Triangles Test

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Triangles Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

If straight-line divides two sides of a triangle proportionally then the straight line is parallel to the third side is called

A
converse of thales theorem

B
pythogoral theorem

C
thales theorem

D
converse of Pythagoras theorem

Solution

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

Converse of Thales theorem states that 'If a straight line divides two sides of triangle in equal ratio, then it is parallel to third side of triangle'
Question 2
1 / -0

The corresponding altitudes of two similar triangles are $$9\,cm$$ and $$15\,cm$$ respectively then the ratio between their areas.

A
$$9 : 15$$

B
$$100 : 25$$

C
$$81 : 225$$

D
$$3 : 15$$

Solution

Let $${ h }_{ 1 }=9\ cm$$
$${ h }_{ 2 }=15\ cm$$

Ratio of areas of two similar triangles is given by,
$$\dfrac { { A }_{ 1 } }{ { A }_{ 2 } } ={ \left( \dfrac { { h }_{ 1 } }{ { h }_{ 2 } } \right) }^{ 2 }$$

$$\dfrac { { A }_{ 1 } }{ { A }_{ 2 } } ={ \left( \dfrac { { h }_{ 1 } }{ { h }_{ 2 } } \right) }^{ 2 }$$

$$\therefore \dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { 81 }{ 225 } $$

$$\therefore { A }_{ 1 }:{ A }_{ 2 }=81:225$$
Question 3
1 / -0

Area of similar triangles are proportional to

A
squares on their corresponding sides

B
squares on their corresponding altitude

C
squares on their corresponding medians

D
all the above

Solution

Ratio of areas of two similar triangles is given by-
1) Ratio of squares of their corresponding sides
2) Ratio of squares of their corresponding altitudes
3) Ratio of squares of their corresponding medians
Question 4
1 / -0

In figure, $$DE\parallel BC$$, $$AD=4cm,DB=6cm$$ and $$AE=5cm$$ then $$EC$$ will be:

A
$$6.5cm$$

B
$$7.0cm$$

C
$$7.5cm$$

D
$$8.0cm$$

Solution

Given,
$$AD=4cm$$
$$DB=6cm$$
$$AE=5cm$$

In $$\triangle ABC$$
$$DE\parallel BC$$
$$\therefore \cfrac{AD}{DB}=\cfrac{AE}{EC}$$ (By basic prop.theorem)
$$\Rightarrow$$ $$\cfrac{4}{6}=\cfrac{5}{EC}$$
$$\Rightarrow$$ $$EC=\cfrac{6\times 5}{4}$$
$$\Rightarrow$$ $$EC=7.5$$

Hence, the length of $$EC$$ is $$7.5\ cm$$
Question 5
1 / -0

If in $$\triangle ABC$$ and $$\triangle DEF$$, $$\angle A={50}^{o}$$, $$\angle {B}={70}^{o}$$, $$\angle {C}={60}^{o}$$, $$\angle D={60}^{o}$$, $$\angle E={70}^{o}$$ and $$\angle F={50}^{o}$$ then which one is correct?

A
$$\triangle ABC\sim \triangle DEF$$

B
$$\triangle ABC\sim \triangle EDF$$

C
$$\triangle ABC\sim \triangle DEF$$

D
$$\triangle ABC\sim \triangle FED$$

Solution

$$\angle A=\angle F={50}^{o}$$
$$\angle B=\angle E={70}^{o}$$
$$\angle C=\angle D={60}^{o}$$
$$\triangle ABC\sim \triangle FED$$ BY AAA
Question 6
1 / -0

If $$\triangle ABC\sim \triangle DEF$$ and $$DE=8cm,AB=10cm$$ then the Ratio of $$ar(\triangle ABC)$$ to $$ar(\triangle DEF)=$$

A
$$25:16$$

B
$$16:25$$

C
$$4:5$$

D
$$5:4$$

Solution

We know that if $$\triangle ABC\sim \triangle DEF$$ then,

$$\cfrac { ar.\left( \triangle ABC \right) }{ ar.\left( \triangle DEF \right) } =\cfrac { { \left( AB \right) }^{ 2 } }{ { \left( DE \right) }^{ 2 } } $$

$$={ \left( \cfrac { 10 }{ 8 } \right) }^{ 2 }=\cfrac{25}{16}$$

$$ar(\triangle ABC):ar(\triangle DEF)=25:16$$
Question 7
1 / -0

$$\triangle ABC$$ and $$\triangle PQR$$ are two similar triangles whose areas are $$100{cm}^{2}$$ and $$144{cm}^{2}$$ and height of $$\triangle ABC$$ is $$6cm$$ then height of $$\triangle PQR$$ will be:

