Question

pls solve this question of maths​

Answer 1

Answer:

QC = 45cm

CD = 40cm

Step-by-step explanation:

$area \: of \: rectangle = l \times b$

$area \: of \: rectangle = 51cm \times 25cm$

$area \: of \: rectangle = 1275c {m}^{2}$

$area \: of \: trapezium = \frac{5}{6} \times area \: of \: rectangle$

$area \: of \: trapezium = \frac{5}{6} \times 1275c {m}^{2}$

$area \: of \: trapezium = \frac{5 \times 1275}{6} c {m}^{2}$

$area \: of \: trapezium = \frac{6375}{6} c {m}^{2}$

$area \: of \: trapezium = 1062.5c {m}^{2}$

Given,

PQ:CD=9:8

Let QC be 9x and CD be 8x

$area \: of \: trapezium = \frac{sum \: of \: parallel \: sides}{2} \times height$

$area \: of \: trapezium = \frac{9x + 8x}{2} \times 25 \\1062.5 = \frac{17x}{2} \times 25 \\ \frac{1062.5}{25} = \frac{17x}{2} \\ 42.5 \times 2 = 17x \\ 85 = 17x \\ x = 85 \div 17 \\ x = 5cm$

$QC = 9x \\ 9x = 9 \times 5cm \\ = 45cm$

$CD = 8x \\ 8x = 8 \times 5cm \\ = 40cm$

Therefore,

• QC = 45cm
• CD = 40cm