मराठी
सी.आई.एस.सी.ई.आयसीएसई ICSE Class 8

A Field is in the Shape of a Quadrilateral Abcd in Which Side Ab = 18 M, Side Ad = 24 M, Side Bc = 40m, Dc = 50 M and Angle a = 90°. Find the Area of the Field. - Mathematics

Advertisements
Advertisements

प्रश्न

A field is in the shape of a quadrilateral ABCD in which side AB = 18 m, side AD = 24 m, side BC = 40m, DC = 50 m and angle A = 90°. Find the area of the field.

बेरीज

उत्तर

Since ∠A = 90°
By Pythagorus Theorem,
In ∆ABD,

Now, the area of ΔABD = `1/2 (18)(24)`

= (18) (12) = 216 m2

In ΔABD

BD = `sqrt("AB"^2 + "AD"^2) = sqrt(18^2 + 24^2)`

= `sqrt(324 + 576) = sqrt(900) = 30`m.

S = `120/20 = 60 cm^2`

 =`sqrt( 60xx10xx20xx30)`

= `sqrt(10xx2xx3xx10xx10xx2xx10xx3)`

= 10 × 10 × 6

= 600 cm2

Area of quadrilateral ABCD

= 816cm2

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Area of a Trapezium and a Polygon - Exercise 20 (A) [पृष्ठ २२४]

APPEARS IN

सेलिना Concise Mathematics [English] Class 8 ICSE
पाठ 20 Area of a Trapezium and a Polygon
Exercise 20 (A) | Q 9 | पृष्ठ २२४
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×