Step-by-step explanation:
Given :-
The length of the garden 8m greater than 2 times the width.
Area of the garden is 280 m²
Let us consider the length as x and width as y.
Sp, we can day length as :-
x = 8 + 2y ---(1)
Now, we know that:-
Area of Rectangle = Length × Breadth
280 = x * y
We can replace the value of x now,
280 = y × ( 8 + 2y)
280 = 8y + 2y²
2y² + 8y - 280 = 0
y² + 4y - 140 = 0
Factorise it.
(y -10)(y + 14)
Cancelling -ve value, we get the width as 10 metres.
<u>Hope</u><u> </u><u>it</u><u> </u><u>helps</u><u> </u><u>:</u><u>)</u>
Answer:
Step-by-step explanation:
A. The measure of RS
As both polygons are similar, corresponding sides will be same
In order to form triangle PQT and quadrilateral TQRS, point T must lie on line PS which is 16 cm. long.
If the ratio of PT to TS is 5:3 and the total length of PS is 16, then PT must be 10 and TS must be 6 (10 + 6 =16) and 10:6 is the same ratio as 5:3. Another way to think about it is 5/3 = 10/6.
Now you have all the lengths that you need to find the areas of the quadrilateral and the triangle.
Make sure you draw a diagram of it!!
first you put 12 over 100 and that is 12/100 and then simplify
