Geometry - 12 marks in PSLE 2021

In PSLE examination 2021, there are 5 geometry questions, totaling 12 marks. That is 12% of the total possible marks of 100. Let's look at what are the concepts that are tested in these 5 questions. Paper 1 Booklet A Question 10 (1 mark) In this question, pupils are tested on the concept of total sum of angles in a triangles, which equals to 180 degrees. There are 3 triangles in this question, 2 smaller triangles joined to form a bigger triangle. Another key point to take note of, is mentioned in the question, angle BDA = angle CDB. This equation allows the pupils to divide angle CDA by 2 in order to find the unknown angle ABD. Step 1: Using the bigger triangle ACD, find angle ADC. Step 2: Divide angle ADC by 2, to get angle ADB.

Step 3: Using triangle ABD, find angle ABD.

Paper 1 Booklet A Question 13 (2 marks) In this question, pupils are tested on the concept of angles in parallel lines and isosceles triangle. In the statement, it is stated that AB and DC are the only pair of parallel lines. Also lines AB = BC = BD, which means triangles ABD and BDC are isosceles triangles. Pupils should know that base angles of isosceles triangles are equal. Below are a few steps to follow to find the unknown angle DBC. Step 1: angle ADB = 62 degrees (base angle of isosceles triangle).

Step 2: Find angle ABD, using sum of angles in a triangle = 180 degrees. Step 3: angle BDC equals angle ABD (alternate angles, parallel lines). Step 4: angle BCD equals angle BDC (base angles of isosceles triangle). Step 5: Find angle CBD using sum of angles in a triangle = 180 degrees.

Paper 1 Booklet B Question 22 (2 marks) In this questions, pupils are tested on the use of a protractor. I always remind pupils to have a protractor in their pencil case. Do not need to buy the whole box of geometrical set. Protractors are available in all school bookshops or community bookshops. Both parts are straight forward questions and no calculation is needed. Part (a) is asking for the smallest angle while part (b) is asking for two angles that is greater than 90 degrees (right angle). In part (a), angles c and e look closely similar and hence, pupils need to use a protractor to find out the answer. in part (b), pupils can use the protractor to see which angles are greater than 90 degrees. They can also use a short ruler (corners are 90 degrees) to find out which angles are more than 90 degrees.

Paper 2 Question 7 (3 marks)

In this question, pupils are tested on the concept of equilateral triangle and rhombus. All three angles in any equilateral triangle is 60 degrees. Opposite angles of any rhombus are equal if a diagonal is drawn, there will be 2 isosceles triangles. Step 1: angle CDA = 102 degrees (opposite angles of rhombus). Step 2: find angle CDG. Step 3: angle EGF = 60 degrees (equilateral triangle). Step 4: find angle EGD (angles at a point). Step 5: angle FEG = 60 degrees (equilateral triangle). Step 6: angle y + 60 degrees = angle CDG + angle DGE. Find y.

Paper 2 Question 13 (4 marks)

In this question, information given AB is parallel to DC. Lines AE = EB, forming an isosceles triangle. Pupils are also tested on the concept of corresponding angles on a pair of parallel lines. part (a) can be found using sum of angles in a triangle. For part (b), Step 1: angle EFA = angle FDC (corresponding angle).

Step 2: angle BFC = 53 degrees (alternate angles). Step 3: Find angle CBF. Step 4: angle EAB = angle EBA (base angle of isosceles triangle) Step 5: Find angle AEF (using triangle AEB, sum of angles in triangle)

Part (c), AF is parallel to DC (it's mentioned in the question), AD is not parallel to FC (as angle AFC is not equal to angle ADC). Hence AFCD is a trapezium.

Reading the sentences in the question is as important as the diagram itself. There will be some information in the sentences that are not shown on the diagram. Pupils are expected to transfer these information into the diagram on their own. Hope this post can be a good guide for you and your child to view these 5 questions better.


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