Two tangents drawn from a point P touch a circle with center O at points Q and R. Points A and B lie on PQ and PR , respectively, such that AB is also a tangent to the same circle. If ∠ AOB = 50∘ , then ∠ APB , in degrees, equals
Official Correct Answer: 80. ### Core Concept
The circle touches PQ, PR and AB, so it is the excircle of triangle PAB opposite P. Its centre O lies on the bisectors of the exterior angles at A and B.
### Step-by-Step Solution
1. Let $\angle PAB = \alpha$ and $\angle PBA = \beta$. Then $\alpha + \beta = 180^\circ - \angle APB$.
2. Tangents from A to the circle are AQ and AB, so AO bisects the exterior angle $\angle QAB = 180^\circ - \alpha$. Hence $\angle OAB = 90^\circ - \dfrac{\alpha}{2}$.
3. Similarly, $\angle OBA = 90^\circ - \dfrac{\beta}{2}$.
4. In triangle OAB:
$$\angle AOB = 180^\circ - \left(180^\circ - \frac{\alpha+\beta}{2}\right) = \frac{\alpha+\beta}{2} = 90^\circ - \frac{\angle APB}{2}$$
5. Setting $90^\circ - \dfrac{\angle APB}{2} = 50^\circ$ gives $\angle APB = 80^\circ$.
### Quick Check
If $\angle APB = 80^\circ$, then $90^\circ - 40^\circ = 50^\circ$, which matches the given $\angle AOB$.
**Correct Answer: 80**
Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD , respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is
Official Correct Answer: A. ### Core Concept
Place the hexagon on coordinates (or use its symmetry) to get the trapezium's parallel sides and height, then compare with the hexagon's area.
### Step-by-Step Solution
1. Take side length 1 and centre at the origin, with $A=(1,0)$, $B=\left(\tfrac12,\tfrac{\sqrt3}{2}\right)$, $C=\left(-\tfrac12,\tfrac{\sqrt3}{2}\right)$, $D=(-1,0)$.
2. The midpoints are $P=\left(\tfrac34,\tfrac{\sqrt3}{4}\right)$ and $Q=\left(-\tfrac34,\tfrac{\sqrt3}{4}\right)$.
3. $PQ$ and $BC$ are both horizontal, so $PBCQ$ is a trapezium with parallel sides $PQ=\tfrac32$ and $BC=1$.
4. The height is $\tfrac{\sqrt3}{2}-\tfrac{\sqrt3}{4}=\tfrac{\sqrt3}{4}$.
5. Area of $PBCQ=\tfrac12\left(\tfrac32+1\right)\cdot\tfrac{\sqrt3}{4}=\tfrac{5\sqrt3}{16}$.
6. Area of the hexagon $=\tfrac{3\sqrt3}{2}=\tfrac{24\sqrt3}{16}$.
7. Ratio $=5:24$.
### Quick Check
The trapezium is a little over one-fifth of the hexagon, and $\tfrac{5}{24}\approx0.208$ fits that.
**Correct Answer: A**
In a △ ABC , points D and E are on the sides BC and AC , respectively. BE and AD intersect at point T such that AD:AT=4:3 , and BE:BT=5:4. Point F lies on AC such that DF is parallel to BE. Then, BD:CD is
Official Correct Answer: C. ### Core Concept
Use the parallel line $DF \parallel BE$ to set up similar triangles. Then relate $DF$ to $BE$ to get the ratio $CD:CB$.
### Step-by-Step Solution
1. From $AD:AT = 4:3$, let $AT=3k$. Then $AD=4k$ and $TD=k$.
2. From $BE:BT=5:4$, let $BT=4m$. Then $BE=5m$ and $TE=m$.
3. In $\triangle ADF$, $TE \parallel DF$ (since $DF \parallel BE$), so $\triangle ATE \sim \triangle ADF$. Then $\dfrac{TE}{DF} = \dfrac{AT}{AD} = \dfrac34$, so $DF = \dfrac{4m}{3}$.
4. In $\triangle CBE$, $DF \parallel BE$, so $\triangle CDF \sim \triangle CBE$. Then $\dfrac{CD}{CB} = \dfrac{DF}{BE} = \dfrac{4m/3}{5m} = \dfrac{4}{15}$.
5. So $CD:CB = 4:15$, which gives $BD = 15-4 = 11$ parts and $CD = 4$ parts.
6. Therefore $BD:CD = 11:4$.
### Quick Check
$\dfrac{BD}{BC} = \dfrac{11}{15}$ and $\dfrac{CD}{BC} = \dfrac{4}{15}$ add to $1$, as they should.
