CAT — Triangles

Given: The vertices of the triangle are
A(1, 2), B(7, 2), and C(1, 10).
Find the radius of the incircle.

Step 1: Find the lengths of the sides
AB = 7 − 1 = 6
AC = 10 − 2 = 8
Using the distance formula,
BC = √[(7 − 1)² + (2 − 10)²]
= √(36 + 64)
= 10
Thus, the triangle is a right-angled triangle with sides 6, 8 and 10.

Step 2: Find the area
Area
= (1/2) × 6 × 8
= 24 sq. units

Step 3: Find the semiperimeter
Semiperimeter
= (6 + 8 + 10)/2
= 12

Step 4: Find the inradius
For any triangle,
Inradius = Area / Semiperimeter
= 24/12
= 2

Answer: 2

Given:
Three equal circles of radius r touch each other externally.
Two more circles, X and Y, are drawn such that each touches all the three circles.
Radius of X is greater than that of Y.
Find the ratio of the radii of X and Y.

Step 1: Locate the centres
The centres of the three equal circles form an equilateral triangle of side
2r.
The centre of both circles X and Y is the common centre (circumcentre) of this equilateral triangle.
The distance from this centre to each vertex is the circumradius of the equilateral triangle.
Circumradius
= (2r)/√3
= 2r/√3

Step 2: Find the radius of the larger circle X
Since X encloses the three circles, the distance between its centre and the centre of each small circle is
R − r,
where R is the radius of X.
Hence,
R − r = 2r/√3
R = r + 2r/√3
= r(1 + 2/√3)

Step 3: Find the radius of the smaller circle Y
Since Y lies inside the gap, the distance between its centre and the centre of each small circle is
r + y,
where y is the radius of Y.
Hence,
r + y = 2r/√3
y = 2r/√3 − r
= r(2/√3 − 1)

Step 4: Find the ratio
R : y
= (1 + 2/√3) : (2/√3 − 1)
Multiply both terms by √3,
= (√3 + 2) : (2 − √3)
Rationalizing,
= (√3 + 2)² : (4 − 3)
= (3 + 4 + 4√3) : 1
= (7 + 4√3) : 1

Answer:
A. (7 + 4√3) : 1

Given:
ABCD is a trapezium with
AB ∥ CD
AB = 2 cm
CD = 1 cm
Perimeter of ABCD = 6 cm.
The extensions of AD and BC meet at E.
Find the perimeter of △AEB.

Step 1: Use similarity of triangles
Since
AB ∥ CD,
triangles AEB and DEC are similar.
Hence,
AB/CD = AE/DE = BE/CE
Substituting the given values,
AE/DE = BE/CE = 2/1
Therefore,
AE = 2DE
and
BE = 2CE

Step 2: Express AD and BC
Since D lies on AE,
AD = AE − DE
= 2DE − DE
= DE
Thus,
AD = DE.
Similarly,
BC = BE − CE
= 2CE − CE
= CE
Thus,
BC = CE.
Hence,
AE = 2AD
and
BE = 2BC.

Step 3: Use the perimeter of the trapezium
Perimeter of ABCD
= AB + BC + CD + AD
6 = 2 + BC + 1 + AD
AD + BC = 3

Step 4: Find the perimeter of △AEB
Perimeter
= AE + BE + AB
= 2AD + 2BC + 2
= 2(AD + BC) + 2
= 2 × 3 + 2
= 8 cm

Answer:
A. 8

Step 1: Find the sides of △ADE
E is the midpoint of CD → DE = 56/2 = 28 cm
AD = 45 cm
AE² = AD² + DE² = 45² + 28² = 2025 + 784 = 2809
AE = 53 cm
Sides of △ADE: 28 cm, 45 cm and 53 cm.

Step 2: Find the area of the triangle
Since AD ⊥ DE:
Area = (1/2) × 45 × 28 = 630 cm²

Step 3: Find the semiperimeter
s = (28 + 45 + 53)/2 = 63 cm

Step 4: Find the inradius
Area = r × s → 630 = r × 63 → r = 10 cm

Step 1: Altitude from A (to BC)
AB = AC = 50 cm, BC = 80 cm → altitude bisects BC (BD = DC = 40).
Altitude from A = √(50² − 40²) = √(2500 − 1600) = √900 = 30 cm

Step 2: Area = (1/2) × 80 × 30 = 1200 cm²

Step 3: Altitude from B (to AC = 50 cm)
1200 = (1/2) × 50 × h₁ → h₁ = 48 cm
By symmetry (AB = AC), altitude from C = 48 cm also.

Step 4: Sum = 30 + 48 + 48 = 126 cm

Since one side is the diameter, the angle opposite it is 90° (Thales' theorem).
The triangle is right-angled with perpendicular sides λa and λb.
Hypotenuse = 2r (diameter).

