Triangles Test

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Triangles Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

If ABC and DEF are similar triangles such that $$\displaystyle \angle A=47^{\circ}$$ and $$\displaystyle \angle B=83^{\circ}$$ then $$\displaystyle \angle F$$ is

A
$$\displaystyle 60^{\circ}$$

B
$$\displaystyle 70^{\circ}$$

C
$$\displaystyle 50^{\circ}$$

D
$$\displaystyle 100^{\circ}$$

Solution

Since triangle ABC and DEF are similar
$$\therefore \angle A=\angle D,\angle B=\angle E and \angle C=\angle F$$
In $$\triangle ABC$$
$$\angle A+\angle B+\angle C=180^\circ$$
$$47^\circ+83^\circ+\angle C=180^\circ$$
$$\angle C=180^\circ-130^\circ$$
$$\angle C=50^\circ$$
$$\therefore \angle F=50^\circ$$
Question 2
1 / -0

In the given figure, $$DE\parallel BC$$, $$AD=2\:cm$$, $$DB=3\:cm$$ and $$AE=1.6\:cm$$. Then EC is equal to

A
1.2 cm

B
2.5 cm

C
2.4 cm

D
4.8 cm

Solution

Given,
$$DE\parallel BC$$
AD=2cm, DB=3cm, AE=1.6cm
as per problem,
$$\frac { AD }{ DB } =\frac { AE }{ EC } $$
$$\frac { 2 }{ 3 } =\frac { 1.6 }{ EC } $$
$$2EC=4.8cm$$
$$EC=2.4cm$$
Question 3
1 / -0

In $$\triangle{ABC}$$, $$\angle{B}=90$$, $$AB=8\:cm$$ and $$BC=6\:cm$$. The length of the median BM is

A
3 cm

B
5 cm

C
4 cm

D
7 cm

Solution

$$AC^2=AB^2+BC^2$$ (because $$\angle{B}=90^\circ$$)
$$=64+36=100$$
$$\therefore AC=10$$

But $$\displaystyle BM=\frac{1}{2}AC=\frac{10}{2}=5\:cm$$
Question 4
1 / -0

Which one of the four hexagons is not similar to the other three?

A
A

B
B

C
C

D
D

Solution

Two shapes are similar if the only difference is size (and possibly the need to turn or flip one around).
Hexagon B is similar to Hexagon D because it has the same shape but it has been rotated by $$\displaystyle { 90 }^{ \circ }$$
Hexagon C is similar to Hexagon D because it has the same shape but a different size.
Only Hexagon A has a different shape to the other three.
Question 5
1 / -0

The areas of two similar triangles are $$\displaystyle 9\ { cm }^{ 2 }$$ and $$\displaystyle 16\ { cm }^{ 2 }$$, respectively. The ratio of their corresponding heights is:

A
$$3 : 4$$

B
$$4 : 3$$

C
$$2 : 3$$

D
$$4 : 5$$

Solution

We know that in similar triangles: -

$${\dfrac{{h_1^2}}{{h_2^2}}} = \dfrac{{S1}}{{S2}}$$

Where $$h_1$$ and $$h_2$$ are the heights of the triangles and $$S1, S2$$ are the areas of similar traingles.

