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Solution for Area of Pqrs = Area of Pqr + Area of Ξ΄Pqs = (6+9.166)π‘π‘š2=15.166π‘π‘š2 - CBSE Class 9 - Mathematics

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Area of PQRS = Area of PQR + Area of ΔPQS = (6+9.166)π‘π‘š2=15.166π‘π‘š2

The sides of a quadrangular field, taken in order are 26 m, 27 m, 7 m are 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.

Solution

The sides of a quadrilateral field taken order as AB = 26m

BC = 27 m
CD = 7m and DA = 24 m
Diagonal AC is joined
Now ΔADC
By applying Pythagoras theorem

⇒`AC^2=AD^2+CD^2`

⇒`AC=sqrt(AD^2 +CD^2)`

⇒`AC=sqrt(24^2+7^2)`

⇒`AC=sqrt625=25 m`

Now area of ΔABC

`S=1/2(AB+BC+CA)=1/2(26+27+25)=78/2=39 m`

By using heron’s formula

Area (ΔABC) = `sqrt(S(S-AD)(S_BC)(S-CA))`

`=sqrt(39(39-26)(39-21)(39-25))`

`sqrt(39xx14xx13xx12xx1)`

`=291.849cm^2`

Now for area of ΔADC

`S=1/2(AD+CD+AC)`

`=1/2(25+24+7)=28cm`

By using heron’s formula

`∴ Area   of   ΔADC = sqrt(S(S-AD)(S-DC)(S-CA))`

`=sqrt(28(28-24)(28-7)(28-25))`

`=84m^2`

∴ Area of rectangular field ABCD = area of ΔABC + area of ΔADC
= `291.849+84`
= `375.8m^2`

 

 

 

 

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APPEARS IN

Solution Area of Pqrs = Area of Pqr + Area of Ξ΄Pqs = (6+9.166)π‘π‘š2=15.166π‘π‘š2 Concept: Application of Heron’S Formula in Finding Areas of Quadrilaterals.
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