Question

From the adjoining sketch, calculate (i)the length of AD (ii)the area of trapezium ABCD (iii)the area of triangle BCD

Answer 1

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Answer 2

Answer:

Length of DA is 9 cm

$\text{Area of trapezium= }247.5 cm^2$

$\text{Area of triangle }=67.5 cm^2$

Step-by-step explanation:

Given the trapezium ABCD in which parallel sides are of length 15 cm and 40 cm. Also, length of BD is 41 cm.

We have to find the

1) length of AD

As DAB is right angled triangle

∴ $DB^2=DA^2+AB^2$

⇒ $41^2=DA^2+40^2$

⇒ $1681-1600=DA^2$

⇒ $DA=\sqrt{81}=9cm$

2)To find area of trapezium

$\text{Area of trapezium= }\frac{1}{2}\times (\text{Sum of parallel sides})\times height$

$=\frac{1}{2}\times (AB+CD)\times AD$

$=\frac{1}{2}\times (40+15)\times 9$

$=\frac{1}{2}\times 55\times 9=247.5 cm^2$

3) Area of triangle BCD

$\text{Area of triangle ABD }=\frac{1}{2}\times base \times height$

$=\frac{1}{2}\times40\times 9=180 cm^2$

Area of ΔBCD=247.5-180=67.5 square centimeter.