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CAT 2024 Slot 1 — QA Question 14

Geometric CentersEasy

ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, in cm, of radius of incircle of ∆ADE is

Answer & solution

Answer: 10

Solution

Medium

Place the rectangle on coordinates to get triangle ADEADE's three side lengths. Then use r=Areasr=\dfrac{\text{Area}}{s} (area over semiperimeter). The right angle at DD makes the area trivial.

A B C D E
1

Find the sides of △ADE\triangle ADE. AD=BC=45AD=BC=45. DE=12CD=12(56)=28DE=\tfrac12 CD=\tfrac12(56)=28. The angle at DD is 90∘90^\circ (rectangle corner), so AEAE is the hypotenuse.

AD=45,DE=28⇒ AE=452+282=2025+784(Pythagoras)⇒ AE=2809=53\begin{aligned} &AD=45,\quad DE=28\\ &\Rightarrow\ AE=\sqrt{45^2+28^2}=\sqrt{2025+784} \quad\text{(Pythagoras)}\\ &\Rightarrow\ AE=\sqrt{2809}=53 \end{aligned}
2

Inradius of a right triangle. With legs a,ba,b and hypotenuse cc, r=a+b−c2r=\dfrac{a+b-c}{2}.

r=AD+DE−AE2⇒ r=45+28−532(from step 1)⇒ r=202=10\begin{aligned} &r=\dfrac{AD+DE-AE}{2}\\ &\Rightarrow\ r=\dfrac{45+28-53}{2} \quad\text{(from step 1)}\\ &\Rightarrow\ r=\dfrac{20}{2}=10 \end{aligned}
r=10 cmr=10\text{ cm}

(45,28,53)(45,28,53) — wait, check it's a clean Pythagorean set: 452+282=2809=53245^2+28^2=2809=53^2. For any right triangle use r=a+b−c2r=\tfrac{a+b-c}{2} directly and skip the area/semiperimeter computation.

CAT 2024 Slot 1 QA Q14: ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. Then, the length, — Solution | TheCATExam