₦airaland Forum

I dey run pass am then come back to solve it... Won't lie... .the thing hard me small.

You can never see agbadorians in this kind of thread... They're all olodos... Just say that T means Peter Obi.

😄😄😄😄😄

C 29 was the ans

The answer already given is correct because no other answer has been given on the contrary. So, I support the answers.

guobe:
One of the most difficult mathematical questions in primary school was finding the perimeter and area of a T sectional object.
Who can solve this.

This was never difficult to me and my friends back then... A very simple Area of rectangles question.

Doyou2019:
I know. Been there too jare.

I made mine permanent though. Just that I made sure my income kept going high and high by moving from unknown to top class schools.

EducasionMe:
C 29 was the ans

Wrong... It's 19cm²
First of all split both rectangles into two (let say A and B)
(Area of rectangle A) + (Area of rectangle B)
=> (6 x 2) + (2 x 3.7) cm²
=> (12) + (7) cm²
=> 19cm²

illicit:
It was like that too at OGSA, the only school in Akure, ondo state

I cant solve any mathematics, I eventually became a Linguist, theres something in Linguistics called Schema, it is actually mathematics in artistic form...


Do you understand?

i rep Oyemekun Grammar School, UP SCHOOL, THE ONLY SCHOOL.... OGSA

L

Oyemekun... school of fighting fighting then with omonuorogbo, Aqiunas. I didn't attend te school but I know d school well.. white on green.

saintbillion:
Wc year? We fit be classmates

That was in 1998 set. Finished around 2003 or thereabouts. I was very famous oh. Very very.

KingWarri:
This was never difficult to me and my friends back then... A very simple Area of rectangles question.

Alot of people didn't have good maths tutors then so you won't blame them

UTILITYMAY:
It was a typo error. Thanks.

Okay ,I thought as much 👌

geoworldedu:
I made mine permanent though. Just that I made sure my income kept going high and high by moving from unknown to top class schools.

Good on you, mate. More wins.

franchasng:
13. Answer is 19cm² na.

The area of a rectangle = L x W

There are two rectangles there with:

Rectangle 1 having dimension: 6 x 2

Rectangle 2 having dimension: 3.5 x 2

Area of rectangle 1 = 6 x 2 = 12cm²

Area of rectangle 2 = 3.5 x 2 = 7cm²

Area of the diagram = area of rectangle 2 + area of rectangle 2 = 12cm² + 7cm² = 19cm²


This is very simple to solve

....area of rectangle 2 + area of rectangle 2??

It doesn't add up.

Perimeter 23cm, area 19cm2

angelboy01:
That was in 1998 set. Finished around 2003 or thereabouts. I was very famous oh. Very very.

2023 was still in primary 4.

saintbillion:
2023 was still in primary 4.

You meant 2003. Me I be Baba now with kids oh. I don marry two wives sef ontop. Lol

The answer is 19sqm and i did it within 2 minutes without pen or paper.

Just offhand.

Split the T sections into 2 rectangles and find area of each rectangle and add I guess.

angelboy01:
The Sec School I went to, if you didn't pass Mathematics from SS1 upwards you will be repeated, I didn't know maths, wished I knew but I still passed. Till date I didn't know how I did it. For the records it's Mayflower School at Ikenne.

What year did u graduates from the school
I've friends and sister-in-law that went to that sch

In those days, this diagram was drawn without due diligence to its dimensions and you dear not ask why. Just take a look at the dimensions. How can the two 2cms on the horizontal portion of the T section which are parts of the length be longer than the 2cm which is the breadth? And on the vertical portion of the T section, you will discover that the 3.5cm which is the length is almost the same dimension as one of the 2cm which forms part of the length of the horizontal portion.

GodfreyOje:
In those days, this diagram was drawn without due diligence to its dimensions and you dear not ask why. Just take a look at the dimensions. How can the two 2cms on the horizontal portion of the T section which are parts of the length be longer than the 2cm which is the breadth? And on the vertical portion of the T section, you will discover that the 3.5cm which is the length is almost the same dimension as one of the 2cm which forms part of the length of the horizontal portion.

Nice observation

guobe:
One of the most difficult mathematical questions in primary school was finding the perimeter and area of a T sectional object.
Who can solve this.

I found it easy by dividing the T into two or three triangles and add their Areas together

For perimeter....just get the missing side and add all sides starting in one direction

A = (2x6) + (3.5 x2) = 19 cm². The diagram dimentions and drawing were poorly done

Perimeter = 6 + (3.5×2) + (2×5) = 23cm

franchasng:
13. Answer is 19cm² na.

The area of a rectangle = L x W

There are two rectangles there with:

Rectangle 1 having dimension: 6 x 2

Rectangle 2 having dimension: 3.5 x 2

Area of rectangle 1 = 6 x 2 = 12cm²

Area of rectangle 2 = 3.5 x 2 = 7cm²

Area of the diagram = area of rectangle 2 + area of rectangle 2 = 12cm² + 7cm² = 19cm²


This is very simple to solve

Pay attention to details

Solve for Area and Perimeter

femi4:
I found it easy by dividing the T into two or three triangles and add their Areas together

For perimeter....just get the missing side and add all sides starting in one direction

A = (2x6) + (3.5 x2) = 19 cm². The diagram dimentions and drawing were poorly done

Perimeter = 6 + (3.5×2) + (2×5) = 23cm

Why 3 5 x2 and 2 x 5

guobe:
Why 3 5 x2 and 2 x 5

5 sides (highlighted in yellow) is 2cm each and the opposite vertical side (highlighted in red) is 3.5cm each

femi4:
5 sides (highlighted in yellow) is 2cm each and the opposite vertical side (highlighted in red) is 3.5cm each

Lovely, best explanation so far,seems you are a lecturer

guobe:
Lovely, best explanation so far,seems you are a lecturer

Well, used to be one

angelboy01:
You meant 2003. Me I be Baba now with kids oh. I don marry two wives sef ontop. Lol

Yeah bro. 2003