Triangles Test

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Triangles Test - 16

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Question 1
1 / -0

In the above figure, if DE || BC, then x equals :

A
6 cm

B
7 cm

C
3 cm

D
4 cm

Solution

In $$\triangle ADE$$ and $$\triangle ABC$$

$$\angle ADE=\angle ABC$$ (As $$DE\parallel BC$$ Corresponding angles)

$$\angle AED=\angle ACB$$ (As $$DE\parallel BC$$ Corresponding angles)
$$\angle DAE=\angle BAC$$ (common angle)

$$\therefore \triangle ADE\sim \triangle ABC $$ by AAA similarity.
If two triangles are similar, then their corresponding sides are proportional.
$$\ \Rightarrow \dfrac { AD }{ AB } =\dfrac { DE }{ BC } \\ \ \;\;\dfrac { 3 }{ 7 } =\dfrac { x }{ 14 } \\ \;\;\;x =\frac{3\times 14}{7}\\ \Rightarrow x=6$$
Question 2
1 / -0

Which of the following cannot be the sides a right angle triangle?

A
9 cm, 15 cm, l2cm

B
2 cm, 1 cm, $$\sqrt{5}$$ cm

C
400 mm, 300 mm, 500 mm

D
9 cm, 5 cm, 7 cm

Solution

In option $$D$$ Pythagoras theorem is not satisfied , So the sides can not be sides of right angled triangle
$${ 9 }^{ 2 }\neq { 5 }^{ 2 }+{ 7 }^{ 2 }\\ 81\neq 25+49\\ 81\neq 74$$
So option $$D$$ is correct.
Question 3
1 / -0

In the given figure, the value of x (in cm) is :

A
4

B
5

C
6

D
8

Solution

$$\dfrac { PA }{ PQ } =\dfrac { 2.4 }{ 6 } =\dfrac { 2 }{ 5 } \\ \dfrac { PB }{ PR } =\dfrac { 3.2 }{ 8 } =\dfrac { 2 }{ 5 } \\ \Rightarrow \dfrac { PA }{ PQ } =\dfrac { PB }{ PR } $$

$$\angle APB=\angle QPR$$
So, By SAS property

$$\therefore \triangle APB\sim \triangle QPR \\ \Rightarrow \dfrac { PA }{ PQ } =\dfrac { PB }{ PR } =\dfrac { AB }{ QR } \\ \dfrac { 2 }{ 5 } =\dfrac { 2 }{ x } \\ \Rightarrow x=5cm$$
Question 4
1 / -0

In the figure given above, if DE $$ \parallel$$ BC, then the value of x equals to:

A
3 cm

B
2 cm

C
4 cm

D
6.7 cm

Solution

In $$\triangle ADE$$ and $$\triangle ABC$$
$$\angle ADE=\angle ABC$$ (As $$DE\parallel BC$$) Corresponding angles
$$\angle AED=\angle ACB$$ (As $$DE\parallel BC$$) Corresponding angles
$$\angle DAE=\angle BAC$$ (common angle)
$$\therefore \triangle ADE\sim \triangle ABC $$ by AAA rule
$$\therefore \dfrac{AD}{AB} = \dfrac{DE}{BC}$$
$$\therefore \dfrac { 3 }{ 2+3 } =\dfrac { 4 }{ x } \\ \Rightarrow x=6.7cm$$
Question 5
1 / -0

In $$\Delta ABC,$$ if $$AB =6\sqrt{3}$$ cm, $$AC=12$$ cm and $$BC=6$$ cm, then angle B is equal to:

A
$$120^{0}$$

B
$$90^{0}$$

C
$$45^{0}$$

D
$$60^{0}$$

Solution

$$AC$$ is the longest side
$$AC=12\\ \Rightarrow { \left( AC \right) }^{ 2 }=144\\ AB=6\sqrt { 3 } ,BC=6\\ \Rightarrow { \left( AB \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }={ \left( 6\sqrt { 3 } \right) }^{ 2 }+{ 6 }^{ 2 }=108+36=144\\ \Rightarrow { \left( AC \right) }^{ 2 }={ \left( AB \right) }^{ 2 }+{ \left( BC \right) }^{ 2 }$$
We can see that pythagoras theorem is applicable in the given triangle.
So, by converse of pythagoras theorem, we have $$ \angle B=90^{\circ}$$ as $$\angle B$$ is opposite to the longest side $$AC$$.
Question 6
1 / -0

The areas of two similar triangles ABC and PQR are $$25\ cm^{2}\ \& \ 49\ cm^{2}$$, respectively. If QR $$=9.8$$ cm, then BC is:

A
9.8 cm

B
7 cm

C
49 cm

D
25 cm

Solution

Given: $$\dfrac { ar(ABC) }{ ar(PQR) } =\dfrac { 25 }{ 49 } $$

In two similar triangles, the ratio of their areas will be equal to the square of the ratio of their sides.

