Question

in the given figure, ABCD is a rectangle of length 51 cm and breadth 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9:8 is cut off from the rectangle, as shown in the figure . If the area of the trapezium PQCD is 5/6 th part of area of the rectangle, find the lengths QC and PD..

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Answer 1

Given

• Rectangle Length = 51 cm
• Rectangle Breadth = 25 cm
• Parallel sides of Trapezium = 9:8
• Trapezium Area = 5/6th of Rectangle Area

Explanation:

Area of the Rectangle ABCD

$\colon\implies{\sf{ Length \times Breadth }} \\ \\ \colon\implies{\sf{ 51 \times 25 }} \\ \\ \colon\implies{\sf{ 1275 \ cm^2 }}$

Let QC and PD be 9x and 8x

$\maltese \ {\pink{\boxed{\underline{\sf{ \dfrac{1}{2} \times (Sum \ of \ Parallel \ Sides) \times Height }}}}}$

$\colon\implies{\sf{ \dfrac{1}{2} \times (QC+PD) \times 25 }} \\ \\ \\ \colon\implies{\sf{ \left[ \dfrac{1}{2} \times (9x+8x) \times 25 \right] cm^2 }} \\ \\ \\ \colon\implies{\sf{ \left[ \dfrac{17 \times 25}{2} \right] x \ cm^2 }}$

Now, Ar( PQCD ) = 5/6th × (Area of Rectangle ABCD) :-

$\colon\implies{\sf{ \left( \dfrac{17 \times 25}{2} \right) x = \left( \dfrac{5}{6} \times 1275 \right) }} \\ \\ \\ \colon\implies{\sf{ x = \left( \dfrac{5}{6} \times 1275 \times \dfrac{2}{17 \times 25 } \right) }} \\ \\ \\ \colon\implies{\boxed{\mathfrak\pink{ x = 5 }}}$

$\maltese \ {\sf{ QC = (9x) cm = (9 \times 5) = 45 \ cm }} \\ \\ \maltese \ {\sf{ PD = (8x) cm = (8 \times 5) = 40 \ cm }}$

Hence,

• The Length of the QC and PD are 45cm and 40cm Respectively.