**Correct Answer: C**
In a circle with center C and radius 62 cm, PQ and SR are two parallel chords separated by one of the diameters. If ∠PQC=45∘ , and the ratio of the perpendicular distance of PQ and SR from C is 3:2 , then the area, in sq. cm, of the quadrilateral PQRS is
Official Correct Answer: C. ### Core Concept
$PQRS$ is a trapezium with parallel sides $PQ$ and $SR$, and the diameter between them gives the height. Find each half-chord from its distance to the centre.
### Step-by-Step Solution
1. $CP=CQ$ (radii), so $\triangle CPQ$ is isosceles with $\angle PQC = 45^\circ$. The perpendicular distance from $C$ to $PQ$ is $6\sqrt2 \sin 45^\circ = 6$.
2. The half-chord is $6\sqrt2 \cos 45^\circ = 6$, so $PQ = 12$.
3. The distances from $C$ are in the ratio $3:2$, so the distance to $SR$ is $\dfrac23 \times 6 = 4$.
4. The half-chord of $SR$ is $\sqrt{(6\sqrt2)^2 - 4^2} = \sqrt{72-16} = \sqrt{56} = 2\sqrt{14}$, so $SR = 4\sqrt{14}$.
5. The chords lie on opposite sides of the centre (separated by a diameter), so the height of the trapezium is $6+4 = 10$.
6. Area $= \dfrac12 (12 + 4\sqrt{14}) \times 10 = 5(12 + 4\sqrt{14}) = 60 + 20\sqrt{14} = 20(3+\sqrt{14})$.
### Quick Check
$20(3+\sqrt{14}) \approx 20 \times 6.74 \approx 135$. The trapezium has parallel sides about $12$ and $15$ and height $10$, so $\tfrac12 \times 27 \times 10 \approx 135$.
**Correct Answer: C**
The (x,y) coordinates of vertices P, Q and R of a parallelogram PQRS are (−3,−2),(1,−5) and (9,1) , respectively. If the diagonal SQ intersects the x - axis at (a,0) , then the value of a is
Official Correct Answer: C. ### Core Concept
In a parallelogram, the diagonals bisect each other. This means the midpoint of diagonal PR is the same as the midpoint of diagonal SQ.
### Step-by-Step Solution
1. **Find Midpoint of PR**:
- P(-3, -2) and R(9, 1)
- Midpoint formula: $\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
- Midpoint of PR: $\left( \frac{-3 + 9}{2}, \frac{-2 + 1}{2} \right) = (3, -0.5)$
2. **Determine Coordinates of S**:
- Midpoint of SQ must also be (3, -0.5)
- Let S = (s_x, s_y)
- Midpoint of SQ: $\left( \frac{1 + s_x}{2}, \frac{-5 + s_y}{2} \right) = (3, -0.5)$
- Solve for s_x and s_y:
- $\frac{1 + s_x}{2} = 3 \Rightarrow s_x = 5$
- $\frac{-5 + s_y}{2} = -0.5 \Rightarrow s_y = 4$
- S = (5, 4)
3. **Find Equation of Line SQ**:
- Points S(5, 4) and Q(1, -5)
- Slope (m): $\frac{-5 - 4}{1 - 5} = \frac{-9}{-4} = \frac{9}{4}$
- Using point-slope form with Q(1, -5):
- $y - (-5) = \frac{9}{4}(x - 1)$
- $y + 5 = \frac{9}{4}(x - 1)$
4. **Find Intersection with x-axis (y = 0)**:
- Substitute y = 0:
- $0 + 5 = \frac{9}{4}(x - 1)$
- $5 = \frac{9}{4}(x - 1)$
- Multiply both sides by 4: $20 = 9(x - 1)$
- $20 = 9x - 9 \Rightarrow 29 = 9x \Rightarrow x = \frac{29}{9}$
### 30-Second Shortcut
Since diagonals bisect each other, find S using midpoint. Then, use two-point formula for SQ to find x-intercept. The result is $\frac{29}{9}$.
### Final Answer
Correct Answer: Option C
If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is
Official Correct Answer: 60. ### Core Concept
The area of a rhombus can be calculated using the formula:
$$
\text{Area} = \frac{d_1 \times d_2}{2}
$$
where \(d_1\) and \(d_2\) are the lengths of the diagonals.
### Step-by-Step Solution
1. **Given**:
- Side length of the rhombus, \(a = 36\) \text{ cm}.
- Area of the rhombus, \(A = 396\) sq. \text{ cm}.