A
$$12cm$$

B
$$6.3cm$$

C
$$7.2cm$$

D
$$4.8cm$$

Solution

We know that ratio of areas of two similar triangles is equal to ratio of square of corresponding heights. Let corresponding height of $$\triangle PQR$$ is $$x\,cm$$
$$\cfrac { ar.\quad \triangle ABC }{ ar.\quad \triangle PQR } =\cfrac { (corresponding\quad height\quad of\quad \triangle ABC)^2}{ (corresponding\quad height\quad of\quad \triangle PQR )^2}$$
$$\Rightarrow \cfrac { 100 }{ 144 } =\cfrac { { 6 }^{ 2 } }{ { x }^{ 2 } } $$
$$\Rightarrow$$ $$\cfrac{100}{144}=\cfrac{36}{{x}^{2}}$$
$$\Rightarrow$$ $$100{x}^{2}=36\times 144$$
$$\Rightarrow$$ $${x}^{2}=\cfrac{36\times 144}{100}$$
$$\Rightarrow$$ $$x=\sqrt { \cfrac { 36\times 144 }{ 100 } } =\cfrac { 6\times 12 }{ 10 } =\cfrac { 72 }{ 10 } =7.2cm$$
Question 8
1 / -0

The point $$D$$ and $$E$$ lies on sides $$AB$$ and $$AC$$ of $$\triangle ABC$$ such that $$DE\parallel BC$$ and $$AD=8cm,AB=12cm$$ and $$AE=12cm$$, then $$CE$$ will be:

A
$$6cm$$

B
$$18cm$$

C
$$9cm$$

D
$$15cm$$

Solution

In $$\triangle ABC$$
$$DE\parallel BC$$
$$\cfrac{AD}{BD}=\cfrac{AE}{CE}$$ (Basic prop. theorem)
$$\cfrac{AD}{AB-AD}=\cfrac{AE}{CE}$$
$$\Rightarrow$$ $$\cfrac{8}{12-8}=\cfrac{12}{CE}$$
$$\Rightarrow$$ $$\cfrac{8}{4}=\cfrac{12}{CE}$$
$$\Rightarrow$$ $$8\times CE=12\times 4$$
$$\Rightarrow$$ $$CE=\cfrac{12\times 4}{8}$$
$$\Rightarrow$$ $$CE=6cm$$
Question 9
1 / -0

In figure, if $$DE\parallel AB$$ then $$AD=2cm$$, $$DC=3cm$$ and $$BE=3cm$$ and then value of $$CE$$

A
$$4.5cm$$

B
$$2cm$$

C
$$18cm$$

D
$$5cm$$

Solution

Since $$DE\parallel AB$$, according to question here $$AD=2cm, DC=3cm$$ and $$BE=3cm$$
Thus by basic Prop. Theorem
$$\cfrac{CD}{AD}=\cfrac{CE}{BE}$$
$$\Rightarrow$$ $$CE=\cfrac{BE\times DC}{AD}$$
$$CE=\cfrac{3\times 3}{2}=4.5cm$$
Question 10
1 / -0

In $$\triangle \text{ABC}$$ if $$\text{AB=5 cm, BC=12 cm}$$ and $$\text{AC=13 cm}$$, then $$\angle \text{B}$$ will be:

A
$${120}^{\circ}$$

B
$${60}^{\circ}$$

C
$${90}^{\circ}$$

D
$${45}^{\circ}$$

Solution

In $$\triangle ABC$$,
$${(AB)}^{2}+{(BC)}^{2}={(5)}^{2}+{(12)}^{2}$$
$$=25+144=169=13^2=(AC)^2$$
$$\implies AC=13$$
The converse of the Pythagoras theorem states that if the square of the longer side of a triangle is equivalent to the sum of its two shorter sides, then it must be a right triangle.
$$\therefore ~~\triangle ABC$$ is right angle triangle, in which $$\angle B={90}^{\circ}$$