \quad =\quad { \left(\dfrac { 1 }{ 3 } \right)\quad }^{ 2 }=\quad \dfrac { 1 }{ 9 } \\ $$