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Each of the numbers from 1 to 9 is placed, one per circle, into the pattern shown. The sums along each of the four sides are equal. How many different.

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Presentation on theme: "Each of the numbers from 1 to 9 is placed, one per circle, into the pattern shown. The sums along each of the four sides are equal. How many different."β€” Presentation transcript:

1 Each of the numbers from 1 to 9 is placed, one per circle, into the pattern shown. The sums along each of the four sides are equal. How many different numbers can be placed in the middle circle to satisfy these conditions? Solution: Because the sums along each side must be equal, therefore sums of the 8 numbers must be divisible by 4. Note that = 45 = 4 Γ— Therefore the numbers which have a remainder of 1 when divided by 4 are the candidates to be placed in the middle. Answer:

2 When Anura was 8 years old his father was 31 years old
When Anura was 8 years old his father was 31 years old. Now his father is twice as old as Anura is. How old is Anura now? Solution: Now, after π‘₯ year, Anura’s father’s age is 31 + π‘₯ and Anura’s age is 8 + π‘₯ Anura’s father’s age is 2 Γ— Anura’s age 31 + π‘₯ = 2 Γ— (8 + π‘₯) = π‘₯ π‘₯ = 15 Anura’s age = =____ Answer: 23

3 In rectangle 𝐴𝐡𝐢𝐷, 𝐴𝐡 = 12π‘π‘š and 𝐴𝐷 = 5π‘π‘š
In rectangle 𝐴𝐡𝐢𝐷, 𝐴𝐡 = 12π‘π‘š and 𝐴𝐷 = 5π‘π‘š. Point P, Q, R and S are all on diagonal AC, so that 𝐴𝑃 = 𝑃𝑄 = 𝑄𝑅 = 𝑅𝑆 = 𝑆𝐢. What is the total area of the shaded region, in cm2? Solution: The area for 𝐴𝐡𝐢𝐷 = 5 Γ— 12 = 60 π‘π‘š2 The area for △𝐴𝐡𝐢 and △𝐴𝐷𝐢 = 60 Γ· 2 = 30 π‘π‘š2 Since 𝐴𝑃=𝐢𝑆= 1 5 𝐴𝐢, therefore the area for △𝐴𝑃𝐷=△𝐴𝑃𝐡=△𝐢𝑆𝐷=△𝐢𝑆𝐡= 1 5 △𝐴𝐡𝐢= 1 5 Γ—30=6 π‘π‘š2 Total area of shaded region = =____ Answer: 24 π‘π‘š2

4 In triangle 𝐴𝐡𝐢, 𝐴𝑃 = 𝐴𝑄 and 𝐡𝑄 = 𝐡𝑅. Determine angle 𝑃𝑄𝑅 in degrees.
Solution: 𝐴𝑃 = π΄π‘„β‡’βˆ π΄π‘ƒπ‘„ = βˆ π΄π‘„π‘ƒ = ∝° 𝐡𝑄 = π΅π‘…β‡’βˆ π΅π‘…π‘„ = βˆ π΅π‘„π‘… =𝛽° βˆ π‘ƒπ΄π‘„ = 180Β° βˆ’ 2∝° βˆ π‘…π΅π‘„ = 180Β° βˆ’ 2𝛽° From △𝐴𝐡𝐢, 70Β° + (180Β° βˆ’ 2∝°) + (180Β° βˆ’ 2𝛽°) = 180Β° 2∝° + 2𝛽° = 250Β° ∝° + 𝛽° = 125Β° βˆ π‘ƒπ‘„π‘… = 180Β° βˆ’ ∝° βˆ’ 𝛽° = 180Β° βˆ’ (∝° + 𝛽°) = 180Β° βˆ’ 125Β° = 55 Β° Answer: 55Β°


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