Question

in triangle ABC A-P-B and A-Q-C such that seg pQ is parallel to aide BC and sed PQ divides triangle ABC in two parts whose areas are equal.find the value of BP/AB

Answer 1

Answer:

$\frac{BP}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}$

Step-by-step explanation:

We are given that in triangle ABC, segment PQ is parallel to side BC and it diveides the trangle ABC in two parts having area equal.

Therefore, Area triangle APQ= area quadrilateral PQCB

⇒Area triangle ABC= Area traingle APQ+ area quadrilateral PQCB

⇒Area triangle ABC=2×Area triangle APQ

⇒$\frac{Area triangle ABC}{Area triangle APQ}=\frac{1}{2}$

Now, by the property of similarity of triangles, we have

$\frac{Area triangle APQ}{Area triangle ABC}=\frac{AP^{2} }{AB^{2}}=\frac{1}{2}$

⇒$\frac{AP}{AB}=\frac{1}{\sqrt{2} }$

Now, we know that AP=AB-BP, therefore,

⇒$\frac{AB-BP}{AB}=\frac{1}{\sqrt{2}}$

⇒$1-\frac{BP}{AB}=\frac{1}{\sqrt{2} }$

⇒$1-\frac{1}{\sqrt{2}}=\frac{BP}{AB}$

⇒$\frac{BP}{AB}=\frac{\sqrt{2}-1}{\sqrt{2} }$

Answer 2

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