Triangles Test

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Triangles Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Given two similar triangles one of which has twice the perimeter of the other, by what factor is the area of the larger triangle bigger than the smaller ?

A
$$2$$

B
$$4$$

C
$$\sqrt{2}$$

D
$$2\sqrt{2}$$

Solution

$$Let\quad \triangle ABC\quad and\quad \triangle PQR\quad be\quad smaller\quad and\quad bigger\quad triangles\quad \\ Given\quad \triangle ABC\sim \triangle PQR\\ 2Perimeter\quad of\quad \triangle ABC=Perimeter\quad of\quad \triangle PQR\\ \Rightarrow \frac { Perimeter\quad of\quad \triangle ABC }{ Perimeter\quad of\quad \triangle PQR } =\frac { 1 }{ 2 } \\ Two\quad similar\quad triangles\quad who\quad have\quad scale\quad factor\quad of\quad a:b\quad then\quad ratio\quad of\quad there\quad areas\quad will\quad be\\ { a }^{ 2 }:{ b }^{ 2 }\\ \Rightarrow \frac { Area\quad of\quad \triangle ABC }{ Area\quad of\quad \triangle PQR } =({ \frac { 1 }{ 2 } ) }^{ 2 }=\frac { 1 }{ 4 } \\ \Rightarrow Area\quad of\quad \triangle PQR=4\times Area\quad of\quad \triangle ABC\\ \therefore Area\quad of\quad larger\quad triangle\quad is\quad 4\quad times\quad bigger\quad than\quad area\quad of\quad smaller\quad triangle.\\ Hence\quad option\quad (B)is\quad correct\quad answer$$
Question 2
1 / -0

In the given figure, PA, OB and RC are each perpendicular to AC. Which of the relations hold true?

A
$$\displaystyle \frac { 1 }{ x } +\frac { 1 }{ y} <\frac { 1 }{ z} $$.

B
$$\displaystyle \frac { 1 }{ x } +\frac { 1 }{ z} =\frac { 1 }{ y} $$.

C
$$\displaystyle \frac { 1 }{ x } +\frac { 1 }{ z } >\frac { 1 }{ y } $$.

D
$$\displaystyle \frac { 1 }{ x} -\frac { 1 }{ y } =\frac { 1 }{ z } $$

Solution

In $$\displaystyle \Delta PAC$$, we have $$\displaystyle BQ\parallel AP$$
$$\displaystyle \Rightarrow \frac { BQ }{ AP } =\frac { CB }{ CA } \quad \left[ \therefore \Delta CBQ\sim \Delta CAP \right] by AAA$$
$$\displaystyle \Rightarrow \frac { y }{ x } =\frac { CB }{ CA } $$.....(i)
In $$\displaystyle \Delta ACR$$, we have $$\displaystyle BQ\parallel CR$$
$$\displaystyle \Rightarrow \frac { BQ }{ CR } =\frac { AB }{ AC } \left[ \therefore \Delta ABQ\sim \Delta ACR \right] $$ by AAA
$$\displaystyle \Rightarrow \frac { y }{ z } =\frac { AB }{ AC } $$......(ii)
Adding (i) and (ii), we get
$$\displaystyle \frac { y }{ x } +\frac { y }{ z } =\frac { CB }{ AC } +\frac { AB }{ AC } $$
$$\displaystyle \Rightarrow \frac { y }{ x } +\frac { y }{ z } =\frac { AB+BC }{ AC } $$
$$\displaystyle \Rightarrow \frac { y }{ x } +\frac { y }{ z } =\frac { AC }{ AC } $$
$$\displaystyle \Rightarrow \frac { y }{ x } +\frac { y }{ z } =1$$
$$\displaystyle \Rightarrow \frac { 1 }{ x } +\frac { 1 }{ z } =\frac { 1 }{ y } $$.
hence Proved
Question 3
1 / -0

In the following figure $$\Delta ABC \sim \Delta APQ\ \ \ ,A(\Delta APQ) = 4A(\Delta ABC)$$. Find the ratio $$\displaystyle \dfrac{BC}{PQ}$$

