The Triangle And Its Properties(7th Grade > Mathematics ) Questions and answers for exam Preparation

7th Grade > Mathematics

THE TRIANGLE AND ITS PROPERTIES MCQs

Total Questions : 111 | Page 7 of 12 pages

Are the following triangles possible? [4 MARKS]
(i) AB = 4 cm, BC = 5.6 cm, AC = 10 cm
(ii) $∠ A$ = 60 $^{o}$ , $∠ B$ = 120 $^{o}$ , $∠ C$ = 30 $^{o}$
(iii) AB = 5 cm, BC = 7 cm, AC = 11 cm
(iv) $∠ A$ = 60 $^{o}$ , $∠ B$ = 90 $^{o}$ , $∠ C$ = 30 $^{o}$

Discuss Question

Answer: Option A. ->
:
Answer with Reason: 1 Mark each
(i) Triangle ABC is not possible. This is because, for any triangle, the sum of two sides should be greater than the third but AB + BC < AC.
(ii) Triangle ABC is not possible. Sum of all angles in a triangle should be 180

^{o}

but here,

∠ A

(60^{o})

∠ B

(120^{o})

∠ C

(30^{o})

= 210

^{o}

(iii) Triangle ABC is possible because it satisfies the condition for the formation of a triangle i.e. the sum of two sides should be greater than the third.
(iv) Triangle ABC is possible. Sum of all angles in a triangle should be 180

^{o}

and here,

∠ A

(60^{o})

∠ B

(90^{o})

∠ C

(30^{o})

= 180

^{o}

Question 62.

(a) For any triangle ABC, draw: [4 MARKS]
(i) All possible medians.
(ii) All possible altitudes.
(b) Find the unknown angles x and y.

Discuss Question

Answer: Option A. ->
:
(a) Each sub-part: 1 Mark
(b) Each angle: 1 Mark
(a)
(i)

(a) For Any Triangle ABC, Draw: [4 MARKS](i) All Possible Me...

(ii)

(b)

∠ B A C = 92^{\circ}

∠ A C B = x^{\circ}

(Equal sides have equal angles opposite to them.)

∠ A C B + ∠ A B C + 92^{\circ} = 180^{\circ}

2 x + 92^{\circ} = 180^{\circ}

x + 46^{\circ} = 90^{\circ}

x = 90^{\circ} - 46^{\circ}

x = 44^{\circ}

∠ A C B + y = 180^{\circ} \Rightarrow x + y = 180^{\circ}

\Rightarrow 44^{\circ} + y = 180^{\circ}

\Rightarrow y = 180^{\circ} - 44^{\circ}

\Rightarrow y = 136^{\circ}

Question 63.

Find the value of x in the following diagram. [4 MARKS]

Discuss Question

Answer: Option A. ->
:
Properties: 1 Mark
Steps: 2 Marks
Result: 1 Mark
According to the question, Sides AB and BC are equal and

∠ B

is 90°
Let

∠ A

be y, then

∠ A C B

will also be y (

∵

angle opposite to the equal sides are equal)
In any triangle, Sum of all angles = 180

^{o}

Or, y + y + 90

^{o}

= 180

^{o}

Or, 2y = 180

^{o}

- 90

^{o}

Or, y =

\frac{90^{o}}{2}

= 45

^{o}

X here is an exterior angle to

∠ A

and

∠ B

Sum of opposite interior angles = exterior angle.
Or, y + 90

^{o}

= X
Or, X = 90

^{o}

+45

^{o}

= 135

^{o}

Question 64.

Find the value of x in the following diagram given that the horizontal lines are parallel to each other:
[4 MARKS]

Discuss Question

Answer: Option A. ->
:
Properties: 2 Marks
Steps: 1 Marks
Result: 1 Mark
In the given triangle,
Two sides are equal and the angle made by joining them is 90

^{o}

.
Let the other angles be y (

∵

equal sides have equal angles opposite to them)
For the given triangle,
Sum of all angles = 180

^{o}

Or, y + y + 90

^{o}

= 180

^{o}

Or, 2y = 180

^{o}

- 90

^{o}

Or, y =

\frac{90^{o}}{2}

= 45

^{o}

∵

The two horizontal lines are parallel to each other.