2. **Using the area formula**:
$$
\frac{d_1 \times d_2}{2} = 396 \implies d_1 \times d_2 = 792
$$
3. **Using the Pythagorean theorem**:
- The diagonals bisect each other at right angles, forming right-angled triangles with legs \(\frac{d_1}{2}\) and \(\frac{d_2}{2}\), and hypotenuse \(a = 36\) \text{ cm}.
$$
\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 36^2
$$
Simplifying:
$$
\frac{d_1^2}{4} + \frac{d_2^2}{4} = 1296 \implies d_1^2 + d_2^2 = 5184
$$
4. **Finding \(|d_1 - d_2|\)**:
- Use the identity \((d_1 - d_2)^2 = d_1^2 + d_2^2 - 2d_1d_2\).
$$
(d_1 - d_2)^2 = 5184 - 2 \times 792 = 5184 - 1584 = 3600
$$
Taking the square root:
$$
|d_1 - d_2| = \sqrt{3600} = 60
$$
### 30-Second Shortcut
- Recognize that the area formula and the Pythagorean theorem with the side length give two equations. Use the identity for the square of the difference of diagonals to find the answer quickly.
### Final Answer Confirmation
Correct Answer: 60
Official Correct Answer: B. ### Core Concept
Complete the square to see the region as a ring (annulus) around a centre, then use the line to cut it.
### Step-by-Step Solution
1. Rewrite the middle expression: $x^2+y^2+4x-4y = (x+2)^2 + (y-2)^2 - 8$.
2. The condition $-4 \le (x+2)^2 + (y-2)^2 - 8 \le 0$ becomes $4 \le (x+2)^2 + (y-2)^2 \le 8$.
3. This is a ring centred at $(-2, 2)$ with inner radius 2 and outer radius $2\sqrt2$.
4. The ring's area is $\pi(8 - 4) = 4\pi$.
5. The boundary line $y = x + 4$ passes through the centre, since $2 = -2 + 4$.
6. A line through the centre splits the ring symmetrically, so $y \ge x+4$ keeps exactly half of it.
7. The required area is $\dfrac{4\pi}{2} = 2\pi$.
### Why Not the Others
- $4\pi$ is the full ring, ignoring the line.
- $\pi$ and $3\pi$ do not match half of $4\pi$.
**Correct Answer: B**
The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is
Official Correct Answer: A. ### Core Concept
A box inscribed in a sphere has its space diagonal equal to the sphere's diameter. Use $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$.
### Step-by-Step Solution
1. Let the edges be $a, b, c$. A box has 4 edges of each length, so $4(a+b+c) = 144$ and $a+b+c = 36$.
2. The surface area gives $2(ab+bc+ca) = 846$.
3. Then $a^2+b^2+c^2 = 36^2 - 846 = 1296 - 846 = 450$.
4. The space diagonal is $d = \sqrt{450} = 15\sqrt2$, and this equals the sphere's diameter.
5. The radius is $r = \dfrac{15\sqrt2}{2}$, so $r^3 = \dfrac{3375 \cdot 2\sqrt2}{8} = \dfrac{3375\sqrt2}{4}$.
6. The volume is $\dfrac43 \pi r^3 = \dfrac43 \pi \cdot \dfrac{3375\sqrt2}{4} = 1125\pi\sqrt2$.
### Quick Check
The diagonal squared is $450$, so $r^2 = \dfrac{450}{4} = 112.5$. Then $r \approx 10.6$ and $\dfrac43\pi r^3 \approx 5000$, which matches $1125\pi\sqrt2 \approx 4998$.
**Correct Answer: A**
Official Correct Answer: 10. 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:
- A: (0, 0)
- D: (0, 45)
- E: (28, 45)
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: \sqrt{(x2 - x1)^2 + (y2 - y1)^2}. 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}.
Three circles of equal radii touch (but not cross) each other externally. Two other circles, X and Y, are drawn such that both touch (but not cross) each of the three previous circles. If the radius of X is more than that of Y, the ratio of the radii of X and Y is
Official Correct Answer: B. ### Core Concept
The centres of the three equal circles form an equilateral triangle. The two circles touching all three lie on its axis of symmetry, at the centroid distance.
### Step-by-Step Solution
1. Let each equal circle have radius $r$. Their centres form an equilateral triangle of side $2r$.
2. The distance from the centroid to each vertex is $\frac{2r}{\sqrt{3}}$.
3. The smaller circle $Y$ sits in the gap between the three circles. It touches each of them externally, so its radius is $y=\frac{2r}{\sqrt{3}}-r$.
4. The larger circle $X$ encloses the three circles and touches each internally, so its radius is $x=\frac{2r}{\sqrt{3}}+r$.
