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

(1)

(2)

MHB - Mathematics (New)