Triangles Test

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Triangles Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

Area of similar triangles are in the ratio $$25:36$$ then ratio of their similar sides is _________?

A
$$5:7$$

B
$$5:6$$

C
$$6:5$$

D
$$6:7$$

Solution

The areas and sides of similar triangles are related as $$\dfrac{Ar(\Delta ABC)}{Ar(\Delta PQR)}=\left(\dfrac {AB}{PQ}\right)^2$$

$$\Rightarrow \dfrac {25}{36}=\left(\dfrac {AB}{PQ}\right)^2$$

$$\Rightarrow \dfrac{AB}{PQ}=\sqrt {\dfrac {25}{36}}$$

$$=\dfrac 56$$
Question 2
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In $$\Delta PQR$$, $$XY\parallel QR$$ and $$\dfrac { PX }{ XQ } =\dfrac { 2 }{ 3 }$$ $$PY=4\ cm$$ then $$YR=$$ ?

A
$$6\ cm$$

B
$$9\ cm$$

C
$$5\ cm$$

D
$$8\ cm$$

Solution

By basic proportionality theorem,
$$\dfrac{PX}{XQ}=\dfrac{PY}{YR}$$

$$\dfrac{2}{3}=\dfrac{4}{YR}$$

$$YR=6 \ cm$$
Question 3
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In the given figure
$$OA \times OB = OC \times OD,$$ then which of the following is correct.

A
$$\angle A = \angle B$$

B
$$\angle A = \angle C$$

C
$$\angle D = \angle C$$

D
none of these

Solution

Given $$ OA \times OB = OC \times OD$$

Then We have

$$\dfrac {OA}{OC}=\dfrac {OD}{OB}$$....(1)

In $$\Delta AOD$$ and $$ \Delta COB$$

$$\dfrac {OA}{OC}=\dfrac {OD}{OB}$$....from (1)

$$\angle AOD = \angle COB [ vertically\, opp. \measuredangle S]$$

$$ \therefore \Delta AOD \sim \Delta COB$$( SAS)

Now, if two triangle are similar. so, corresponding angles are equal.

$$\angle A = \angle C$$

and $$ \angle D = \angle B$$
Question 4
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In the adjoining figure, $$PQ\bot BC,AD\bot BC,PQ=4,AD=6$$
Write down the following ratios:
$$(i)~\dfrac { A(\triangle PQB) }{ A(\triangle ADB) } $$
$$(ii)~\dfrac { A(\triangle PBC) }{ A(\triangle ABC) } $$

A
$$(i)\frac{2}{9}, (ii)~\frac{2}{3}$$

B
$$(i)\frac{4}{9}, (ii)~\frac{4}{3}$$

C
$$(i)\frac{4}{9}, (ii)~\frac{2}{3}$$

D
$$(i)\frac{1}{9}, (ii)~\frac{1}{3}$$

Solution
Question 5
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The area of two similar triangles are $$16\ cm^{2}$$ and $$36\ cm^{2}$$ respectively, If the altitudes are in the ratio $$2:x$$, then the value of $$x$$ is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$6$$

Solution

REF.Image
Consider two trianles ABC and PQR with AM and PN as altitudes respectively.
$$\frac{area of \Delta ABC}{area of \Delta PQR}= \frac{16}{36}$$
We know $$\frac{area of \Delta _{1}}{area of \Delta _{2}}$$ = square of radius of proportional sides.
so $$\frac{BC}{QR}=\frac{16}{36}=\frac{4}{9}$$ _______ (1)
$$\frac{Area of \Delta ABC}{Area of \Delta PQR}=\frac{\frac{1}{2}BC\times AM}{\frac{1}{2}QR\times PN}=\frac{16}{36}=\frac{4}{9}$$
$$\Rightarrow \frac{BC}{QR}\times \frac{AM}{PN}=\frac{4}{9}$$
Given $$\frac{BC}{QR}=\frac{2}{3}\Rightarrow \frac{2}{3}\times \frac{AM}{PN}=\frac{4}{9}$$
$$\Rightarrow \frac{AM}{PN}=\frac{2}{3}$$
$$\therefore x=3$$
Question 6
1 / -0

The triangle $$\Delta$$ ABC is such that AB=9 cm, BC=6 cm, AC=7.5 cm. Triangle DEF is similar to $$\Delta ABC$$. If EF=12 cm then DE is :

A
6 cm

B
16 cm

C
18 cm

D
15 cm

Solution

Given: $$\Delta ABC $$ similar to $$\Delta DEF$$

$$\therefore \dfrac{AB}{BC} = \dfrac{DE}{EF}$$

$$\therefore \dfrac{9}{6} = \dfrac{x}{12}$$

$$DE = x=18cm$$
Question 7
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Two trees $$7 \,m$$ and $$4 \,m$$ high stand upright on a ground. If their bases (roots) are $$4 \,m$$ apart, then the distance between their tops is

A
$$3 \,m$$

B
$$5 \,m$$

C
$$4 \,m$$

D
$$11 \,m$$

Solution

Let $$AB = 7 \,m$$ and $$CD = 4 \,m$$
$$AC = DE = 4 \,m$$
Now, in triangle $$BED$$ is right-angled triangle.
$$\Longrightarrow (BD)^2 = (BE)^2 + (DE)^2$$
$$\Longrightarrow (BD)^2 = 3^2 + 4^2$$
$$\Longrightarrow (BD)^2 = 25$$
$$\Longrightarrow (BD)^2 = 5$$
Threfore, the distance the tops of trees in $$5$$ meter.
Question 8
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In figure, $$PQ\bot RQ,\quad PQ=5cm,\quad QR=5cm$$
Then $$\triangle PQR$$ is

A
A right triangle but not isosceles

B
An isosceles right triangle

C
Isosceles but not a right triangle

D
Neither isosceles nor right triangle

Solution

Given that
$$PQ\bot RQ,\quad PQ=5cm,\quad QR=5cm$$
$$\Rightarrow\angle PQR={ 90 }^{ o }$$ and $$PQ=QR=5cm$$
$$\therefore \triangle PQR$$ is an isosceles right angle triangle

Hence, the correct answer is option [B].
Question 9
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Tick the correct answer and justify:
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

If we have two similar triangles having ratio of there sides $$4:9$$
According to similar triangle property ratio of there area is square of the ratio
i.e $$4^2:9^2$$
Therefore$$16:81$$
Question 10
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If straight line is drawn parallel to one side of a triangle then it divides the other two sides proportionally is called.

A
thales theorem

B
converse of thales theorem

C
Corollary of thales

D
Pythagoras theorem

Solution

Thales stated that ' If a straight line is drawn parallel to one side of a triangle, then it divides other two sides in equal ratio.'