Subscribe

Mathematics Magazine for Grades 1-12  

 

Grade 10

 

 

 
Solutions from the Previous Issue:

1.        Find the perimeter of a right triangle with shorter sides measuring 5 and 12 units.

Solution:

Since the sides of a right triangle satisfy the Pythagorean Theorem, a2 + b2 = c2, we have 52 + 122 = 25 + 144 = 169 = 132. So, the three sides are 5, 12 and 13 and the perimeter is 5 + 12 + 13 = 30.

 

2.        Why could a right triangle with integers for lengths of sides not have shorter sides measuring 5 and 10 units?

Solution:

By the Pythagorean Theorem, 52 + 102 = 25 + 100 = 125 which is not a perfect square [125 ~ (11.1803...)2]. Thus, the third side cannot have an integer length.

 

3.         

Observe the two squares in the figure above with side lengths of 4 and 10 units respectively. What is the area of the shaded region?

Solution:

Let H the height of the red triangle.

;       H =2.85

 

4.         

The four circles in the figure above each have a radius of 3 units. The centers of each circle form a square. Find the area of the green shaded region.

        Solution:

        Areagreen=Area square-Area circle

        Areagreen= (3 + 3)2 - π∙32 = 36 28.26 = 7.74 units2

 

5.       

In the above figure, the area of triangle CDE is 3 square units. Find the area of triangle ABC.

        Solution:

        From Pythagoras AB = = 6 units

        AreaABC= square units

 

Proposed Problems

1.        The line segments joining the vertices of triangle ABC to the midpoints of the opposite edges are called medians and are concurrent at P.

Prove that P splits each median in the ratio 2:1.

2.        A second order recurrence relation is defined by un+1 = (un+un−1)/2; that is, each new term is the mean of the previous two terms.

For example, when u1=2 and u2=5, we generate the following sequence:

2, 5, 3.5, 4.25, 3.875, ... .

Find the limit of the sequence for different starting values.

3.        Four digits are selected from the set {1,2,3,4,5} to form a 4-digit number. Find the sum of all possible permutations.

4.        An arrow is formed in a 2 2 square by joining the bottom corners to the midpoint of the top edge and the centre of the square.

Find the area of the arrow.