Triangles Test

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Triangles Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of the areas of two similar triangles is equal to:

A
The ratio of corresponding medians

B
The ratio of corresponding sides

C
The ratio of the squares of corresponding sides

D
None of these

Solution

$${\textbf{Step -1: Concept of similar triangle.}}$$
$${\text{The ratio of the areas of two similar triangles is equal to the}}$$
$${\text{squares of the ratio of their corresponding sides.}}$$
$${\text{As we know if 3 sides of one triangle are equal to the 3 sides of another triangle,}}$$
$${\text{then the two triangles are congruent.}}$$
$${\textbf{Thus, option C is correct.}}$$
Question 2
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The areas of two similar triangles are $$49 \ {cm}^{2}$$ and $$64 \ {cm}^{2}$$ respectively. The ratio of their corresponding sides is:

A
$$49:64$$

B
$$7:8$$

C
$$64:49$$

D
none of these

Solution

Areas of two similar triangles are $$49 $$ cm $$^2$$ and $$64$$ cm $$^2.$$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding sides.
Hence, $$\dfrac{A_1}{A_2} = \dfrac{(s_1)^2}{(s_2)^2}$$
$$\Longrightarrow \dfrac{49}{64} = \dfrac{(s_1)^2}{(s_2)^2}$$
$$\Longrightarrow\dfrac{s_1}{s_2} = \dfrac{7}{8}$$
Question 3
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$$\Delta ABC \sim \Delta PQR$$ and $$\displaystyle\frac{A( \Delta ABC)}{A( \Delta PQR)}=\dfrac{16}{9}$$. If $$PQ=18$$ cm and $$BC=12$$ cm, then $$AB$$ and $$QR$$ are respectively:

A
$$9$$ cm, $$24$$ cm

B
$$24$$ cm, $$9$$ cm

C
$$32$$ cm, $$6.75$$ cm

D
$$13.5$$ cm, $$16$$ cm

Solution

$$\displaystyle\frac{16}{9}=\left[\frac{AB}{PQ}\right]^2=\left[\frac{BC}{QR}\right]^2$$
$$\displaystyle\Rightarrow \frac{16}{9}=\left[\frac{AB}{18}\right]^2$$ and $$\displaystyle\frac{16}{9}=\left[\frac{12}{QR}\right]^2$$
$$\displaystyle \Rightarrow \frac{4}{3}=\frac{AB}{18}$$ and $$\displaystyle \frac{4}{3}=\frac{12}{QR}$$
$$\Rightarrow AB=24$$ cm, $$QR=9$$ cm.
Question 4
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In figure, two lines segments $$AC$$ and $$BD$$ intersect each other at the point $$P$$ such that $$PA=6cm$$, $$PB=3cm$$, $$PC=2.5cm$$, $$PD=5cm$$, $$\angle APB={50}^{o}$$ and $$\angle CDP={30}^{o}$$. Then $$\angle PBA$$ is equal to:

A
$${50}^{o}$$

B
$${30}^{o}$$

C
$${60}^{o}$$

D
$${100}^{o}$$

Solution

Given: $$PA = 6$$ cm, $$PB = 3$$ cm, $$PC = 2.5$$ cm, $$PD=5$$ cm
$$\dfrac{PA}{PD} = \dfrac{6}{5}$$
$$\dfrac{PB}{PC}$$ = $$\dfrac{3}{2.5}$$ = $$\dfrac{6}{5}$$
Thus, $$\dfrac{PA}{PD} = \dfrac{PB}{PC}$$
In $$\triangle APB$$ and $$\triangle DPC$$,
$$\dfrac{PA}{PD} = \dfrac{PB}{PC}$$      ...Given
$$\angle APB = \angle DPC$$        ...Vertically opposite angles
Hence, $$\triangle APB \sim \triangle DPC$$           ....S.A.S test of similarity
Hence, $$\angle A = \angle D = 30^{o}$$
Now, In $$\triangle APB$$,
Sum of angles $$= 180^o$$
$$\angle A + \angle B + \angle APB = 180$$
$$30 + 50 + \angle B = 180^{o}$$
$$\angle B = 100^{\circ}$$
Question 5
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In the adjoining figure, $$XY$$ is parallel to $$AC$$. If $$XY$$ divides the triangle into equal parts, then the value $$\cfrac{AX}{AB}=$$

