Two metal strips are cut to a width of 4.5 cm. The strips are laid across one another at an angle of 40 °. Determine the area of the region that overlaps.

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Let srips are overlaped in such a way that one end of ne strip lies on the side of another strip. So one edge inclined to another side by 40 degree. Another sde of strip produced and so parallelogram formed.

Thus overlaped area is difference between area of parallelogram and triangle so frmed by extending the side

Length of the side of parallelogram= 4.5 Sec(40)=7

moreover this parallelogram is rhombus.

Thus area of parallelogram= 4.5 x 7 sq cm

=31.5 sq cm

Area of the triangle s formed= (1/2)x 4.5 cos(40) x7

=12.07 sq cm

**Thus overlaped area=31.5 - 12.07 = 19.43 sq cm.**

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