Q. 13.7( 33 Votes )

Find the areas of given plots. (All measures are in meters.)

(1)

(2)

Answer :

(1)



Given that,


PA = 30m, AC = 30m, and CT = 30m


PC = PA + AC = 30 + 30 = 60m


∆PCT is right angled triangle at C


Area of ∆PCT = 1/2 × PC × CT



= 900m…………(1)


In, ∆SCT is right angled triangle at C


SB = 60m, BC = 30m, and CT = 30m


Area of ∆SCT = 1/2× base × height



= ×30×90


= 1350m…………….(2)


In ∆SBR is right angled triangle at B


SB = 60m, BR = 25m


Area of ∆SBR = 1/2 × base × height


= × SB × BR


= × 60 × 25


= 750m…………..(3)


In ∆APQ is right angled triangle at A


AP = 30m, AQ = 50m


Area of ∆APQ = × base × height


= ×AP×AQ


= ×50×30


= 750m…………(4)


Now, in trapezium ABRQ


AQ and RB are the 2 parallel sides


Also, AQ = 50m and BR = 25m


AQ + BR = 75m


The distance between AQ and BR = 60m


Hence,



= 2250 sq m………….(5)


Now area of quadrilateral PQRST = (1)+(2)+(3)+(4)+(5)


= 900+1350+750+750+2250


= 6000 sq m


(2) the data for this question is inadequate.


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