A
$$\cfrac{1}{2}$$

B
$$\cfrac{1}{\sqrt{2}}$$

C
$$\cfrac { \sqrt { 2 } +1 }{ \sqrt { 2 } } $$

D
$$\cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } } $$

Solution

We have, Area of $$\triangle BXY$$ = Area of AXYC
Now adding Area of $$\triangle BXY$$ on both sides, we get,
$$2\times$$ Area of $$\triangle BXY$$=Area of $$\triangle ABC$$
$$\Rightarrow \frac {Area\quad of\quad \triangle ABC}{ Area\quad of\quad \triangle BXY }=2$$ ------(i)

In $$\triangle ABC$$ and $$\triangle XBY$$
$$\angle ABC=\angle XBY$$
$$\angle BAC=\angle BXY$$ ---------(corresponding angles)
By AA Similarity, $$\triangle ABC \sim \triangle XBY$$
From (i), we have,
$$\dfrac { { AB }^{ 2 } }{ { BX }^{ 2 } }=2$$

$$\Rightarrow\dfrac { { AB } }{ { BX } }=\sqrt { 2 } $$

$$\Rightarrow \dfrac { { BX } }{ { AB } }=\frac { 1 }{ \sqrt { 2 } } $$

$$\Rightarrow 1-\dfrac { { BX } }{ { AB } }=1-\dfrac { 1 }{ \sqrt { 2 } }$$

$$\Rightarrow \dfrac { { AB-BX } }{ { AB } }=\dfrac {\sqrt { 2 }- 1 }{ \sqrt { 2 } }$$
$$\Rightarrow \dfrac { { AX } }{ { AB } }=\dfrac {\sqrt { 2 }- 1 }{ \sqrt { 2 } }$$
Question 6
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State True or False
The two triangles are similar .$$EB||DC$$

A
true

B
false

C
only $$\angle AEB$$ is equal to $$\angle ADC$$

D
incomplete data

Solution

In $$\triangle ABE$$ and $$\triangle ACD$$

$$\Rightarrow$$ $$\angle EAB=\angle DAC$$ [ Common angle ]

$$\Rightarrow$$ $$\angle ABE=\angle ACD$$ [ Corresponding angles ]

$$\Rightarrow$$ $$\angle BEA=\angle CDA$$ [ Corresponding angles ]

$$\therefore$$ $$\triangle ABE\sim\triangle ACD$$ [ By $$AAA$$ similarity rule ]

$$\therefore$$ We have proved that, the two triangles are similar.
Question 7
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In the figure, $$PQRS$$ is a parallelogram with $$PQ=15cm$$ and $$RQ=10cm$$. $$L$$ is a point on $$RP$$ such that $$RL:LP=2:3$$. $$QL$$ produced meets $$RS$$ at $$M$$ and $$PS$$ produced at $$N$$. Find the lengths of $$PN$$ and $$RM$$.

