__1.Area of the rectangular wire=32 * 18 m²=576 m²__

__Perimeter of the wire=2(32+18) m=100 m__

__which is equal to the perimeter of the square (Since the rectangle is bent to form this).__

__Measure of each side=100/4 m =25 m__

__Area of the square=25 *25 m²=625 m²__

__Area of the square>Area of the rectangle__

__2.Area of the 4 walls=2(25+10)5 m² =350 m² __

__Therefore, total area to be whitewashed=350 m²__

__3.Area of the room=500 *35 dm²__

__Breadth of carpet=7 dm__

__Req.Length of carpet=Area of the room/Breadth__

__=500*35/7 dm=2500 dm=50 m__

__4.Area of rectangle ABCD=26 *15 cm²__

__=390 cm²__

__Base of ΔEDA=15 cm (opposite sides are equal)__

__Height of ΔEDA=8 cm(given)__

__therefore, Area of ΔEDA=bh/2 cm=60 cm².__

__Area of shaded part=390 cm²-60 cm²=330 cm²__

__5.Area of the quadrilateral=__

__Sum of areas of 2 triangles formed.__

__=14.2/2*84+5.8*84/2 cm²__

__=840 cm²__

__6.a>The unshaded Δ's are right angled.__

__Their total area=7*4/2+6*3/2 cm²__

__=23 cm²__

__Area of the rectangle=(4+3)*( 7+6) cm²__

__=7*13 cm²__

__=91 cm²__

__Therefore, area of shaded part=91-23=68 cm²__

__b>ΔYPX and ΔXMN are right-angled.__

__XM=√NX²+MN² (By pythagoras theorem, in a right angled Δ)__

__=√225 cm²+64 cm²__

__=√289 cm²__

__=13 cm__

__And,__

__XY=√YP²+PX² (By pythagoras theorem, in a right angled Δ)__

__=√100+36 cm²__

__=√136 cm²__

__=11.66 cm (approximated , since √136 is irrational)__

__Area of the shaded portion=11.66*13 / 2 cm²__

__=75.79 cm² (approx)__