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secondary 4 | E Maths
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Can someone please help me with qns 2? Sry if the qns seem a bit blurry. I really appreciate it if u took the time to help me out :)) thank uu
Date Posted: 3 years ago
Views: 524
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A) CD = Perimeter - the rest of the length
= (20x - 3) - (2x + 6) - (4x+3) - (5x-3)
= 9x - 9 cm//
B) AD = BC
hence 2x + 6 = 5x - 3
3x = 9
x = 3cm
:. perimeter
= 20(3) - 3
= 57cm //
C) to find area of trapezium,
We need 2 parallel lengths & perpendicular breadth.
In other words, we need to find AX OR BY
Focusing on AX:
(1) We will find DX first
DC = AB + DX + CY
= AB + 2 DX
from (a):
9x - 9 = 4x + 3 + 2DX
Since x = 3:
27 - 9 = 12 + 3 + 2DX
DX = 1.5cm
Since DX = 1.5cm, AD = 12cm,
By Pythagoras theorem:
AX = square root (12^2 - 1.5^2)
= 11.906cm
:. Area of trapezium
= 0.5 * 11.906 * (AB + CD)
= 0.5 * 11.906 * 33
= 196cm^2 (3sf) //
= (20x - 3) - (2x + 6) - (4x+3) - (5x-3)
= 9x - 9 cm//
B) AD = BC
hence 2x + 6 = 5x - 3
3x = 9
x = 3cm
:. perimeter
= 20(3) - 3
= 57cm //
C) to find area of trapezium,
We need 2 parallel lengths & perpendicular breadth.
In other words, we need to find AX OR BY
Focusing on AX:
(1) We will find DX first
DC = AB + DX + CY
= AB + 2 DX
from (a):
9x - 9 = 4x + 3 + 2DX
Since x = 3:
27 - 9 = 12 + 3 + 2DX
DX = 1.5cm
Since DX = 1.5cm, AD = 12cm,
By Pythagoras theorem:
AX = square root (12^2 - 1.5^2)
= 11.906cm
:. Area of trapezium
= 0.5 * 11.906 * (AB + CD)
= 0.5 * 11.906 * 33
= 196cm^2 (3sf) //
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