A
$$PN=15cm; RM=10cm$$

B
$$PN=10cm; RM=10cm$$

C
$$PN=25 cm; RM=15cm$$

D
$$PN=25cm; RM=10cm$$

Solution

Given: $$PQRS$$ is a parallelogram. $$PQ = 15$$ and $$RQ = 10$$, $$\frac{RL}{LP} = \frac{2}{3}$$
In $$\triangle RLQ$$ and $$\triangle NLP$$
$$\angle RLQ = \angle NLP$$ (vertically opposite angles)
$$\angle LRQ = \angle LPN$$ (Alternate angles)
$$\angle LQR = \angle LNP$$ (Alternate angles)
Thus, $$\triangle RLQ \sim \triangle PLN$$ (AAA rule)
Hence, $$\dfrac{RL}{PL} = \dfrac{RQ}{PN}$$ (Corresponding sides of similar triangles)
$$PN = \dfrac{3 \times 10}{2}$$
$$PN = 15 cm$$
Similarly, $$\triangle RLM \sim \triangle PLQ$$
$$\dfrac{RL}{LP} = \dfrac{RM}{PQ}$$ (Corresponding sides of similar triangles)
$$RM = \dfrac{15 \times 2}{3}$$
$$RM = 10$$ cm.
Question 8
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Find the length $$JN$$ in the given figure.

A
$$12$$

B
$$\frac{16}{3}$$

C
$$27$$

D
$$14$$

Solution

Given: $$KN \parallel ML$$

Use the basic proportionality theorem

$$\cfrac{JN}{NM}=\cfrac{JK}{KL}$$

$$\cfrac{JN}{8}=\cfrac{18}{12}$$ Find the cross products

$$12\times JN=8\times 18$$

$$JN=\cfrac{144}{12}$$

$$JN=12$$
Question 9
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If $$\widetilde{BAD} \cong \widetilde{ADC}$$, then

A
AB = CD

B
$$AB \neq CD $$

C
AD = BC

D
$$AB \neq BC$$

Solution

Since the given two triangles are congruent , then corresponding sides and the angles will be equal.
This way:
$$BA=AD$$ (i)
$$AD=DC$$ (ii)
$$DB=CA$$ (iii)
Using (i) and (iii)
$$AB=CD$$
Question 10
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$$P$$ and $$Q$$ are points on the sides $$AB$$ and $$AC$$ respectively of $$\triangle ABC$$ such that $$PQ\parallel BC$$ and divides the triangle into two parts of equal area. Find $$PB:AB$$.

A
$$1:\sqrt 2$$

B
$$(\sqrt 2):(\sqrt 2 - 1)$$

C
$$(\sqrt 2-1):\sqrt 2$$

D
$$1:(\sqrt 2 - 1)$$

Solution

It is given that $$PQ$$ divides $$\triangle ABC$$ into two parts of equal area. So,
$$\begin{aligned}{}ar\left( {APQ} \right) &= ar\left( {PBCQ} \right)\\ar\left( {APQ} \right) + ar\left( {APQ} \right) &= ar\left( {PBCQ} \right) + ar\left( {APQ} \right)\\2ar\left( {APQ} \right) &= ar\left( {ABC} \right)\\\frac{{ar\left( {APQ} \right)}}{{ar\left( {ABC} \right)}}& = \frac{1}{2}\quad\quad\quad\quad\dots(i)\end{aligned}$$

Now, in $$\triangle APQ$$ and $$\triangle ABC$$, we have
$$\angle PAQ=\angle BAC$$ [Common]
$$\angle APQ=\angle ABC$$ [Corresponding angles]

So, by $$AAA$$ criteria of similarity,
$$\triangle APQ \sim \triangle ABC$$

We know that the areas of similar triangles are proportional to the squares of their corresponding sides.
$$\therefore$$ $$\cfrac{ar(\triangle APQ)}{ar(\triangle ABC)}=\cfrac{{AP}^{2}}{{AB}^{2}}$$
$$\Rightarrow $$ $$\cfrac{AP^2}{AB^2}=\cfrac{1}{2}$$ [By using $$(i)$$]
$$\Rightarrow AB=\sqrt 2\cdot AP$$
$$\Rightarrow$$ $$AB=\sqrt 2(AB-PB)$$
$$\Rightarrow$$ $$\sqrt 2 PB=(\sqrt{2} -1)AB$$
$$\Rightarrow$$ $$\cfrac{PB}{AB}=\cfrac{(\sqrt 2-1)}{\sqrt 2}$$
$$\therefore$$ $$PB:AB=(\sqrt 2-1):\sqrt 2$$