Free The Triangle and Its Properties Subjective Test 02 Practice Test - 7th grade

Question 1

In triangle ABC, which side will lie opposite to $∠ A$ ? [1 MARK]

SOLUTION

Solution : $∠ A$ will be formed by the sides BA and AC. Hence, the side opposite to $∠ A$ will be BC.

Question 2

Find the value of the unknown. [2 MARKS]

(a)

(b)

SOLUTION

Solution : Each part: 1 Mark each

(a) $50^{\circ} + x = 115^{\circ}$ (Exterior angle of a triangle is equal to the sum of its interior opposite angles)
$x = 115^{\circ} - 50^{\circ}$
$x = 65^{\circ}$

(b) $30^{\circ} + x = 80^{\circ}$ (Exterior angle of a triangle is equal to the sum of its interior opposite angles)
$x = 80^{\circ} - 30^{\circ}$
$x = 50^{\circ}$

Question 3

Explain the three types of triangles based on sides. [2 MARKS]

SOLUTION

Solution : Explanation: 2 Marks

Three types of triangles categorized by sides are:
(i) If all the three sides of the triangle have unequal length, the triangle is called a Scalene triangle.
(ii) If two out of three sides are equal in length, it's called an Isosceles triangle.
(iii) If all three sides of a triangle are equal in length, it's called an Equilateral triangle.

Question 4

Explain the three types of triangles based on their angles. [2 MARKS]

SOLUTION

Solution : Types of triangle: 2 Marks

The three types of triangles based on angles are:

(i) A triangle is called Obtuse angled when one of the angles has an angle greater than 90 $^{o}$ .

(ii) If all angles of the triangle are less than 90 $^{o}$ , it's called an Acute angled triangle.

(iii) If one of the angles is equal to 90 $^{o}$ , it's called a Right angled triangle.

Question 5

What is the difference between median and altitude? [2 MARKS]

SOLUTION

Solution : Each difference: 1 Mark

1. Median is a line segment formed between a vertex and the mid-point of the opposite side.
Altitude is the perpendicular distance of a vertex from its opposite side.

2. Median always bisects the opposite side of the vertex irrespective of the type of triangle.
Altitude may or may not bisect the opposite side of the vertex depending on the type of triangle.

In the figure below, CH is the altitude from vertex C on side AB and CM is median from C on side AB.

Question 6

Find the value of x in the following triangles. [3 MARKS]

(a)

(b)

SOLUTION

Solution : Each part: 1.5 Marks

(a) Let the triangle be ABC with $∠ A$ being x and $∠ B$ being the right angle.

In a triangle, Sum of opposite interior angles = exterior angle
Here,
x + 90 $^{o}$ = 150 $^{o}$
Or, x = 150 $^{o}$ - 90 $^{o}$ = 60 $^{o}$

(b) $∠ B A C = 70^{\circ}$ (Vertically opposite angles)

$∠ A B C = 40^{\circ}$ (Alternate angles)

$x = 180 - (40 + 70)$ (Sum of angles in a triangle is $180^{\circ}$

$x = 180 - 110 = 70^{\circ}$

Question 7

Find all the angles of the following triangle. [3 MARKS]

SOLUTION

Solution : Each angle: 1 Mark

The three angles in the given triangle are x, 2x and 3x.
For any triangle,
Sum of all the angles = 180 $^{o}$
Or, x + 2x + 3x = 180 $^{o}$
Or, 6x = 180 $^{o}$
Or x = $\frac{180^{o}}{6}$ = 30 $^{o}$

The angles of the triangle will be 30 $^{o}$ (x), 60 $^{o}$ (2x) and 90 $^{o}$ (3x).

Question 8

Find the value of $∠ C$ in the triangle given below if the value of $∠ C A B$ = 25 $^{o}$ . If AB = BD, what can be the value of $∠ D$ ? [3 MARKS]

SOLUTION

Solution : Steps: 1 Mark
Solution for $∠ C$ and $∠ D$ : 2 Marks

In the given triangle CAB
Sum of opposite interior angles = exterior angle
Or, $∠ C$ + $∠ C A B$ = 90 $^{o}$
Or, $∠ C$ = 90 $^{o}$ - 25 $^{o}$
Or, $∠ C$ = 65 $^{o}$

$∵$ AB = BD $\Rightarrow ∠ B A D = ∠ D = y (s a y)$ (Equal sides have equal angles opposite to them.)
$\Rightarrow ∠ B A D + ∠ D + 90 = 180$
$\Rightarrow ∠ B A D + ∠ D = 90$
$\Rightarrow 2 y = 90$
$∵ ∠ B A D = ∠ D = y$
$\Rightarrow y = \frac{90}{2}$
$\Rightarrow y = 45$

$∴ ∠ B A D = ∠ D = 45^{\circ}$

Question 9

(a) What will be the value of x in the following diagram? [3 MARKS]

(b) If a $△$ QPR is right angled at P, which is the longest side of the triangle?

SOLUTION

Solution : (a) Steps: 1 Mark
Result: 1 Mark
(b) Answer with reason: 1 Mark

(a) According to the question, the three sides of the triangle are equal.
Hence, all angles will also be equal (say x).

Sum of angles of a triangle = 180 $^{o}$
Or, x + x + x = 180 $^{o}$
Or, 3x = 180 $^{o}$
Or x = $\frac{180^{o}}{3}$ = 60 $^{o}$

(b) In a triangle, the side opposite to the angle with the largest value is the longest.
Here the biggest angle is the right angle, hence QR is the longest side in $△$ QPR, right angled at P.

Question 10

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 $^{o}$

SOLUTION

Solution : 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 11

(a) For any triangle ABC, draw: [4 MARKS]
(i) All possible medians.
(ii) All possible altitudes.

(b) Find the unknown angles x and y.

SOLUTION

Solution : (a) Each sub-part: 1 Mark
(b) Each angle: 1 Mark

(a)

(i) (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 12

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

SOLUTION

Solution : 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 13

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

SOLUTION

Solution : 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 14

(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]

SOLUTION

Solution : 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 15

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

SOLUTION

Solution : 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.

Free The Triangle and Its Properties Subjective Test 02 Practice Test - 7th grade

Question 1

SOLUTION

Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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