A
$$1:1$$

B
$$1:2$$

C
$$1:3$$

D
$$1:4$$

Solution

$$\Delta ABC \sim \Delta APQ$$ (Given)

Apply the properties of similar triangles

$$\displaystyle \therefore \dfrac{A(\Delta ABC)}{A(\Delta APQ)} = \dfrac{BC^2}{PQ^2}$$ ......$$(i)$$

[$$\therefore$$ Ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides]

But $$ A(\Delta APQ) = 4A(\Delta ABC)$$ (given)

$$\displaystyle \therefore \frac{A(\Delta ABC)}{A(\Delta APQ)} = \frac{1}{4}$$ [From $$(i)$$ and $$(ii)$$

$$\displaystyle \therefore \frac{BC}{PQ} = \frac{1}{2}$$
Question 4
1 / -0

In $$\Delta ABC,$$ $$D,$$ $$E$$ and $$F$$ are the mid points of the sides $$BC,\ CA$$ and $$AB$$ respectively. The area of $$\triangle ABC$$ is $$24\ cm^2,$$ then the area of $$\triangle DEF$$ is

A
$$7\ cm^2$$

B
$$6\ cm^2$$

C
$$18\ cm^2$$

D
$$16\ cm^2$$

Solution

In $$\triangle ABC$$,
$$D$$ and $$E$$ are midpoints of sides $$BC$$ and $$AC$$ respectively.
By midpoint theorem,
$$DE = \dfrac 12 AB$$ and $$DE \parallel AB$$ $$...(1)$$
In $$\triangle CAB$$ and $$\triangle CED$$,
$$\angle C$$ is the common angle.
$$\angle CAB = \angle CED$$ $$[$$alternate angles since $$DE \parallel AB]$$
$$\therefore \triangle CAB \sim \triangle CED$$ $$[$$AA test of similarity$$]$$
$$\therefore \dfrac {CA}{CE} = \dfrac {AB}{DE} = \dfrac {BC}{DC} = \dfrac 21$$ $$...(2)$$ [C.S.S.T and from (1)]

Similarly, we can prove $$\triangle ABC \sim \triangle AFE \sim \triangle FBD$$ [AA test of similarity]
Hence, $$\dfrac {EF}{BC} = \dfrac {DE}{AB} = \dfrac {DF}{AC} = \dfrac 12$$ ....C.S.S.T
$$\therefore \triangle ABC \sim \triangle DEF$$ [SSS test of similarity]
By theorem on ratio of areas of similar triangles, we get
$$\Rightarrow \dfrac {A(\triangle DEF)}{A(\triangle ABC)} = \left(\dfrac {DE}{AB}\right)^2$$
$$\Rightarrow A(\triangle DEF) = \dfrac 14 \times 24 $$
$$= 6\ cm^2$$
Question 5
1 / -0

If the ratio of the corresponding sides of two similar triangles is $$2:3$$ then the ratio of their corresponding altitude is

A
$$3 : 5$$

B
$$16 : 81$$

C
$$4 : 9$$

D
$$2 : 3$$

Solution

If two triangles are similar ,then the ratio of their corresponding sides and altitude are also same. Therefore the ratio of the altitude is $$2:3.$$
Question 6
1 / -0

Two poles of height $$10$$ m and $$15$$ m, stand on a plane ground. If the distance between their feet is $$12$$m, the distance between their tops is:

A
$$12$$ m

B
$$13$$ m

C
$$12.5$$ m

D
$$13.5$$ m

Solution

Since the two poles of height $$10$$ m and $$15$$ m stand on a plane ground, therefore, their difference will be the perpendicular of the plane that is $$15 - 10 = 5$$ m.

It is given that the distance between their feet is $$12$$ m which means the base $$= 12$$ m.

