Question

abcd is a rhombus whose diagonals ac and bd intersect at o.if sides ab =10cm. and diagonal bd=16cm.,find the area of rhombus​

Answer 1

Answer:

96 cm²

Step-by-step explanation:

The diagonals of a rhombus meet at right angles and bisect each other.

So rhombus abcd is divided into 4 congruent triangles oab, obc, ocd, oda.

So...

         area of rhombus abcd = 4 × area of triangle oab.

Triangle oab has a right angle at o.

Side ab = 10 cm.

Side ob = half of bd = 8 cm.

Side oa = √( 10² - 8² ) = √( 100 - 64 ) = √36 = 6 cm.   [ Pythagoras' Theorem ]

Area of right angled triangle oab = (1/2) × ob × oa = (1/2) × 8 × 6 = 24 cm².

So finally,

        area of rhombus abcd = 4 × 24 = 96 cm².

Hope this helps!