$${{\rm{}}\dfrac{{h_1^2}}{{h_2^2}}} = \dfrac{{9c{m^2}}}{{16c{m^2}}}$$

$$\dfrac{{h_1}}{{h_2}} = \sqrt {\dfrac{9}{{16}}} $$

$$\dfrac{{h_1}}{{h_2}} = \dfrac{3}{4}\ or\ {3:4}$$
Question 6
1 / -0

Two poles of heights 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

Let us draw a perpendicular from $$B$$ on $$CD$$ which meets $$CD$$ at $$P$$.
It is clear that $$BP = 12 \space\mathrm{m}$$ because it is given that distance between feet of the two poles is $$12\space\mathrm{m}$$.
After drawing the perpendicular we get a rectangle $$BACP$$ such that $$AB = PC$$ and $$BP = AC$$.
Because of this construction we also obtained a right angled triangle $$BPD$$.
Now we will use Pythagoras theorem,
$$BD^2=BP^2+DP^2$$
Let us substitute the values of $$BP$$ and $$PD$$ we get,
$$BD^2=12^2+5^2\quad[\because DP=DC-PC=15-10=5\space\mathrm{m}]$$
$$\Rightarrow BD^2=144+25$$
$$\Rightarrow BD^2=169$$
$$\Rightarrow BD=13$$
Therefore, distance between the top of the two poles is $$13\space\mathrm{m}$$.
So, $$\text{B}$$ is correct option.
Question 7
1 / -0

Which one of the four pentagons is not similar to the other three?

A
A

B
B

C
C

D
D

Solution

Two shapes are similar if the only difference is size (and possibly the need to turn or flip one around).
Pentagon A is similar to Pentagon B because it has the same shape but has been rotated.
Pentagon C is similar to Pentagon B because it has the same shape but a different size.
Only Pentagon D has a different shape to the other three.
Question 8
1 / -0

When the ratios of the lengths of their corresponding sides are equal, then the two figures are:

A
similar

B
congruent

C
equal

D
none of these

Solution

When the ratios of the lengths of their corresponding sides are equal, then the two figures are then those figure are similar
Question 9
1 / -0

A............ can never be made up of all odd numbers or two even numbers and one odd number.

A
Pythagorean Double

B
Pythagorean Triple

C
Pythagorean single

D
None

Solution

Because:
The square of an odd number is an odd number and the square of an even number is an even number.The sum of two even numbers is an even number and the sum of an odd number and an even number is in odd number.
Therefore, B is the correct answer.
Question 10
1 / -0

The Pythagoras theorem , In the right triangle, the square of the hypotenuse is equal to the sum of other two sides. What are we proving here?

A
$$\displaystyle { a }^{ 2 }\div { b }^{ 2 }={ c }^{ 2 }$$

B
$$\displaystyle { a }^{ 2 }-{ b }^{ 2 }={ c }^{ 2 }$$

C
$$\displaystyle { a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }$$

D
$$\displaystyle { a }^{ 2 }x{ b }^{ 2 }={ c }^{ 2 }$$

Solution

Let $$a,b$$ and $$c$$ be the sides of triangle with $$c$$ as hypotenuse.

Then according to Pythagoras theorem,

$$a^2+b^2=c^2$$

Option $$C$$ is correct.

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Re-Attempt Weekly Quiz Competition

Triangles Test - 20

Result

Triangles Test - 20

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SHARING IS CARING

Question 1

$$\displaystyle 60^{\circ}$$

$$\displaystyle 70^{\circ}$$

$$\displaystyle 50^{\circ}$$

$$\displaystyle 100^{\circ}$$

Solution

Question 2

1.2 cm

2.5 cm

2.4 cm

4.8 cm

Solution

Question 3

3 cm

5 cm

4 cm

7 cm

Solution

Question 4

A

B

C

D

Solution

Question 5

$$3 : 4$$

$$4 : 3$$

$$2 : 3$$

$$4 : 5$$

Solution

Question 6

12 m

13 m

12.5 m

13.5 m

Solution

Question 7

A

B

C

D

Solution

Question 8

similar

congruent

equal

none of these

Solution

Question 9

Pythagorean Double

Pythagorean Triple

Pythagorean single

None

Solution

Question 10

$$\displaystyle { a }^{ 2 }\div { b }^{ 2 }={ c }^{ 2 }$$

$$\displaystyle { a }^{ 2 }-{ b }^{ 2 }={ c }^{ 2 }$$

$$\displaystyle { a }^{ 2 }+{ b }^{ 2 }={ c }^{ 2 }$$

$$\displaystyle { a }^{ 2 }x{ b }^{ 2 }={ c }^{ 2 }$$

Solution

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