5. The ratio is $\frac{x}{y}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$.
6. Multiply the numerator and denominator by $2+\sqrt{3}$. The denominator becomes $4-3=1$, so $\frac{x}{y}=(2+\sqrt{3})^2=7+4\sqrt{3}$.
### Why Not the Others
- A: $4+2\sqrt{3}$ drops the square that comes from rationalising.
- C: $2+\sqrt{3}$ is only one factor of the ratio.
- D: $4+\sqrt{3}$ does not arise from the geometry.
**Correct Answer: B**
Official Correct Answer: 2. ### Core Concept
The radius of the in-circle (inradius) of a triangle is given by the formula:
$$ r = \frac{A}{s} $$
where \( A \) is the area of the triangle and \( s \) is the semi-perimeter.
### Step-by-Step Solution
1. **Identify the Triangle Type and Calculate Sides**:
- The triangle has vertices at \( (1, 2) \), \( (7, 2) \), and \( (1, 10) \).
- Calculate side lengths using the distance formula:
- \( AB = \sqrt{(7-1)^2 + (2-2)^2} = 6 \)
- \( AC = \sqrt{(1-1)^2 + (10-2)^2} = 8 \)
- \( BC = \sqrt{(7-1)^2 + (2-10)^2} = 10 \)
- The triangle is right-angled with legs 6 and 8, and hypotenuse 10.
2. **Calculate the Area (A)**:
- For a right-angled triangle, area \( A = \frac{1}{2} \times \text{base} \times \text{height} \).
- \( A = \frac{1}{2} \times 6 \times 8 = 24 \).
3. **Calculate the Semi-Perimeter (s)**:
- \( s = \frac{a + b + c}{2} = \frac{6 + 8 + 10}{2} = 12 \).
4. **Compute the Inradius (r)**:
- \( r = \frac{A}{s} = \frac{24}{12} = 2 \).
### 30-Second Shortcut
For right-angled triangles, the inradius can be quickly calculated using:
$$ r = \frac{a + b - c}{2} $$
where \( c \) is the hypotenuse. Plugging in:
$$ r = \frac{6 + 8 - 10}{2} = \frac{4}{2} = 2 $$
### Final Answer
Correct Answer: 2
ABCD is a trapezium in which AB is parallel to CD. The sides AD and BC when extended, intersect at point E. If AB = 2 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of △AEB is
Official Correct Answer: A. ### Core Concept
The problem involves similar triangles formed by the sides of a trapezium. When the non-parallel sides of a trapezium are extended, they form similar triangles with the bases of the trapezium. The ratio of similarity is equal to the ratio of the lengths of the parallel sides.
### Step-by-Step Solution
1. **Identify Similar Triangles**:
- Triangles \( \triangle EAB \) and \( \triangle ECD \) are similar because \( AB \parallel CD \).
- Ratio of similarity \( k = \frac{AB}{CD} = \frac{2}{1} = 2 \).
2. **Express Sides in Terms of Ratio**:
- Let \( EA = x \) and \( EB = y \).
- Since \( \triangle EAB \) is similar to \( \triangle ECD \) with ratio 2, we have:
\[
\frac{EA}{ED} = 2 \quad \text{and} \quad \frac{EB}{EC} = 2
\]
- Let \( AD = m \) and \( BC = n \). Then:
\[
ED = EA - AD = x - m \quad \text{and} \quad EC = EB - BC = y - n
\]
- From the similarity ratio:
\[
x = 2m \quad \text{and} \quad y = 2n
\]
3. **Use Perimeter Information**:
- Perimeter of trapezium \( ABCD \):
\[
AB + BC + CD + DA = 2 + n + 1 + m = 6 \implies m + n = 3
\]
4. **Calculate Perimeter of \( \triangle AEB \)**:
- Perimeter \( P \) of \( \triangle AEB \):
\[
P = EA + EB + AB = 2m + 2n + 2 = 2(m + n) + 2 = 2(3) + 2 = 8 \text{ \text{ cm}}
\]
### 30-Second Shortcut
Since the ratio of similarity is 2, the perimeter of \( \triangle AEB \) is twice the sum of the non-parallel sides of the trapezium plus the length of \( AB \). Given the perimeter of the trapezium is 6, the non-parallel sides sum to 3, making the perimeter of \( \triangle AEB \) \( 2 \times 3 + 2 = 8 \).
### Final Answer
Correct Answer: A
A circular plot of land is divided into two regions by a chord of length 103 meters such that the chord subtends an angle of 120∘ at the center. Then, the area, in square meters, of the smaller region is
Official Correct Answer: C. ### Core Concept
The smaller region cut off by a chord is a circular segment. Its area equals the sector area minus the triangle formed by the two radii and the chord.