Using Pythagoras:
(λa)² + (λb)² = (2r)²
λ²(a² + b²) = 4r² → λ² = 4r²/(a² + b²)

Area = (1/2) × λa × λb = (1/2) × λ²ab
= (1/2) × (4r²/(a² + b²)) × ab
= 2abr²/(a² + b²)

Step 1: Total area of rectangle
AB = 9 cm, BC = 6 cm → Area = 9 × 6 = 54 sq cm

Step 2: Determine areas using geometric progression
Let A1 = Area(ABP), A2 = Area(APQ), A3 = Area(AQCD).
A1, A2, A3 are in GP with common ratio r, and A3 = 4 × A1.
A1 × r² = 4 × A1 → r² = 4 → r = 2
So: A1 = a, A2 = 2a, A3 = 4a

Step 3: Solve for a
a + 2a + 4a = 54 → 7a = 54 → a = 54/7 sq cm

Step 4: Find segment lengths
Triangles ABP and APQ have height AB = 9 cm.
BP: Area(ABP) = (1/2) × BP × 9 → 54/7 = 4.5 × BP → BP = 12/7 cm
PQ: Area(APQ) = (1/2) × PQ × 9 → 108/7 = 4.5 × PQ → PQ = 24/7 cm
QC: BC − BP − PQ = 6 − 12/7 − 24/7 = 42/7 − 36/7 = 6/7 cm

Step 5: Ratio
BP : PQ : QC = 12/7 : 24/7 : 6/7 = 12 : 24 : 6 = 2 : 4 : 1

Given,
● Altitude AB = 5 cm
● Base BC = 12 cm
Area of △ABC = (1/2) × 12 × 5 = 30 cm²
Given, Area of △ABC = 1.5 × Area of △ABP
So, Area of △ABP = 30 ÷ 1.5 = 20 cm²
Let the area of △ABQ be x.
Since the areas of △ABP, △ABQ and △ABC are in arithmetic progression, 20, x, 30 are in AP.
Therefore, 2x = 20 + 30 → 2x = 50 → x = 25 cm²
Now, Area of △ABP = (1/2) × AB × BP → 20 = (1/2) × 5 × BP → 20 = 2.5 × BP → BP = 8 cm
Similarly, Area of △ABQ = (1/2) × 5 × BQ → 25 = 2.5 × BQ → BQ = 10 cm
Hence, PQ = BQ − BP = 10 − 8 = 2 cm
Answer: 2

Step 1: △ABC is isosceles with AB = AC = 12 cm. Let BC = b.
Midpoint of BC is M; AM² = 144 − b²/4.

Step 2: AD = 8 cm.
Using the condition ∠AEB = ∠ACB, points B, C, D, E are concyclic
(since ∠AEB = ∠ACB means E lies on the circumcircle of △ABC, or by
power of a point / similar triangles).
By Power of a Point: AD × AE = AB × AC
8 × AE = 12 × 12
8 × AE = 144
AE = 18 cm

Given:
In △ABC,
● M, N and P are the midpoints of AB, BC and AC, respectively.
● The medians from A, B and C intersect MP, MN and NP at X, Y and Z, respectively.
● Area of △ABC = 1440 sq cm.
Find the area of △XYZ.

Step 1: Choose convenient coordinates
Since area ratios are preserved under affine transformations, take
A = (0, 0)
B = (2, 0)
C = (0, 2)
Then,
Area of △ABC
= (1/2) × 2 × 2
= 2 sq units.

Step 2: Find the midpoints
M = (1, 0)
N = (1, 1)
P = (0, 1)

Step 3: Find X
Median from A passes through N.
Equation:
y = x
Line MP joins (1, 0) and (0, 1).
Equation:
x + y = 1
Solving,
x = y
2x = 1
x = y = 1/2
Hence,
X = (1/2, 1/2)

Step 4: Find Y
Median from B joins B(2, 0) and P(0, 1).
Equation:
y = 1 − x/2
Line MN is
x = 1
Hence,
Y = (1, 1/2)

Step 5: Find Z
Median from C joins C(0, 2) and M(1, 0).
Equation:
y = 2 − 2x
Line NP is
y = 1
Hence,
Z = (1/2, 1)

Step 6: Find the area of △XYZ
XY = 1/2
YZ = 1/2
Thus,
Area of △XYZ
= (1/2) × (1/2) × (1/2)
= 1/8 sq units.
Since
Area of △ABC = 2 sq units,
Area ratio
= (1/8)/2
= 1/16

Step 7: Find the required area
Area of △XYZ
= (1/16) × 1440
= 90 sq cm

Final Answer
90

Step 1: Use the given ratios on the cevians. Since AD : AT = 4 : 3, we have AT : TD = 3 : 1. By the Mass Points Theorem, Mass at D : Mass at A = 3 : 1. Let the mass at A be x. Then, Mass at D = 3x. Similarly, BE : BT = 5 : 4, so BT : TE = 4 : 1. By the Mass Points Theorem, Mass at E : Mass at B = 4 : 1. Let the mass at B be y. Then, Mass at E = 4y.

Step 2: Express the masses at D and E. Since D lies on BC: Mass at D = Mass at B + Mass at C → 3x = y + Mass at C → Mass at C = 3x − y. Since E lies on AC: Mass at E = Mass at A + Mass at C → 4y = x + Mass at
C. Substituting Mass at C = 3x − y: 4y = x + 3x − y → 5y = 4x → y = 4x/5.

Step 3: Find the mass at
C. Mass at C = 3x − 4x/5 = 11x/5. Thus, Mass at B : Mass at C = 4x/5 : 11x/5 = 4 : 11.

Step 4: Find BD : CD. A point on a line divides the segment in the inverse ratio of the masses at the endpoints. Therefore, BD : CD = Mass at C : Mass at B = 11 : 4.