Question

ABCD is a rectangle, F is the midpoint of AB, and BC is extended to X, and BC =
14
5
cm. 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?

Answer 1

Answer:

नमस्ते वहाँ मेरे दोस्त have मैं एक समय के लिए आपके पास रहा हूं नतीजतन मेरे पास एक महान सप्ताहांत है मुझे कम से कम कोई कल्पना नहीं है मैं एक महान सप्ताहांत के लिए सही समय भेजूंगा सही जगह मैं पी होगा

Answer 2

According the question,

ABCD is a rectangle,

1. F is the midpoint of AB.
2. BC is extended to X,
3. BC =14/5cm.
4. Area of Δ AFX = 5/8 of area of rectangle ABCD.

We have to find out the length of BX ( in cm).

Area of (ΔAFX) = Area of rectangle ABCD          

we know that:-

Area of ΔAFX = $\frac{1}{2}$ x B x H

Area of rectangle ABCD = BC x AB

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

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

Hence BX= (14 ×2)/8cm = $\frac{7}{2}$cm= 3.5 cm

Ans :- The length of BX will be 3.5cm.

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