CAT 2024 Slot 1 QA Question 17

Consider. ABCD means AB and CD are the lengths, each 56 \text{ cm}, and BC and AD are the breadths, each 45 \text{ cm}. So, if I imagine the rectangle, A is the bottom-left corner, B is the bottom-right, C is the top-right, and D is the top-left. E is the midpoint of CD, so that would be halfway along the top side.

Now, triangle ADE is formed by points A, D, and E. Place point A at (0, 0). Then, since AB is 56 \text{ cm}, point B would be at (56, 0). BC is 45 \text{ cm}, so point C is at (56, 45), and point D is at (0, 45). E is the midpoint of CD, so CD goes from (56, 45) to (0, 45). The midpoint would be at ((56+0)/2, (45+45)/2) = (28, 45).

So, the coordinates are:

Now, We find the lengths of the sides of triangle ADE. Computing each side one by one.

First, side AD is from A(0,0) to D(0,45). That's straightforward; it's just the vertical distance, which is 45 \text{ cm}.

Next, side AE is from A(0,0) to E(28,45). To find this length, Using the distance formula: \. Plugging in the values, it's \$\sqrt{(28 - 0)^2 + (45 - 0)^2}$ = \$\sqrt{784 + 2025}$ = \$\sqrt{2809}$$. Hmm, 2809 is 53 squared because 53*53 is 2809. So, AE is 53 \text{ cm}.

Then, side DE is from D(0,45) to E(28,45). Since both points are on the same horizontal line, the distance is just the difference in the x-coordinates. So, DE = 28 - 0 = 28 \text{ cm}.

Triangle ADE has sides AD = 45 \text{ cm}, AE = 53 \text{ cm}, and DE = 28 \text{ cm}.

AD is vertical, so yes, 45 \text{ cm}. AE is from (0,0) to (28,45), which we calculated as 53 \text{ cm}. DE is horizontal from (0,45) to (28,45), which is 28 \text{ cm}. Now, We find the radius of the incircle of triangle ADE. the radius of the incircle (r) of a triangle is given by the formula:

r = A / s

where A is the area of the triangle, and s is the semi-perimeter.

So, first, let's find the semi-perimeter (s). The semi-perimeter is half the sum of all sides.

s = (AD + AE + DE) / 2 = (45 + 53 + 28) / 2

Compute that: 45 + 53 is 98, plus 28 is 126. So, s = 126 / 2 = 63 \text{ cm}.

Next, The area is given by (A) of triangle ADE. Since I have the coordinates of all three points, Using the shoelace formula to find the area.

The shoelace formula is given by:

A = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2|

Plugging in the coordinates: x1 = 0, y1 = 0 (point A) x2 = 0, y2 = 45 (point D) x3 = 28, y3 = 45 (point E)

So,

A = |(0(45 - 45) + 0(45 - 0) + 28(0 - 45)) / 2| = |(0 + 0 + 28(-45)) / 2| = |( -1260 ) / 2| = | -630 | = 630 \text{ cm}²

Alternatively, since triangle ADE is a right triangle? Is angle at D a right angle? AD is vertical, DE is horizontal, so yes, angle at D is 90 degrees. So, triangle ADE is a right-angled triangle at D.

Therefore, the area can also be calculated as (base * height) / 2. Here, base is DE = 28 \text{ cm}, height is AD = 45 \text{ cm}.

So, area A = (28 * 45) / 2 = (1260) / 2 = 630 \text{ cm}². Now, using the formula for the inradius:

r = A / s = 630 / 63 = 10 \text{ cm}.

So, the radius of the incircle is 10 \text{ cm}.