Since we know the base and height, we can calculate the hypotenuse $$h$$, that is:

$${ h }^{ 2 }={ 5 }^{ 2 }+{ 12 }^{ 2 }$$

$$\Rightarrow $$$${ h }^{ 2 }=25+144$$

$$\Rightarrow $$$${ h }^{ 2 }=169$$

$$\Rightarrow $$$${ h }=\sqrt { 169 } $$

$$\Rightarrow $$$${ h }=13$$ m

So the distance between their tops is $$13$$ m.
Question 7
1 / -0

Two isosceles triangles have equal vertical angles and their areas are in the ratio of 9 : 16. Then their heights are in the ratio of:

A
9 : 16

B
16 : 9

C
4 : 3

D
3 : 4

Solution

Let the two isosceles triangles be $$\triangle ABC$$ and $$\triangle DEF$$.
Given: $$\angle A = \angle D$$       ....vertical angles of triangles
$$\dfrac {AB}{DE} = \dfrac {BC}{EF}$$
So, $$\triangle ABC \sim \triangle DEF$$        ....S.A.S test of similarity
Hence, $$\dfrac {A(\triangle ABC)}{A(\triangle DEF)} = \left(\dfrac {h_1}{h_2}\right)^2$$
where, $$h_1$$ and $$h_2$$ are heights of the two triangles respectively.
$$\dfrac {9}{16} = \dfrac {(h_1)^2}{(h_2)^2}$$
$$\dfrac {h_1}{h_2} = \dfrac 34$$.
Question 8
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P and Q are the mid points of the sides AB and BC respectively of the triangle ABC, right-angled at B, then:

A
$$AQ^2+CP^2=AC^2$$

B
$$\displaystyle AQ^2+CP^2=\frac{3}{5}AC^2$$

C
$$\displaystyle AQ^2+CP^2 =\frac{4}{5}AC^2$$

D
$$\displaystyle AQ^2+CP^2=\frac{5}{4}AC^2$$

Solution

$$AQ^2=AB^2+BQ^2$$

since Q is midpoint on $$BC$$, $$BQ=\dfrac{BC}{2}$$

$$\displaystyle AQ^2=AB^2+\left(\frac{BC}{2}\right)^2$$

$$\displaystyle=AB^2+\frac{BC^2}{4}$$ ...(i)

Similarly $$CP^2=BC^2+BP^2$$

since P is midpoint on AB, $$BP =\dfrac{AB}{2}$$

$$\displaystyle CP^2=BC^2+\left(\frac{AB}{2}\right)^2$$

$$\displaystyle\ \ \ \ \ \ \ =\frac{AB^2}{4}+BC^2$$ ...(ii)

$$\displaystyle\therefore AQ^2+CP^2= AB^2+\frac{BC^2}{4}+\frac{AB^2}{4}+BC^2$$ ( from equations (i) and (ii) )
$$ \displaystyle =AB^2\left(1+\frac{1}{4}\right)+BC^2\left(1+\frac{1}{4}\right)$$

$$\displaystyle=\frac{5}{4}(AB^2+BC^2)=\frac{5}{4}AC^2$$
Question 9
1 / -0

If two triangles are similar then, ratio of corresponding sides are:

A
unequal

B
equal

C
zero

D
none of these

Solution

Similar triangles have $$:$$
$$i)$$ All their angles equal
$$ii)$$ Corresponding sides have the same ratio
So, option $$B$$ is correct.
Question 10
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In a $$\triangle ABC$$, a straight line parallel to $$BC$$ intersects $$AB$$ and $$AC$$ at point D and E respectively. If the area of $$\triangle ADE$$ is one-fifth of the area of $$\triangle ABC$$ and $$BC = 10$$ cm, then DE equals:

A
$$2$$ cm

B
$$ 2 \sqrt5$$ cm

C
$$4$$ cm

D
$$4 \sqrt5$$ cm

Solution

$$\displaystyle \frac{\mbox {Area of} \Delta ADE}{\mbox {Area of} \Delta ABC} = \frac{DE^2}{BC^2}$$
$$\displaystyle \frac{\displaystyle \frac{1}{5} \times \mbox{Area of} \Delta ABC}{\mbox {Area of } \Delta ABC} = \frac{DE^2}{10^2}$$
$$\displaystyle \frac{1}{5} = \frac{DE^2}{100}$$
$$\displaystyle DE^2 = \frac{100}{5} = 20$$
$$\therefore DE = \sqrt{20} = 2 \sqrt 5 cm$$