Geometry & Mensuration: CAT previous year questions

Solution

$CP = CQ = 6\sqrt{2}$ and $\angle PQC = 45^\circ$, so triangle $CPQ$ is right-angled at C: $PQ = 12$, and the distance from C to PQ $= 6$.

The distances are in ratio $3 : 2$, so SR is 4 cm from C: half of SR $= \sqrt{72 - 16} = 2\sqrt{14}$, so $SR = 4\sqrt{14}$.

The chords lie on opposite sides of C, so they are $6 + 4 = 10$ cm apart.

Area $= \dfrac{1}{2}(12 + 4\sqrt{14}) \times 10 = 20(3 + \sqrt{14})$.

Solution

Diagonals $d_1, d_2$: area $= \dfrac{d_1 d_2}{2} = 396 \Rightarrow d_1 d_2 = 792$.

Half-diagonals and the side form a right triangle: $d_1^2 + d_2^2 = 4 \times 36^2 = 5184$.

$(d_1 - d_2)^2 = 5184 - 1584 = 3600 \Rightarrow |d_1 - d_2| = 60$.

Solution

For tangents from P with a third tangent AB, the angle at the centre is $\angle AOB = 90^\circ - \dfrac{\angle APB}{2}$.

$50 = 90 - \dfrac{\angle APB}{2} \Rightarrow \angle APB = 80^\circ$.

Solution

Take side 2, centre at the origin: $B(1, \sqrt{3})$, $C(-1, \sqrt{3})$, $P(1.5, \tfrac{\sqrt{3}}{2})$, $Q(-1.5, \tfrac{\sqrt{3}}{2})$.

PBCQ has parallel sides $PQ = 3$ and $BC = 2$, height $\dfrac{\sqrt{3}}{2}$: area $= \dfrac{1}{2}(5)\dfrac{\sqrt{3}}{2} = \dfrac{5\sqrt{3}}{4}$.

Hexagon area $= \dfrac{3\sqrt{3}}{2} \times 4 = 6\sqrt{3}$.

Ratio $= \dfrac{5\sqrt{3}/4}{6\sqrt{3}} = 5 : 24$.

Solution

$AT : TD = 3 : 1$ and $BT : TE = 4 : 1$. Use mass points with mass 1 at A.

On AD: $1 \times 3 = m_D \times 1 \Rightarrow m_D = 3$, so $m_B + m_C = 3$.

On BE: $m_B \times 4 = m_E \times 1$, and $m_E = m_A + m_C = 1 + m_C$, so $1 + m_C = 4m_B$.

Solving: $m_B = 0.8$, $m_C = 2.2$.

$BD : DC = m_C : m_B = 2.2 : 0.8 = 11 : 4$.

Solution

Altitude to BC $= \sqrt{50^2 - 40^2} = 30$. Area $= \dfrac{1}{2} \times 80 \times 30 = 1200$.

Altitudes to AB and AC $= \dfrac{2 \times 1200}{50} = 48$ each.

Sum $= 30 + 48 + 48 = 126$.

Solution

The height AD equals the diameter: $AD = 6$.

A tangential quadrilateral has $AB + DC = AD + BC$, so with $AB = 3DC$: $BC = 4DC - 6$.

Also $BC^2 = 6^2 + (AB - DC)^2 = 36 + 4DC^2$.

$(4DC - 6)^2 = 36 + 4DC^2 \Rightarrow 12DC^2 = 48DC \Rightarrow DC = 4$, $AB = 12$.

Area $= \dfrac{1}{2}(12 + 4) \times 6 = 48$.

Solution

Triangles ABD and AEB share $\angle BAD$, and $\angle ABD = \angle ACB = \angle AEB$ (since $AB = AC$).

So they are similar: $\dfrac{AB}{AE} = \dfrac{AD}{AB}$.

$AE = \dfrac{AB^2}{AD} = \dfrac{144}{8} = 18$.

Solution

Triangle ADE is right-angled at D with $AD = 45$ and $DE = 28$.

$AE = \sqrt{45^2 + 28^2} = \sqrt{2809} = 53$.

Inradius of a right triangle $= \dfrac{45 + 28 - 53}{2} = 10$.

Solution

Let the edges be $a, b, c$: $4(a + b + c) = 144 \Rightarrow a + b + c = 36$, and $2(ab + bc + ca) = 846$.

$a^2 + b^2 + c^2 = 36^2 - 846 = 450$, so the space diagonal $= \sqrt{450} = 15\sqrt{2}$ = sphere's diameter.

$R = \dfrac{15\sqrt{2}}{2}$. Volume $= \dfrac{4}{3}\pi R^3 = \dfrac{4}{3}\pi \times \dfrac{3375 \times 2\sqrt{2}}{8} = 1125\sqrt{2}\,\pi$.

