Question

Triangles A and B have vertical heights x cm and (x+3) cm respectively. The areas of Triangles A and B are 30cm² and 32cm² respectively. (a) Find, in terms of x, an expression for the base of: (i) Triangle A (ii) Triangle B (b) Given that the base of Triangle B is 4 cm less than the base of Triangle A, form an equation and show that it reduces to x² + 4x - 25 = 0 (c) Solve the equation and hence, find the height of Triangle B.​

Answer 1

Answer:

Hi,

Step-by-step explanation:

Let say a the base of the triangle A and b the base of the triangle B.

$a)\\\left\{\begin{array}{ccc}\dfrac{a*x}{2}&=&30\\\dfrac{b*(x+3)}{2}&=&32\\b=a-4\\\end{arraqy}\right.\\\\(i): a=\dfrac{60}{x}\\\\(ii): b=\dfrac{64}{x+3}\\\\b)\\\dfrac{64}{x+3}=\dfrac{60}{x}-4\\\\64x=(60-4x)*(x+3)\\\\4x^2+16x-180=0\\\\x^2+4x-45=0\\\\\Delta=16+4*45=196=14^2\\x=\dfrac{-4+14}{2}=5\ or\ x=\dfrac{-4-14}{2}=-9\ (impossible)\\So\ x=5,\ a=\dfrac{60}{5}=12, b=\dfrac{64}{5+3}=8\\\\Area\ of\ A=12*5/2=30\\Area\ of\ B=8*8/2=32\\\\Height\ of\ triangle\ B=x+3=5+3=8.$