Question 13 Marks
A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
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Question 23 Marks
Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Question 33 Marks
Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Answer
The triangle drawn is an obtuse-angled and isosceles triangle.
View full question & answer→The triangle drawn is an obtuse-angled and isosceles triangle.
Question 43 Marks
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from $6$ to $3$
Answer
View full question & answer→By looking at the clock we can see when the hour hand goes from $6$ to $3$ it basically covers three right angles which is of $270^\circ .$
Therefore, required Fraction = $\frac{270}{360}$ = $\frac{3}{4}$
Therefore, required Fraction = $\frac{270}{360}$ = $\frac{3}{4}$
Question 53 Marks
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from $1$ to $10$
Answer
View full question & answer→By looking at the clock we can see when the hour hand goes from $1$ to $10$ it basically covers three right angles which is of $270^\circ .$
So, required Fraction = $\frac{270}{360}$ = $\frac{3}{4}$
So, required Fraction = $\frac{270}{360}$ = $\frac{3}{4}$
Question 63 Marks
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from $12$ to $9$
Answer
View full question & answer→By looking at the clock we can see when the hour hand goes from $12$ to $9$ it basically covers three right angles which is of $= 90 + 90 + 90 = 270^\circ .$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Question 73 Marks
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from $7$ to $10$
Answer
View full question & answer→By looking at the clock we can see when the hour hand goes from $12$ to $9$ it basically covers three right angles which is of $= 90 + 90 + 90 = 270^\circ .$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Question 83 Marks
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from $4$ to $7$
Answer
View full question & answer→By looking at the clock we can see when the hour hand goes from $12$ to $9$ it basically covers three right angles which is of $= 90 + 90 + 90 = 270^\circ .$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Question 93 Marks
What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from $3$ to $9$
Answer
View full question & answer→By looking at the clock we can see when the hour hand goes from $12$ to $9$ it basically covers three right angles which is of $= 90 + 90 + 90 = 270^\circ .$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Question 103 Marks
If B is the mid point of $\overline{A C}$ and $C$ is the point of $\overline{\mathrm{BD}}$ . where $A, B, C, D$ lie on a straight line, say why $AB = CD?$
View full question & answer→Question 113 Marks
Draw any line segment, say $\overline{A B}$ . Take any point $C$ lying in between $A$ and $B.$ Measure the lengths of $AB, BC$ and $AC$. Is $AB = AC + CB?$
[Note: If $A, B, C$ are any three points on a line such that $AC + CB = AB$, then we can be sure that $C$ lies between $A$ and $B.]$
[Note: If $A, B, C$ are any three points on a line such that $AC + CB = AB$, then we can be sure that $C$ lies between $A$ and $B.]$
Answer
View full question & answer→By looking at the clock we can see when the hour hand goes from $12$ to $9$ it basically covers three right angles which is of $= 90 + 90 + 90 = 270^\circ .$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$
Therefore, required Fraction = $\frac{270}{260}$ = $\frac{3}{4}$