∴

X = 45

^{o}

+ y (

∵

interior alternate angles are equal)
Or, X = 45

^{o}

+ 45

^{o}

= 90

^{o}

Question 65.

(a) Find the value of x in the below diagram given that, $∠ A C B$ = 25 $^{o}$ , $∠ C B A$ = 15 $^{o}$ and CD is the height of C from the extended line segment BA:

(b) The lengths of two sides of a triangle are 13 cm and 16 cm. The third side should lie between 'a' cm and 'b' cm for the triangle to be formed. Find the value of a + b? [4 MARKS]

Discuss Question

Answer: Option A. ->
:
Each part: 2 Marks
In the given triangle ABC,
(a) Sum of all angles of a triangle = 180

^{o}

Or,

∠ C A B

∠ A B C

∠ B C A

= 180

^{o}

Or,

∠ C A B

+ 15

^{o}

+ 25

^{o}

= 180

^{o}

Or,

∠ C A B

= 180

^{o}

- 40

^{o}

= 140

^{o}

For the triangle ACD,

∠ D

+ X =

∠ C A B

(

∵

sum of interior opposite angles = exterior angle)
90

^{o}

+ X = 140

^{o}

(

∵

CD is an altitude from C to extended BA,

∴

∠ D

= 90

^{o}

)
Or, X = 50

^{o}

(b) The third side of a triangle must be greater than the difference between the other two sides.

That is, third side $>$ (16 - 13) which is 3 cm.

Also, Sum of lengths of any two sides of a triangle is always greater than the third side.

i.e., third side $<$ (16+13) which is 29 cm.

Hence, a + b = 3 + 29 = 32.

Question 66.

If in the below triangle $∠ B$ = 60 $^{o}$ , which type of triangle is it and why?
[4 MARKS]

Discuss Question

Answer: Option A. ->
:
Correct answer: 1 Mark
Reason: 1 Mark
Steps: 2 Marks
From the figure, we can see that
BO is perpendicular to AC and AO = OC

∵

BO is cutting the base into half and is also perpendicular to it.

∴

BO is acting both as an altitude and median.
An altitude acts as a median only in an Equilateral or Isosceles triangle with AB and BC being equal in length.
For the triangle ABC,
Sum of all angles = 180

^{o}

Or,

∠ A

∠ B

∠ C

= 180

^{o}

Or,

∠ A

∠ C

+ 60

^{o}

= 180
Or, 2

∠ A

= 180

^{o}

- 60

^{o}

(

∵

angles opposite to equal sides are equal)
Or,

∠ A

\frac{120^{o}}{2}

= 60

^{o}

∵

∠ A

∠ B

∠ C

= 60

^{o}

∴

The above triangle is an Equilateral triangle.

Question 67.

Two angles of a triangle are $50^{\circ}$ and $70^{\circ}$ . The third angle is:

$50^{\circ}$
$70^{\circ}$
$60^{\circ}$
$120^{\circ}$

Discuss Question

Answer: Option C. ->

60^{\circ}

:
C

The sum of all the angles of a triangle is $180^{\circ}$ .

Therefore,
$Third angle = 180^{\circ} - (50^{\circ} + 70^{\circ}) = 180^{\circ} - (120^{\circ}) = 60^{\circ}$

Question 68.

Pythagoras' theorem holds good for right angled triangles.

True
False
$60^{\circ}$
$120^{\circ}$

Discuss Question

Answer: Option A. -> True
:
A

In a right angled triangle, square of the hypotenuse (the side opposite to the right angle) is equal to the sum of the squares of the other two sides. This is Pythagoras' theorem.

Question 69.

Median and altitude of an isosceles triangle are one and the same.