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Question
The figure shows the cross section of 0.2 m a concrete wall to be constructed. It is 0.2 m wide at the top, 2.0 m wide at the bottom and its height is 4.0 m, and its length is 40 m. Calculate the cross sectional area
Solution
Complete the diagram as shown:
Ler AD = x m
AB = AD + DB
= ( x +4)m
BC = `(1)/(2) xx "QC"`
= `(1)/(2) xx 2`
= 1m
DE = `(1)/(2) xx "PE"`
= `(1)/(2) xx 0.2`
= 0.1m
In ΔADE and ΔABC,
∠ADE = ∠ABC ...(90°each)
∠DAE - ∠BAC ...(Common angle)
∴ ΔADE ∼ ΔABC by AA test
∴ `"AD"/"AB" = "DE"/"BC"` ...(C.S.S.T.)
`(x)/(x + 4) = (0.1)/(1)`
`(10x)/(x + 4) = (10 xx 0.1)/(1)` ...Multiply by 10 on both sides
10x = x + 4
9x = 4
x = `(4)/(9)"m"`
∴ AB = `(4)/(9) + 4`
= `(40)/(9)"m"`
Area of the cross section of the wall
= A(ΔAQC) - A(ΔAPE)
= `(1)/(2) xx "QC" xx "AB" - (1)/(2) xx "PE" xx "AD"`
= `(1)/(2) xx 2 xx (40)/(9) - (1)/(2) xx 0.2 xx (4)/(9)`
= `(40)/(9) - (0.4)/(9)`
= `(39.6)/(9)`
= 4.4
∴ The area of the cross section of the wall is 4.4sq.m.
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