### Step-by-Step Solution
1. The chord subtends $120^\circ$ at the centre, so by the chord formula $2r\sin 60^\circ = 10\sqrt{3}$.
2. This gives $r\sqrt{3} = 10\sqrt{3}$, so $r = 10$ m.
3. Sector area: $\dfrac{120}{360}\pi r^2 = \dfrac{100\pi}{3}$.
4. Triangle area: $\dfrac{1}{2} r^2 \sin 120^\circ = \dfrac{1}{2}\cdot 100\cdot \dfrac{\sqrt{3}}{2} = 25\sqrt{3}$.
5. Segment area, which is the smaller region:
$$\frac{100\pi}{3} - 25\sqrt{3} = 25\left(\frac{4\pi}{3} - \sqrt{3}\right)$$
### Why Not the Others
- A and B: they have the wrong coefficient (20) and the wrong $\frac{4\pi}{\sqrt{3}}$ form, which does not match the sector area.
- D: it adds the triangle instead of subtracting it. That would give the larger part of the sector plus triangle, not the segment.
### Quick Check
Numerically, $104.72 - 43.30 \approx 61.4$ square metres, which is well below the circle's area of about $314$. So it is the smaller region.
**Correct Answer: C**
The midpoints of sides AB , BC , and AC in △ABC are M , N , and P , respectively. The medians drawn from A , B , and C intersect the line segments MP,MN and NP at X , Y , and Z , respectively. If the area of △ABC is 1440sq cm, then the area, in sq cm, of △XYZ is
Official Correct Answer: 90. ### Core Concept
The medians of $\triangle ABC$ cut the sides of the medial triangle $MNP$ at their midpoints, so $\triangle XYZ$ is the medial triangle of $\triangle MNP$. Each medial triangle has $\tfrac14$ of the area of its parent.
### Step-by-Step Solution
1. $M$, $N$, $P$ are the midpoints of the sides, so $\triangle MNP$ is the medial triangle of $\triangle ABC$. Its area is $\tfrac14 \times 1440 = 360$.
2. By the midpoint theorem, $MP \parallel BC$. The median from $A$ meets $BC$ at its midpoint and bisects every segment parallel to $BC$ between $AB$ and $AC$. So $X$ is the midpoint of $MP$.
3. By the same argument, the median from $B$ bisects $MN$, so $Y$ is the midpoint of $MN$. The median from $C$ bisects $NP$, so $Z$ is the midpoint of $NP$.
4. $X$, $Y$, $Z$ are therefore the midpoints of the sides of $\triangle MNP$, so $\triangle XYZ$ is the medial triangle of $\triangle MNP$.
5. $[XYZ] = \tfrac14 \times 360 = 90$.
### Quick Check
$[XYZ] = \tfrac{1}{16} \times 1440 = 90$.
**Correct Answer: 90**
Official Correct Answer: A. ### Core Concept
The square ACEG has the diagonal AC of the octagon as its side, so its area is $AC^2$. Find AC with the cosine rule in triangle ABC.
### Step-by-Step Solution
1. Each interior angle of a regular octagon is $\dfrac{(8-2)\times180^\circ}{8}=135^\circ$, so $\angle ABC = 135^\circ$.
2. In triangle ABC, $AB = BC = 6$. By the cosine rule:
$$AC^2 = 6^2+6^2-2\cdot6\cdot6\cos135^\circ = 72+72\cdot\frac{\sqrt2}{2}=72+36\sqrt2$$
3. A, C, E, G are alternate vertices of the regular octagon, so ACEG is a square with side $AC$.
4. Area $= AC^2 = 72+36\sqrt2 = 36(2+\sqrt2)$ sq. cm.
### Why Not the Others
- B and D: $72(1+\sqrt2)$ and $72(2+\sqrt2)$ are too large. They are about 174 and 246, while $AC^2\approx123$.
- C: $36(1+\sqrt2)$ drops the extra $36$ from the $72$ term of $AC^2$.
### Quick Check
$AC^2\approx72+50.9=122.9$, and $36(2+1.414)\approx122.9$.
**Correct Answer: A**
Official Correct Answer: B. ### Core Concept
The problem involves a cyclic quadrilateral where the diagonals intersect. The key theorem used is the Ratio Lemma for cyclic quadrilaterals, which states that the ratio of the segments of one diagonal is equal to the ratio of the products of the adjacent sides.
