CAT 2025 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.
Question 1 : A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the sum of the lengths, in cm, of all three altitudes of the triangle ABC is
Since \(\triangle ABC\) is isosceles with \(AB=AC=50\) cm, dropping altitude \(AD\) to base \(BC\) bisects \(BC\):
\(BD=DC=\frac{80}{2}=40\) cm.
Using Pythagoras theorem in right-angled \(\triangle ABD\):
\(AD=\sqrt{50^2-40^2}=\sqrt{2500-1600}=\sqrt{900}=30\) cm.
Area of \(\triangle ABC=\frac{1}{2}\times80\times30=1200\text{ cm}^2\).
Since \(AB=AC=50\) cm, the corresponding altitudes to \(AB\) and \(AC\) are equal:
\(h_B=h_C=\frac{2\times1200}{50}=48\) cm.
Sum of all three altitudes \(=30+48+48=126\) cm.
The question is " A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the sum of the lengths, in cm, of all three altitudes of the triangle ABC is "
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