Question

in the given figure ABCD is a trapezium in इन व्हिच AE इक्वल टू 3 सेंटीमीटर EB इक्वल टू 5 सेंटीमीटर BC इज इक्वल टू 5 सेंटीमीटर एंड AE is वन एंड हॉफ टाइम एस लोंग एस DC
Find,

1. द लेंथ ऑफ डीसी
2.द रेश्यो ऑफ AECD to that of abcd ​

Answer 1

hi bro here is your ANS..

For better understanding of the solution see the attached figure of the problem :

Given : m∠ABC = 90°, AD = 16 cm, AC = 41 cm and BC = 40 cm

To find : Area of trapezium ABCD

Solution : Since, m∠ABC is 90°

So, AB is the height of the trapezium ABCD

Now, to find AB : By using Pythagoras theorem in ΔABC

AC² = AB² + BC²

41² = AB² + 40²

⇒ AB² = 41² - 40²

⇒ AB = 9 cm

Now, Area of the trapezium is given by the formula :

$$\begin{lgathered}Area = \frac{1}{2}\times Height\times \text{(Sum of parallel sides)}\\\\\implies Area = \frac{1}{2}\times AB\times (AD+BC)\\\\\implies Area = \frac{1}{2}\times 9\times (40+16)\\\\\implies Area = \frac{1}{2}\times 9\times 56\\\\\implies Area =9\times 28\\\\\bf Area = 252\thinspace cm^2\end{lgathered}$$

Hence, The area of the given trapezium ABCD = 252 cm²

please mark me brainliest please...