### Step-by-Step Solution
1. **Given Ratios**:
- \( AB : CD = 2 : 1 \)
- \( BC : AD = 5 : 4 \)
2. **Let**:
- \( AB = 2x \), \( CD = x \)
- \( BC = 5y \), \( AD = 4y \)
3. **Apply the Ratio Lemma**:
\[
\frac{AE}{EC} = \frac{AB \times AD}{BC \times CD}
\]
4. **Substitute the Values**:
\[
\frac{AE}{EC} = \frac{2x \times 4y}{5y \times x} = \frac{8xy}{5xy} = \frac{8}{5}
\]
5. **Simplify**:
\[
AE : EC = 8 : 5
\]
### 30-Second Shortcut
For cyclic quadrilaterals, use the Ratio Lemma: \( \frac{AE}{EC} = \frac{AB \times AD}{BC \times CD} \). Plug in the given ratios directly to find the segment ratio.
### Final Answer
Correct Answer: Option B
Let C be the circle x2+y2+4x−6y−3=0 and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to 60°. Then, the point at which L touches the line x=6 is
Official Correct Answer: D. ### Core Concept
The point from which two tangents make a fixed angle lies at a fixed distance from the centre, so the locus is a concentric circle.
### Step-by-Step Solution
1. Complete the squares: $(x+2)^2 + (y-3)^2 = 3 + 4 + 9 = 16$. The centre is $(-2, 3)$ and the radius is $r = 4$.
2. Let $P$ be the intersection of the tangents and $d$ its distance from the centre. The line from $P$ to the centre bisects the $60°$ angle, so $\sin 30° = \frac{r}{d}$.
3. This gives $d = \frac{4}{1/2} = 8$.
4. So $L$ is the circle with centre $(-2, 3)$ and radius 8: $(x+2)^2 + (y-3)^2 = 64$.
5. The line $x = 6$ is vertical and lies at distance $6 - (-2) = 8$ from the centre, which equals the radius. So it is tangent to $L$, at the point with the centre's $y$-coordinate, $(6, 3)$.
### Why Not the Others
- A, B, C: for each of these points, $(6-(-2))^2 + (y-3)^2 = 64 + (y-3)^2$, which exceeds 64 for $y \ne 3$, so none of them lies on $L$.
**Correct Answer: D**
In a right-angled triangle △ A B C , the altitude AB is 5cm , and the base BC is 12cm. P and Q are two points on BC such that the areas of △ABP,△ABQ and △ABC are in arithmetic progression. If the area of △ABC is 1.5 times the area of △ABP , the length of PQ in cm , is
Official Correct Answer: 2. ### Core Concept
In a right-angled triangle, the area can be calculated using the formula:
$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$
### Step-by-Step Solution
1. **Identify the triangle and given values**:
- Triangle \( \triangle ABC \) is right-angled at \( B \).
- \( AB = 5 \, \text{\text{ cm}} \) (altitude), \( BC = 12 \, \text{\text{ cm}} \) (base).
- The area of \( \triangle ABC \) is:
$$ \text{Area}_{ABC} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 5 \times 12 = 30 \, \text{\text{ cm}}^2 $$
2. **Set up the areas in arithmetic progression**:
- Let the areas of \( \triangle ABP \), \( \triangle ABQ \), and \( \triangle ABC \) be \( A_1 \), \( A_2 \), and \( A_3 \) respectively.
- Given \( A_3 = 1.5 \times A_1 \) and \( A_1, A_2, A_3 \) are in AP.
- Therefore, \( 2A_2 = A_1 + A_3 \) and substituting \( A
Official Correct Answer: 126. ### Core Concept
All three altitudes of a triangle follow from its area: each altitude equals $2\times\text{Area}$ divided by the side it falls on.
### Step-by-Step Solution
1. The triangle is isosceles with $AB=AC=50$ and $BC=80$. The altitude from $A$ meets $BC$ at its midpoint, which is 40 from $B$.
2. That altitude is $\sqrt{50^2-40^2} = \sqrt{900} = 30$.
3. The area is $\tfrac12\times 80\times 30 = 1200$.
4. The altitude on $AB$ is $\dfrac{2\times1200}{50} = 48$. By symmetry, the altitude on $AC$ is also 48.
5. The sum of the altitudes is $30+48+48 = 126$.
### Quick Check
The area from the altitude on $AB$ is $\tfrac12\times50\times48 = 1200$, which matches step 3.
**Correct Answer: 126**
ABCD is a trapezium in which AB is parallel to DC, AD is perpendicular to AB, and AB = 3DC. If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm, then the area, in sq. cm, of the trapezium is
Official Correct Answer: A. ### Core Concept
A tangential quadrilateral has equal sums of opposite sides (Pitot theorem). A right trapezium with an incircle also has height equal to the diameter.
