The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 16

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Weekly Quiz Competition

Question 1
1 / -0

In $$\Delta ABC, BC=AB$$ and $$\angle B=80^{o}.$$ Then $$\angle A$$ is equal to

A
$$80^{o}$$

B
$$40^{o}$$

C
$$50^{o}$$

D
$$100^{o}$$

Solution

Given, $$BC=AB$$ and $$\angle B=80^o$$
In $$\triangle ABC$$, we have
$$\angle A+ \angle B+ \angle C=180$$.....(1) (Sum of all the angles in triangle is $$180^o$$)
$$BC=AB$$
$$\Rightarrow \angle A= \angle C$$ .......... [Angles opposite to equal sides are equal]
and $$\angle B= 80^o$$.....(2)
From (1) and (2), we get
$$2\angle A= 100$$
$$\therefore \angle A= 50^{o}$$
Question 2
1 / -0

In $$\Delta PQR, \angle R=\angle P$$ and $$QR$$ $$=4$$ cm and $$PR$$ $$= 5$$ cm. then what is the length of $$PQ$$ ?

A
$$4$$ cm

B
$$5$$ cm

C
$$2$$ cm

D
$$2.5$$ cm

Solution

Given, $$\angle R= \angle P$$ and $$QR=4$$ cm, $$PR=5$$ cm
In an isosceles triangle two sides are congruent and their base angles are also congruent.
Here, $$\angle R=\angle P$$
$$\therefore QR=PQ$$.
It is given that $$QR=4$$cm, therefore $$PQ=4$$ cm.
Question 3
1 / -0

Two sides of a triangle are of lengths $$4$$ cm and $$1.5$$ cm. The length of the third side of the triangle cannot be

A
$$3.6$$ cm

B
$$4.1$$ cm

C
$$3.8$$ cm

D
$$5.8$$ cm

Solution

Sum of given sides of the triangle is $$5.5$$ cm.
The length of the third side cannot be greater than the sum of the two given sides.
Here, one option gives length as $$5.8$$ cm is not possible.
Question 4
1 / -0

The construction of a triangle $$ABC$$, given that $$BC =$$ $$6$$ cm, $$B =$$ $$45 ^{\circ}$$ is not possible when difference of $$AB$$ and $$AC$$ is equal to:

A
$$6.9$$ cm

B
$$5.2$$ cm

C
$$5.0$$ cm

D
$$4.0$$ cm

Solution

According to the theorem of inequalities, the sum of any two sides of the triangle is greater than the third side.
Therefore, $$AC+BC>AB$$
$$\Rightarrow BC>AB-AC$$
Therefore, only the first option that is $$6.9$$ cm does not satisfy the above equation. Rest all the options satisfy the equation.
Question 5
1 / -0

If lengths of sides of a triangle are 6, 10, 8, then ABC is .........triangle.

A
isosceles

B
equilateral

C
right angled

D
none of these
Question 6
1 / -0

Which of the following is NOT the length of a

median in triangle ABC with vertices A(1,3), B(1,1) and C(5,1)?

A
$$2\sqrt{6}$$

B
$$\sqrt{26}$$

C
$$2\sqrt{5}$$

D
$$\sqrt{2}$$
Question 7
1 / -0

From the adjoining figure, calculate the values of $$a$$.

A
$$46^o$$

B
$$38^o$$

C
$$56^o$$

D
none of the above

Solution

We know, by angle sum property, the sum of angles of triangle $$= 180^o$$
Then, $$a + 110^o + 32^o = 180^o $$
$$\implies$$ $$a + 142^o = 180^o $$
$$a =180^o-142^o=38^{\circ}$$.

Thus measure of angle $$a$$ is $$38^o$$.

Hence, option $$B$$ is correct.
Question 8
1 / -0

In $$\Delta$$ ABC, $$\angle B = 90 ^\circ$$, AB = 6 cm and BC = 8 cm. The length of AC is

A
3 cm

B
10 cm

C
4 cm

D
7 cm

Solution

Given that in the $$\triangle ABC, \angle B = 90 ^\circ $$
$$\therefore \triangle ABC$$ is a right angle triangle.
So we can apply pythagoras theorem here.
$$\implies AC^2 = AB^2 + BC^2$$
$$\therefore $$Side, AC = $$\sqrt{(6^2 + 8^2)}$$ = 10 cm
Question 9
1 / -0

If length of the largest side of a triangle is 12 cm then other two sides of triangle can be

A
4.8 cm, 8.2 cm

B
3.2 cm, 7.8 cm

C
6.4 cm, 2.8 cm

D
7.6 cm, 3.4 cm

Solution

Sum of any two sides of a triangle is greater than the third side.
Here the sum must be greater than $$12\ \ cm$$
In option $$A$$
$$4.8\ \ cm+8.2\ \ cm=13\ \ cm$$
$$\Rightarrow 13\ \ cm>12 \ \ cm$$
In rest of the options sum is less than $$12\ \ cm$$
So option $$A$$ is correct.
Question 10
1 / -0

In the given figure, $$ABC$$ is a triangle, side $$CB$$ is produced to $$E$$ and $$\angle A$$: $$\angle B$$: $$\angle C$$ $$= 2: 1: 3$$. Find $$\angle DBE$$, if $$DB$$ is perpendicular to $$AB$$.

A
$$66^o$$

B
$$46^o$$

C
$$60^o$$

D
none of the above

Solution

In triangle $$ABC$$, $$\angle A : \angle B : \angle C = 2: 1: 3$$
Let the angles be $$2x, x, 3x$$
By angle sum property, the sum of the angles $$2x + x+ 3x = 180^o$$
$$\implies$$ $$6x = 180^o$$
$$\implies$$ $$x = 30^o$$.

Hence, $$\angle B = 30^o$$.

Now,
$$\angle DBE + \angle DBA + \angle ABC = 180^o$$ ...[Straight angle property]
$$\angle DBE + 90^o + 30^o = 180^o $$
$$\angle DBE = 180^o - 120^o $$
$$\angle DBE = 60^{\circ}$$.

Therefore, option $$C$$ is correct.

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Re-Attempt Weekly Quiz Competition

The Triangle and Its Properties Test - 16

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The Triangle and Its Properties Test - 16

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Question 1

$$80^{o}$$

$$40^{o}$$

$$50^{o}$$

$$100^{o}$$

Solution

Question 2

$$4$$ cm

$$5$$ cm

$$2$$ cm

$$2.5$$ cm

Solution

Question 3

$$3.6$$ cm

$$4.1$$ cm

$$3.8$$ cm

$$5.8$$ cm

Solution

Question 4

$$6.9$$ cm

$$5.2$$ cm

$$5.0$$ cm

$$4.0$$ cm

Solution

Question 5

isosceles

equilateral

right angled

none of these

Question 6

$$2\sqrt{6}$$

$$\sqrt{26}$$

$$2\sqrt{5}$$

$$\sqrt{2}$$

Question 7

$$46^o$$

$$38^o$$

$$56^o$$

none of the above

Solution

Question 8

3 cm

10 cm

4 cm

7 cm

Solution

Question 9

4.8 cm, 8.2 cm

3.2 cm, 7.8 cm

6.4 cm, 2.8 cm

7.6 cm, 3.4 cm

Solution

Question 10

$$66^o$$

$$46^o$$

$$60^o$$

none of the above

Solution

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