The Triangle and Its Properties Test

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The Triangle and Its Properties Test - 18

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Question 1
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In figure, sides QP and RQ of $$\Delta PQR$$ are produced to points S and T respectively. If $$\angle SPR=135^o$$ and $$\angle PQT=110^o$$, find $$\angle PRQ$$.

A
$$85^o$$

B
$$65^o$$

C
$$45^o$$

D
$$35^o$$

Solution

Here, $$\angle QPR+135^o=180^o$$ ....[Linear pair]

$$\Rightarrow \angle QPR=45^o$$.

Now, $$\angle QRP+\angle QPR=\angle PQT$$ ....[Exterior angle property]

$$\Rightarrow \angle QRP+45^o=110^o$$

$$\Rightarrow \angle QRP=65^o$$.

Therefore, option $$B$$ is correct.
Question 2
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In fig, then $$\angle A+\angle B+\angle C+\angle D+\angle E+\angle F=$$

A
$$180^o$$

B
$$270^o$$

C
$$360^o$$

D
$$540^o$$

Solution

In $$\triangle AEC$$,
$$ \angle A + \angle E + \angle C = 180^o $$ .....[Angle sum property].

Also, in $$\triangle BDF$$,
$$ \angle B + \angle D + \angle F = 180^o $$ .....[Angle sum property].

Adding both the equations, we get,
$$ \angle A+\angle B+\angle C+\angle D+\angle E+\angle F= 180^o + 180^o = 360^o $$.

Therefore, option $$C$$ is correct.
Question 3
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Which of the following is true?
(i) A triangle can have two right angles.
(ii) A triangle can have all angles less than $$60^o$$.
(iii) A triangle can have two acute angles.

A
Only (ii)

B
Only (i)

C
Only (iii)

D
All are true

Solution

(i) If the triangle has two right angles, the sum of angles of the triangle will be more than $$180^{\circ}$$, which is not possible.
(ii) If the triangle has all the angles less than $$60^{\circ}$$, the sum of the triangle will be less than $$180^{\circ}$$, which is not possible.
(iii) Thus, the only possibility is the triangle has two acute angles.
Question 4
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In $$\Delta\, ABC$$, if $$AB = BC$$ and $$\angle\, B\, =\, 80^{\circ},$$ then $$\angle C\, =$$

A
$$50^{\circ}$$

B
$$100^{\circ}$$

C
$$130^{\circ}$$

D
None

Solution

Given, in $$\triangle ABC, AB=BC, \angle B=80^o$$
$$\Rightarrow\, \angle A\, =\, \angle C$$
We know $$\angle A\, +\, \angle B\, +\, \angle C\, =\, 180^{\circ}$$
$$\Rightarrow 2\, \angle C\, +\, 80^{\circ}=\, 180^{\circ}$$
$$\Rightarrow \angle\,C =\, 50^{\circ}$$.
Question 5
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Which of the following statement(s) is/are false with respect to quadrilateral?

A
Each diagonal of a quadrilateral divides it into two triangles.

B
Each side of a quadrilateral is less than the sum of the remaining three sides.

C
A quadrilateral can at-most have three obtuse angles.

D
A quadrilateral has four diagonals.

Solution

Option A is true. As each diagonal of a quadrilateral divides it into two triangles.
Option B is also correct as it is not possible to construct a quadrilateral with one side greater than the sum of three other sides.
Option C is also true. A quadrilateral can at most have three obtuse angles. All the four angles cannot be obtuse and sum of 4 angles of a quadrilateral is $$360^\circ$$.
Option D is incorrect. A quadrilateral does not have four diagonals.
A quadrilateral has exactly two diagonals.
So, option (D) is false
Question 6
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In the given figure, value of $$x$$ is .........

A
45$$^o$$

B
34$$^o$$

C
64$$^o$$

D
109$$^o$$

Solution

In triangle $$\Delta ABC$$,
$$\angle A+\angle ABC+\angle BCA=180^o$$
$$34^o+ \angle ABC +30^o=180^o$$
$$\angle ABC=116^o$$
So, $$\angle ABC+\angle CBD=180^o$$
$$116^o+\angle CBD=180^o$$
$$\angle CBD=180^o-116^o$$
$$\angle CBD=64^o$$
Now use remote interior angle theorem.
So, $$\angle X = \angle CBD+ \angle D=64^o+45^o=109^o$$
Question 7
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In a right angled triangle $$ABC,\,\angle B=90^{\circ}$$ such that $$AB=6\;cm,\;BC=8\;cm$$. Then find $$AC$$.

A
$$13\ cm$$

B
$$14\ cm$$

C
$$10\ cm$$

D
$$12\ cm$$

Solution

By using Pythagoras theorem in $$\triangle ABC,$$
$$AC^2=AB^2+BC^2$$
$$\Rightarrow AC^2=6^2+8^2$$
$$\Rightarrow AC^2=36+64$$
$$\Rightarrow AC^2=100$$
$$\Rightarrow AC=10\ cm$$
Question 8
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Rohit used the Pythagorean theorem to see how much time he would save taking a shortcut to home from football practice. He usually walked $$6$$ blocks south and $$9$$ blocks east. Which picture shows his shortcut?

A

B

C

D

Solution

According to the Pythagorean theorem
$$(AB)^2+(BC)^2=(AC)^2$$
$$(6)^2+(9)^2=(AC)^2$$
$$(AC)^2=36+81$$
$$AC=\sqrt{117}$$
Hence, option $$A$$ is answer.
Question 9
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The length of side of an equilateral triangular garden whose perimeter is $$66 \ m$$ is:

A
$$66 \ m$$

B
$$11 \ m$$

C
$$3 \ m$$

D
$$22 \ m$$

Solution

Let the side of the equilateral triangular garden be $$a$$,
Then, $$3a = 66$$
$$\Rightarrow a\, =\, \displaystyle \frac {66}{3}\, =\, 22\, \ m$$
Hence length of side of garden $$= 22 \ m.$$
Question 10
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In a right angled triangle $$ABC,\,\angle B=90^{\circ}$$ such that $$AC=13\;cm,\;BC=5\;cm$$. Then, find $$AB$$.

A
$$12 cm$$

B
$$17 cm$$

C
$$15 cm$$

D
$$14 cm$$

Solution

In a right angled triangle $$ ABC $$, if $$ \angle B={ 90 }^{ 0 }$$

using Pythagoras theorem, we have

$${AC}^{2} = {AB}^{2} + {BC}^{2} $$

$$ AC=13cm,BC=5cm $$

$$\Rightarrow {13}^{2} = {AB}^{2} + {5}^{2} $$

$$\Rightarrow 169 = {AB}^{2} + 25 $$

$$\Rightarrow 169 - 25 = {AB}^{2} $$

$$\Rightarrow AB = \sqrt {144} = 12 cm$$