### Step-by-Step Solution
1. $AD \perp AB$ and $AB \parallel DC$, so $AD$ is the height. The circle touches both parallel sides, so $AD = 2r = 6$.
2. Let $DC = a$, so $AB = 3a$. The slant side $BC$ has horizontal offset $3a-a = 2a$ and vertical rise 6, so $BC^2 = 36+4a^2$.
3. By the Pitot theorem, $AB+DC = AD+BC$. This gives $4a = 6+BC$, so $BC = 4a-6$.
4. Equate the two expressions for $BC^2$: $(4a-6)^2 = 36+4a^2$. This gives $16a^2-48a+36 = 36+4a^2$, so $12a^2 = 48a$ and $a = 4$.
5. The parallel sides are $DC=4$ and $AB=12$.
6. Area $= \tfrac12(4+12)\times 6 = 48$.
### Quick Check
$BC = 4(4)-6 = 10$, and $\sqrt{36+8^2} = 10$. This is consistent.
**Correct Answer: A**
In △ABC,AB=AC=12 cm and D is a point on side BC such that AD=8 cm. If AD is extended to a point E such that ∠ACB=∠ AEB , then the length, in cm, of AE is
Official Correct Answer: B. ### Core Concept
Equal angles on a shared vertex give similar triangles. The similarity turns the given lengths into the relation $AB^2 = AD \cdot AE$.
### Step-by-Step Solution
1. Since $AB = AC$, the base angles are equal: $\angle ABC = \angle ACB$.
2. It is given that $\angle ACB = \angle AEB$. Hence $\angle ABD = \angle ABC = \angle AEB$.
3. Compare $\triangle ABD$ and $\triangle AEB$. Both contain the angle at $A$, because $D$ lies on $AE$ and so $\angle BAD = \angle BAE$.
4. They also have $\angle ABD = \angle AEB$. So by AA, $\triangle ABD \sim \triangle AEB$.
5. Corresponding sides are proportional: $\frac{AB}{AE} = \frac{AD}{AB}$, so $AB^2 = AD \cdot AE$.
6. Substituting, $144 = 8 \cdot AE$, so $AE = 18$ cm.
### Why Not the Others
- 16, 14 and 20 do not satisfy $AD \cdot AE = 144$, since $8\times16=128$, $8\times14=112$ and $8\times20=160$.
**Correct Answer: B**
A triangle is drawn with its vertices on the circle C such that one of its sides is a diameter of C and the other two sides have their lengths in the ratio a:b. If the radius of the circle is r , then the area of the triangle is
Official Correct Answer: C. ### Core Concept
When a triangle is inscribed in a circle with one side as the diameter, it is a right-angled triangle (Thales' theorem). The area of a right-angled triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
### Step-by-Step Solution
1. **Identify the Triangle Type**:
- The triangle is right-angled with the hypotenuse as the diameter of the circle.
- Hypotenuse length = $2r$.
2. **Assign Variables to Sides**:
- Let the other two sides be $a \cdot k$ and $b \cdot k$ where $k$ is a scaling factor.
3. **Apply Pythagoras Theorem**:
\[
(a \cdot k)^2 + (b \cdot k)^2 = (2r)^2
\]
\[
k^2(a^2 + b^2) = 4r^2
\]
\[
k = \frac{2r}{\sqrt{a^2 + b^2}}
\]
4. **Calculate the Area**:
\[
\text{Area} = \frac{1}{2} \times (a \cdot k) \times (b \cdot k)
\]
\[
= \frac{1}{2} \times a \cdot b \cdot \left(\frac{2r}{\sqrt{a^2 + b^2}}\right)^2
\]
\[
= \frac{1}{2} \times a \cdot b \cdot \frac{4r^2}{a^2 + b^2}
\]
\[
= \frac{2abr^2}{a^2 + b^2}
\]
### 30-Second Shortcut
- Recognize the triangle is right-angled.
- Use the area formula for right-angled triangles.
- Substitute the sides using the given ratio and solve.
### Final Answer
Correct Answer: Option C
In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is
Official Correct Answer: B. ### Core Concept
All three regions share the height $AB=9$ along $BC$, so the areas are proportional to the lengths $BP$, $PQ$ and $QC$ in the way set out below. The geometric progression gives the area ratios, and the total area fixes the actual values.