Hence in $$\triangle ABC$$ and $$\triangle PQR,$$
$$\Rightarrow {\left( \dfrac{BC}{QR} \right)}^2=\dfrac{ar \triangle ABC}{ar \triangle PQR}$$
$$\Rightarrow { \left( \dfrac { BC }{ QR } \right) }^{ 2 }=\dfrac { 25 }{ 49 }$$
$$\Rightarrow \,\,\,\,\,\,\,\,\,\,\dfrac { BC }{ QR } =\dfrac { 5 }{ 7 }$$
$$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\dfrac { BC }{ 9.8 } =\dfrac { 5 }{ 7 }$$
$$\Rightarrow\,\,\,\,\,\,\,\,\,\,\,\,\, BC=\dfrac { 5 }{ 7 } \times 9.8=7$$
Question 7
1 / -0

Let $$\triangle$$ABC ~ $$\triangle$$PQR. If area(ABC) = 2.25 $$m^{2}$$, area(PQR) = 6.25 $$m^{2}$$, PQ = 0.5 $$m$$, then length of AB is:

A
30 cm

B
0.5 m

C
50 cm

D
3 m

Solution

$$\dfrac { ar(ABC) }{ ar(PQR) } =\dfrac { 2.25 }{ 6.25 } $$

In two similar triangles, the ratio of their areas is the square of the ratio of their sides

$$\Rightarrow { \left( \dfrac { AB }{ PQ } \right) }^{ 2 }=\dfrac { 225 }{ 625 } \\ \Rightarrow \dfrac { AB }{ PQ } =\dfrac { 15 }{ 25 } \\ \Rightarrow \dfrac { AB }{0 .5 } =\dfrac { 3 }{ 5 } \\ \Rightarrow AB=0.3m\\ \Rightarrow AB=0.3\times 100=30cm$$
Question 8
1 / -0

In figure, DE||BC, then the value of x equals to:

A
2.5 cm

B
2 cm

C
1.4 cm

D
4 cm

Solution

Given that,
$$DE\parallel BC$$.
To find out,
The value of $$x$$.
In $$\triangle ADE$$ and $$\triangle ABC$$
$$\angle ADE=\angle ABC$$ [Corresponding angles]
$$\angle AED=\angle ACB$$ [Corresponding angles]
$$\angle DAE=\angle BAC$$ (common angle)
Hence, by AAA criteria
$$\triangle ADE\sim \triangle ABC\\ \Rightarrow \dfrac { AD }{ AB } =\dfrac { AE }{ AC } \\ \dfrac { 1.5 }{ 4.5 } =\dfrac { 1 }{ 1+x } \\ 1+x=3$$
$$x=2$$.

Hence, the value of $$x$$ is $$2$$.
Question 9
1 / -0

In figure , if $$DE\parallel BC$$, then the value of x is equal to

A
3 cm

B
4 cm

C
7 cm

D
4.7 cm

Solution

In $$\triangle ADE$$ and $$\triangle ABC$$

$$\angle ADE=\angle ABC$$ (As $$DE\parallel BC$$) Corresponding angles
$$\angle AED=\angle ACB$$ (As $$DE\parallel BC$$) Corresponding angles
$$\angle DAE=\angle BAC$$ (common angle)
$$\therefore \triangle ADE\sim \triangle ABC $$ by AAA
$$ \Rightarrow \dfrac { 3 }{ 7 } =\dfrac { 2 }{ x } \\ \Rightarrow x=4.7cm$$
Question 10
1 / -0

In a $$\Delta$$PQR, right angled at Q, PQ$$ =$$ 24 cm and QR$$ =$$ 7 cm. S is the mid point of PR, then RS is :

A
3.5cm

B
12cm

C
25cm

D
12.5cm

Solution

In $$\triangle PQR$$
$${ \left( PQ \right) }^{ 2 }+{ \left( QR \right) }^{ 2 }={ { \left( PR \right) } }^{ 2 }\\ { \left( 24 \right) }^{ 2 }+{ \left( 7 \right) }^{ 2 }={ { \left( PR \right) } }^{ 2 }\\ \Rightarrow { { \left( PR \right) } }^{ 2 }=576+49=625\\ \Rightarrow PR=25$$
Now $$S$$ is the mid point of $$PR$$
$$\therefore RS=\dfrac { PR }{ 2 } =\dfrac { 25 }{ 2 } =12.5cm$$