Question

Find the area of a trapezium with parallel side of 40 & 57 & non-parallel sides of 26cm & 25cm.

Answer 1

Solution -

For a trapezium , the parallel sides are 40 cm and 50 cm respectively .

The non parallel sides are 25cm and 26 cm respectively.

We have to find the area of the trapezium .

$\setlength{\unitlength}{1.1 cm}\begin{picture}(0,0)\linethickness{0.4mm}\qbezier(0,0)(0,0)(1.5,3)\qbezier(7,0)(7,0)(5.5,3)\qbezier(1.5,3)(1.5,3)(5.5,3)\qbezier(7,0)(7,0)(0,0)\put(1.3,3.2){ \sf A }\put(-0.35,-0.35){ \sf B }\put(1.5,-0.4){ \sf C }\put(5.3,-0.4){ \sf D }\put(5.45,3.15){ \sf F }\put(7.2,-0.34){ \sf E }\qbezier(5.5,0)(5.5,0)(5.5,3)\qbezier(1.5,0)(1.5,0)(1.5,3)\put(3.2, 3.2){ \sf 40 \: cm }\put(2, - 0.3){\vector(1,0){3}}\put(2, - 0.3){\vector(-1,0){0}}\put(3.2,-0.7){ \sf 40 \: cm }\put(5.7,-0.3){\vector(1,0){1.2}}\put(5.7,-0.3){\vector(-1,0){0}}\put(6,-0.7){ \sf 8.5 \: cm }\put(0.2, - 0.3){\vector(1,0){1.2}}\put(0.2,-0.3){\vector(-1,0){0}}\put(0.55,-0.7){ \sf 8.5 \: cm }\put(0.2, - 1){\vector(1,0){6.7}}\put(0.2, - 1){\vector(-1,0){0}}\put(3,-1.45){ \sf 57 \: cm }\put(1.7, 1.4){ \sf h \: cm }\put(4.55,1.4){\sf h \:cm}\end{picture}$

We have to find the height of the trapezium .

Consider any of the right angled triangles .

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf h \: cm}\put(2.8,.3){\large\bf 8.5 \: cm }\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \Theta }\end{picture}$

h² + (8.5)² = 25×25

> h² + 72.25 = 625

> h² = 552.75

> h ≈ 23.5 cm

Area of the trapezium = ½( a + b) h

> ½ × 97 × 23.5

> 929 cm² approximately .

This is the required answer .

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