### Step-by-Step Solution
1. Let the areas of $ABP$, $APQ$ and $AQCD$ be $a$, $ar$ and $ar^2$.
2. The condition $AQCD=4\times ABP$ gives $ar^2=4a$, so $r=2$. The areas are $a, 2a, 4a$.
3. The three regions together make up the whole rectangle, so $7a=9\times6=54$ and $a=\dfrac{54}{7}$.
4. Triangle $ABP$ has area $\tfrac12\cdot9\cdot BP=a$, so $BP=\dfrac{2a}{9}=\dfrac{12}{7}$.
5. Triangle $APQ$ has area $\tfrac12\cdot9\cdot PQ=2a$, so $PQ=\dfrac{4a}{9}=\dfrac{24}{7}$.
6. $QC=6-\dfrac{12}{7}-\dfrac{24}{7}=\dfrac{6}{7}$.
7. The ratio is $BP:PQ:QC=12:24:6=2:4:1$.
### Why Not the Others
- A) 1:1:2 would make $PQ=BP$, so $APQ$ would equal $ABP$, but $r=2$ requires it to be double.
- C) 1:2:1 gives $QC=BP$, but $AQCD$ contains the rectangle part $QCD$ and must be 4 times $ABP$, so this ratio does not fit.
- D) 1:2:4 puts the largest segment at $QC$, which does not match the computed lengths.
**Correct Answer: B**
Let △ABC be an isosceles triangle such that AB and AC are of equal length. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that ∠AOB=105∘, then AD/BE equals
Official Correct Answer: C. ### Core Concept
$O$ is the orthocentre, so $\angle AOB$ is linked to an angle of the triangle. Once the triangle's angles are known, both altitudes can be written in terms of $AB$.
### Step-by-Step Solution
1. In right triangle $ADC$, $\angle DAC = 90^\circ - C$. In right triangle $AEO$, $\angle AEO = 90^\circ$ and $\angle OAE = 90^\circ - C$. So $\angle AOE = C$.
2. Hence $\angle AOB = 180^\circ - C = 105^\circ$, which gives $C = 75^\circ$.
3. Since $AB = AC$, $B = C = 75^\circ$ and $A = 180^\circ - 150^\circ = 30^\circ$.
4. In right triangle $ADB$, $AD = AB\sin B = AB\sin 75^\circ$.
5. In right triangle $AEB$, $BE = AB\sin A = AB\sin 30^\circ = \tfrac{AB}{2}$.
6. So $\dfrac{AD}{BE} = \dfrac{\sin 75^\circ}{\sin 30^\circ} = 2\sin 75^\circ = 2\cos 15^\circ$.
### Why Not the Others
- A, B, D: the ratio is $2\sin 75^\circ$, which equals $2\cos 15^\circ$. It is not $\cos 15^\circ$, $2\sin 15^\circ$ or $\sin 15^\circ$.
**Correct Answer: C**
A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is
Official Correct Answer: B. ### Core Concept
Maximise the area of a rectangle whose base lies on the diameter and whose top corners touch the arc.
### Step-by-Step Solution
1. Place the centre at the origin. Let the rectangle have half-width $x$ and height $h$. The top corners lie on the circle, so $x^2 + h^2 = 4$.
2. Area $= 2xh$. Since $2xh \le x^2 + h^2 = 4$, the area is at most 4. Equality holds when $x = h$.
3. Then $2x^2 = 4$, so $x = h = \sqrt{2}$.
4. The sides are width $2x = 2\sqrt{2}$ and height $h = \sqrt{2}$.
5. The ratio of the largest to the smallest side is $2\sqrt{2} : \sqrt{2} = 2 : 1$.
### Why Not the Others
- A: $\sqrt{5}:1$ does not match the sides $2\sqrt{2}$ and $\sqrt{2}$.
- C: $\sqrt{2}:1$ would need width $= \sqrt{2} \times$ height, but here the width is twice the height.
- D: $1:1$ would be a square, which has a smaller area than the maximum.
**Correct Answer: B**
+67 More Verified Questions Available
Practice the complete collection of 92 questions in our interactive CBT practice interface with personal progress tracking.
Frequently Asked Questions about CAT Geometry Previous Year Questions (PYQs)
How many Geometry questions appear in CAT each year?
Historically, Geometry forms a significant core of the Quantitative Aptitude section, accounting for high-weightage questions in every single slot from 2017 to 2025.
What is the best strategy to master Geometry for CAT?
Start by solving every official past question from 2017 to 2025. Focus on identifying option elimination cues, core conceptual traps, and algebraic/logical shortcuts.
Related CAT Papers & Topic Mastery Hubs
Explore other slots, sections, and high-weightage topic collections
