Question

ABCD is a rectangle, F is the midpoint of AB, and BC is extended to X, and BC = 14 /5cm. What is the
length of BX (in cm) for which the area of triangle AFX is 5/8 of the area of the rectangle ABCD?
11
X​

Answer 1

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Answer 2

Step-by-step explanation:

Given that,

• F is midpoint of AB.
• BC=14/5cm
• Area (triangle AFX) = 5/8 x Area (rectangle ABCD

To find-

• Length of BX i.e. l(BX)

Solution:

As it is given that,

A(ΔAFX) = A(rect ABCD)

we know that,

A(ΔAFX) = $\frac{1}{2}$ x B x H and

A(rect ABCD) = BC x AB

∴  $\frac{1}{2}$ x BX x AB =  $\frac{5}{8}$ x $\frac{14}{5}$ x AB

Solving this,

BX = $\frac{5}{8}$ x $\frac{14}{5}$ x 2

∴ length(BX) = $\frac{7}{2}$cm