Let s be the measure of the measure of the sides of the square such that its area is s². From the conditions above, the length of the rectangle is s + 5 and its width is 0.5s. The area of the rectangle is (s+5)(0.5s)
s² = (s + 5)(0.5s)
The value of s is 5. Thus the dimension of the rectangle in w x h is 2.5 x 10.
Using the Pythagoras theorem since it is given that it is a right angled triangle
let us name the triangle as A, B, C with B as right angle
AB^2 + BC^2 = AC^2 ( Pythagoras theorem)
where AB = 4 , BC = 7 , AC = x
Answer:
Step-by-step explanation:
![\frac{15}{\sqrt{31} - 4}\\\\=\frac{15}{\sqrt{31} - 4} \times \frac{\sqrt{31} + 4}{\sqrt{31}+ 4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ rationalizing \ the \ denominator \ ]\\\\=\frac{15( \sqrt{31} + 4 )}{(\sqrt{31})^2 - (4)^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ (a-b)(a+b) = a^2 - b^2 \ ]\\\\=\frac{15 ( \sqrt{31} + 4)}{31 - 16}\\\\=\frac{15 (\sqrt{31} + 4)}{15}\\\\= \sqrt{31} + 4](https://tex.z-dn.net/?f=%5Cfrac%7B15%7D%7B%5Csqrt%7B31%7D%20-%204%7D%5C%5C%5C%5C%3D%5Cfrac%7B15%7D%7B%5Csqrt%7B31%7D%20-%204%7D%20%5Ctimes%20%5Cfrac%7B%5Csqrt%7B31%7D%20%2B%204%7D%7B%5Csqrt%7B31%7D%2B%204%7D%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%20%5B%20%20%5C%20rationalizing%20%5C%20the%20%5C%20denominator%20%5C%20%5D%5C%5C%5C%5C%3D%5Cfrac%7B15%28%20%5Csqrt%7B31%7D%20%2B%204%20%29%7D%7B%28%5Csqrt%7B31%7D%29%5E2%20-%20%284%29%5E2%7D%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%5C%20%5B%20%5C%20%28a-b%29%28a%2Bb%29%20%3D%20a%5E2%20-%20b%5E2%20%5C%20%5D%5C%5C%5C%5C%3D%5Cfrac%7B15%20%28%20%5Csqrt%7B31%7D%20%2B%204%29%7D%7B31%20-%2016%7D%5C%5C%5C%5C%3D%5Cfrac%7B15%20%28%5Csqrt%7B31%7D%20%2B%204%29%7D%7B15%7D%5C%5C%5C%5C%3D%20%5Csqrt%7B31%7D%20%2B%204)
(5.1x10^8) - (3.61x10^8)
(5.1-3.61) x 10^8
1.49 x 10^8 = Final answer
