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
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.
6.a>The unshaded Δ's are right angled.
Their total area=7*4/2+6*3/2 cm²
Area of the rectangle=(4+3)*( 7+6) 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
XY=√YP²+PX² (By pythagoras theorem, in a right angled Δ)
=11.66 cm (approximated , since √136 is irrational)
Area of the shaded portion=11.66*13 / 2 cm²
=75.79 cm² (approx)