## BdMO National Secondary 2020 P4

Mursalin
Posts:68
Joined:Thu Aug 22, 2013 9:11 pm
BdMO National Secondary 2020 P4
$$ABCD$$ একটি বর্গক্ষেত্র। $$P$$ এবং $$Q$$ যথাক্রমে $$BC$$ এবং $$CD$$ রেখাংশের উপর দুইটি বিন্দু যাতে করে $$\angle APQ = 90^\circ$$ হয়। দেওয়া আছে যে, $$AP = 4$$ এবং $$PQ= 1$$। যদি $$AB$$-এর দৈর্ঘ্যকে লঘিষ্ঠ আকারে $$\frac{m}{n}$$ হিসেবে লেখা যায়, তবে $$m+10n$$-এর মান বের কর।

$$ABCD$$ is a square. $$P$$ and $$Q$$ are two points in segment $$BC$$ and $$CD$$ respectively such that $$\angle APQ = 90^\circ$$. It is given that $$AP = 4$$ and $$PQ = 1$$. If we express the length of segment $$AB$$ as $$\frac{m}{n}$$ in lowest terms, compute $$m+10n$$.
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Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm

### Re: BdMO National Secondary 2020 P4

This problem can be solved by making three equations with three variables. But I am still looking for an easier solution.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Ridwan056
Posts:2
Joined:Sun Jan 17, 2021 7:51 pm

### Re: BdMO National Secondary 2020 P4

I think we can use similar triangles. Triangle APB and triangle PCQ are similar. Hence, AB= 4CP, which implies BP=3CP; and BP=4CQ. so, AP^2= 16= AB^2 +BP^2. Similarly, applying like this on PCQ gives BP=12/5. Therefore, AB=16/5. The final answer is m+10n= 16+50=66.