Polygons Worksheet Grade 5 contains 8 MCQ questions. Answers to Polygons Worksheet Grade 5 are available after clicking on the answer. Maths Worksheets for Class 5Â help to check the concept you have learnt from detailed classroom sessions and application of your knowledge.
| Category | Maths Worksheets for Class 5 |
| Subject | Maths |
| Chapter | Polygons |
Polygons Worksheet Grade 5
Q.1) Determine, the type of Triangle for the given figure ?​
a) Equilateral Triangle
b) Isosceles Triangle
c) Scalene Triangle
Answer
Answer: c) Scalene Triangle
Explanation
In the given figure
Sides of the Triangle are
ABÂ Â Â =Â Â Â 7 cm
BCÂ Â Â =Â Â Â 5 cm
ACÂ Â Â =Â Â Â 6 cm
We know that,
If all the sides of a Triangle are unequal, it is called a Scalene Triangle.
In the given figure all the sides of the Triangle are unequal.
Therefore, the given Triangle is a Scalene Triangle.
Q.2) Determine, the type of Triangle for the given figure ?
a) Equilateral Triangle
b) Isosceles Triangle
c) Scalene Triangle
Answer
Answer: b) Isosceles Triangle
Explanation
In the given figure
The Sides of the triangle are
ABÂ Â Â =Â Â Â 4 cm
BCÂ Â Â =Â Â Â 4 cm
ACÂ Â Â =Â Â Â 6 cm
We know that,
If two of the sides of a Triangle are equal, it is called an Isosceles Triangle.
In the given figure, sides AB and BC of the Triangle are equal.
Therefore, the given Triangle is an Isosceles Triangle.
Q.3) Determine, the type of Triangle for the given figure ?
a) Equilateral Triangle
b) Isosceles Triangle
c) Scalene Triangle
Answer
Answer: a) Equilateral Triangle
Explanation
In the given figure
Sides of triangle are
ABÂ Â Â =Â Â Â 5 cm
BCÂ Â Â =Â Â Â 5 cm
ACÂ Â Â =Â Â Â 5 cm
We know that,
If all the sides of a Triangle are equal, it is called an Equilateral Triangle.
In the given Triangle all the sides i.e. AB, BC and AC are equal.
Therefore, the given Triangle is an Equilateral Triangle.
Q.4) Determine, the type of Triangle for the given figure ?
a) Acute Angled Triangle
b) Obtuse Angled Triangle
c) Right Angled Triangle
Answer
Answer: c) Right Angled Triangle
Explanation
In the given figure
The three angles of the Triangle are, 30Ëš, 60Ëš and 90Ëš
We know that,
If one angle of a triangle is equal to 90Ëš, it is called a Right Angled Triangle.
In the given figure one of the angles in the Triangles is equal to 90Ëš
Therefore, the given Triangle is a Right Angled Triangle.
Q.5) Determine, the type of Triangle for the given figure ?
a) Obtuse Angled Triangle
b) Acute Angled Triangle
c) Right Angled Triangle
Answer
Answer: a) Obtuse Angled Triangle
Explanation
In the given figure
The three angles of the Triangle are 30Ëš, 30Ëš and 120Ëš
We know that,
If one angle of a triangle is more than 90Ëš (known as Obtuse angle), It is called an Obtuse Angled Triangle.
In the given figure one of the angles is 120Ëš, which is more than 90Ëš
Therefore, the given Triangle is an obtuse angled Triangle.
Q.6) Determine whether the given figure is a Square or Rectangle ?
a) Square
b) Â Rectangle
Answer
Answer: b) Â Rectangle
Explanation
Square – Square is a 2D shape in which all the sides are equal and each angle is of 90˚.
Rectangle – Rectangle is 2D shape is which Opposite sides are equal and each angle is 90˚.
In the given figure,
AB = CD = 6 cm and AD = BC = 4 cm and each angle is Right angle
Therefore, the given figure is a Rectangle (Since all the sides are not equal, it is not a square).
Polygons Worksheet Grade 5
Q.7) If the two angles of the Triangle are ∠A = 40˚ , ∠C = 65˚, find ∠B ?
a) ∠B = 65˚
b) ∠B = 85˚
c) ∠B = 75˚
Answer
Answer: c) ∠B = 75˚
Explanation
In the given figure,
∠A = 40˚
∠C = 65˚
∠B = ?
We know that,
Sum of all the angles of Triangle = 180Ëš
Therefore,
∠A + ∠B + ∠C = 180˚
40˚ + 65˚ + ∠B = 180˚
105˚ + ∠B = 180˚
∠B = 180Ëš – 105Ëš
∠B = 75˚
Q.8) In the given figure , find xËš and yËš ?
a) xËš = 50Ëš , yËš = 85Ëš
b) xËš = 70Ëš , yËš = 65Ëš
c) xËš = 60Ëš , yËš = 75Ëš
Answer
Answer: c) xËš = 60Ëš , yËš = 75Ëš
Explanation
In the given figure,
In Δ ABC
∠A = 45˚
∠ABC = x˚
∠ACB = 75˚
We know that,
Sum of all the three angles of Traingle = 180Ëš
∠A + ∠ABC + ∠ACB = 180˚
45Ëš + xËš + 75Ëš = 180Ëš
120Ëš + xËš = 180Ëš
xËš = 180Ëš – 120Ëš
xËš = 60Ëš
Similarly,
In Δ BCD
∠BCD = 35˚
∠CBD = y˚
∠D = 70˚
∠BCD + ∠CBD + ∠D = 180˚
35Ëš + yËš + 70Ëš = 180Ëš
105Ëš + yËš = 180Ëš
yËš = 180Ëš – 105Ëš
yËš = 75Ëš
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