Calculate the area of a trapezium in cm²,lengths of whose parallel si

1.	Calculate the area of a trapezium in cm²,lengths of whose parallel sides (represented by a and b ) and distance between them (h) are: (1.)a=45cm,b=32cm=11cm. (explain in detail) with formula
Answer» ${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$ GIVEN: Length of parallel sides = 32CM & 45cm Height = 11cm To FIND: AREA of trapezium Solution: $\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){$\bf A $}\put(3,2.4){$\bf D $}\put(-0.3,-0.3){$\bf B$}\put(4,-0.3){$\bf C$}\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){$\bf 11\ cm$}\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){$\bf 45\ cm $}\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){$\bf 32\ cm $}\end{picture}\\\\$ $\text{We know that,}\\\longrightarrowArea\: of \:a \:trapezium = \dfrac{1}{2} * (sum\:of\:parallel\:sides) * height. \\So,\\\\\implies Area\: of \:the\:trapezium = \dfrac{1}{2} * (45+32) * 11 \\\\\implies Area\: of \:the\:trapezium = \dfrac{1}{2} * 77 * 11 \\\\\implies Area\: of \:the\:trapezium = 847/2 \\\\\bf{\implies Area\: of \:the\:trapezium = 423.5cm^2}$ HOPE it helps!

Calculate the area of a trapezium in cm²,lengths of whose parallel sides (represented by a and b ) and distance between them (h) are: (1.)a=45cm,b=32cm=11cm. (explain in detail) with formula

Answer»

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

GIVEN:

Length of parallel sides = 32CM & 45cm
Height = 11cm

Solution:

$\setlength{\unitlength}{1.5cm}\begin{picture}\thicklines\qbezier(0,0)(0,0)(1,2.2)\qbezier(0,0)(0,0)(4,0)\qbezier(3,2.2)(4,0)(4,0)\qbezier(1.5,2.2)(0,2.2)(3,2.2)\put(0.8,2.4){$\bf A $}\put(3,2.4){$\bf D $}\put(-0.3,-0.3){$\bf B$}\put(4,-0.3){$\bf C$}\put(4.4,0){\vector(0,0){2.2}}\put( 4.4, 0){\vector(0,-1){0.1}}\put(4.6,1){$\bf 11\ cm$}\put(0, -0.5){\vector(1,0){4}}\put(0, -0.5){\vector( - 1, 0){0.1}}\put(1.7, - 0.9){$\bf 45\ cm $}\put(0.8, 2.8){\vector(1,0){2.5}}\put(0.8, 2.8){\vector( - 1, 0){0.1}}\put(1.7, 3){$\bf 32\ cm $}\end{picture}\\\\$

$\text{We know that,}\\\longrightarrowArea\: of \:a \:trapezium = \dfrac{1}{2} * (sum\:of\:parallel\:sides) * height. \\So,\\\\\implies Area\: of \:the\:trapezium = \dfrac{1}{2} * (45+32) * 11 \\\\\implies Area\: of \:the\:trapezium = \dfrac{1}{2} * 77 * 11 \\\\\implies Area\: of \:the\:trapezium = 847/2 \\\\\bf{\implies Area\: of \:the\:trapezium = 423.5cm^2}$