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∘+x=115∘ (Exterior angle of a triangle is equal to the sum of its interior opposite angles)
x=115∘−50∘
x=65∘
(b) 30∘+x=80∘ (Exterior angle of a triangle is equal to the sum of its interior opposite angles)
x=80∘−30∘
x=50∘
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 90o.
(ii) If all angles of the triangle are less than 90o, it's called an Acute angled triangle.
(iii) If one of the angles is equal to 90o, 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.
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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 + 90o = 150o
Or, x = 150o - 90o = 60o
(b) ∠BAC=70∘ (Vertically opposite angles)
∠ABC=40∘ (Alternate angles)
x=180−(40+70) (Sum of angles in a triangle is 180∘
x=180−110=70∘
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 = 180o
Or, x + 2x + 3x = 180o
Or, 6x = 180o
Or x = 180o6 = 30o
The angles of the triangle will be 30o(x), 60o(2x) and 90o(3x).
Question 8
Find the value of ∠C in the triangle given below if the value of ∠CAB = 25o. 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 + ∠CAB = 90o
Or, ∠C = 90o - 25o
Or, ∠C = 65o
∵ AB = BD ⇒∠BAD=∠D=y (say) (Equal sides have equal angles opposite to them.)
⇒∠BAD+∠D+90=180
⇒∠BAD+∠D=90
⇒2y=90
∵ ∠BAD=∠D=y
⇒y=902
⇒y=45
∴ ∠BAD=∠D=45∘
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 = 180o
Or, x + x + x = 180o
Or, 3x = 180o
Or x = 180o3 = 60o
(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 = 60o, ∠B = 120o, ∠C = 30o
(iii) AB = 5 cm, BC = 7 cm, AC = 11 cm
(iv) ∠A = 60o, ∠B = 90o, ∠C = 30o
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 180o but here,
∠A(60o) + ∠B(120o) + ∠C(30o) = 210o
(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 180o and here,
∠A(60o) + ∠B(90o) + ∠C(30o) = 180o
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)
∠BAC=92∘
∠ACB=x∘ (Equal sides have equal angles opposite to them.)
∠ACB+∠ABC+92∘=180∘
2x+92∘=180∘
x+46∘=90∘
x=90∘−46∘
x=44∘
∠ACB+y=180∘⇒x+y=180∘
⇒44∘+y=180∘
⇒y=180∘−44∘
⇒y=136∘
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 ∠ACB will also be y (∵ angle opposite to the equal sides are equal)
In any triangle, Sum of all angles = 180o
Or, y + y + 90o = 180o
Or, 2y = 180o - 90o
Or, y = 90o2 = 45o
X here is an exterior angle to ∠A and ∠B
Sum of opposite interior angles = exterior angle.
Or, y + 90o = X
Or, X = 90o +45o = 135o
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 90o.
Let the other angles be y (∵ equal sides have equal angles opposite to them)
For the given triangle,
Sum of all angles = 180o
Or, y + y + 90o = 180o
Or, 2y = 180o - 90o
Or, y = 90o2 = 45o
∵ The two horizontal lines are parallel to each other.
∴ X = 45o + y (∵ interior alternate angles are equal)
Or, X = 45o + 45o = 90o
Question 14
(a) Find the value of x in the below diagram given that, ∠ACB = 25o, ∠CBA = 15o 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 = 180o
Or, ∠CAB + ∠ABC + ∠BCA = 180o
Or, ∠CAB + 15o + 25o = 180o
Or, ∠CAB = 180o - 40o = 140o
For the triangle ACD,
∠D + X = ∠CAB (∵ sum of interior opposite angles = exterior angle)
90o + X = 140o (∵ CD is an altitude from C to extended BA, ∴ ∠D = 90o)
Or, X = 50o
(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 = 60o, 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 = 180o
Or, ∠A + ∠B + ∠C = 180o
Or, ∠A + ∠C + 60o = 180
Or, 2∠A = 180o - 60o (∵ angles opposite to equal sides are equal)
Or, ∠A = 120o2 = 60o
∵ ∠A = ∠B = ∠C = 60o
∴ The above triangle is an Equilateral triangle.