Solution

Take the three equal radii as 1. Their centres form an equilateral triangle of side 2, with circumradius $\dfrac{2}{\sqrt{3}}$.

Small circle Y (in the middle): radius $\dfrac{2}{\sqrt{3}} - 1$. Large circle X (around them): radius $\dfrac{2}{\sqrt{3}} + 1$.

Ratio $= \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = (2 + \sqrt{3})^2 = 7 + 4\sqrt{3}$.

Solution

Triangle EDC is similar to EAB with ratio $\dfrac{CD}{AB} = \dfrac{1}{2}$, so D and C are the midpoints of EA and EB.

So $EA = 2AD$ and $EB = 2BC$.

Perimeter of ABCD: $2 + 1 + AD + BC = 6 \Rightarrow AD + BC = 3$.

Perimeter of AEB $= 2 + 2(AD + BC) = 8$.

Solution

Radius: $\dfrac{10\sqrt{3}}{2\sin 60^\circ} = 10$ m.

Smaller region $=$ sector $-$ triangle $= \dfrac{120}{360}\pi(100) - \dfrac{1}{2}(100)\sin 120^\circ$.

$= \dfrac{100\pi}{3} - 25\sqrt{3} = 25\left(\dfrac{4\pi}{3} - \sqrt{3}\right)$.

Solution

Each interior angle of a regular octagon is $135^\circ$. In triangle ABC:

$AC^2 = 6^2 + 6^2 - 2 \times 36 \cos 135^\circ = 72 + 36\sqrt{2}$.

Area of square ACEG $= AC^2 = 36(2 + \sqrt{2})$.

Solution

The median from A bisects MP (the midline), so X is the midpoint of MP. Similarly Y and Z are the midpoints of MN and NP.

So XYZ is the medial triangle of MNP, and MNP is the medial triangle of ABC.

Area $= \dfrac{1}{4} \times \dfrac{1}{4} \times 1440 = 90$.

Solution

In a cyclic quadrilateral, triangles ABE and DCE are similar: $\dfrac{AE}{DE} = \dfrac{AB}{DC} = 2$.

Triangles AED and BEC are similar: $\dfrac{DE}{CE} = \dfrac{AD}{BC} = \dfrac{4}{5}$.

$\dfrac{AE}{CE} = 2 \times \dfrac{4}{5} = \dfrac{8}{5}$, i.e. $8 : 5$.

Solution

Area of ABC $= \dfrac{1}{2} \times 5 \times 12 = 30$, so area of ABP $= 20 \Rightarrow BP = 8$.

AP: $20, [ABQ], 30 \Rightarrow$ ABQ $= 25 \Rightarrow BQ = 10$.

$PQ = 10 - 8 = 2$.

Solution

A triangle on a diameter is right-angled. Let the legs be $ak$ and $bk$; the hypotenuse is $2r$.

$k^2(a^2 + b^2) = 4r^2$.

Area $= \dfrac{1}{2}abk^2 = \dfrac{1}{2}ab \times \dfrac{4r^2}{a^2 + b^2} = \dfrac{2abr^2}{a^2 + b^2}$.

Solution

Rectangle area $= 54$. In a GP with third term $= 4 \times$ first, the ratio is 2: areas are $x, 2x, 4x$.

$7x = 54$, and area of ABP $= \dfrac{9 \times BP}{2}$, area of APQ $= \dfrac{9 \times PQ}{2}$, so $PQ = 2BP$.

$BP = \dfrac{2 \times 54/7}{9} = \dfrac{12}{7}$, $PQ = \dfrac{24}{7}$, $QC = 6 - \dfrac{36}{7} = \dfrac{6}{7}$.

$BP : PQ : QC = 2 : 4 : 1$.

Solution

Let $\angle B = \angle C = \beta$, so $\angle A = 180^\circ - 2\beta$.

$\angle OAB = 90^\circ - \beta$ (AD bisects $\angle A$) and $\angle OBA = 90^\circ - \angle A = 2\beta - 90^\circ$.

In triangle AOB: $105 + (90 - \beta) + (2\beta - 90) = 180 \Rightarrow \beta = 75^\circ$, $\angle A = 30^\circ$.

$AD = AB\cos 15^\circ$ and $BE = AB\sin 30^\circ = \dfrac{AB}{2}$.

$\dfrac{AD}{BE} = 2\cos 15^\circ$.

Solution

Let the rectangle be $2x$ wide and $y$ tall with $x^2 + y^2 = 4$.

Area $= 2xy$ is largest when $x = y = \sqrt{2}$.

Sides: $2\sqrt{2}$ and $\sqrt{2}$. Ratio $= 2 : 1$.

Solution

Interior + exterior $= 180^\circ$ and interior $-$ exterior $= 120^\circ$, so exterior $= 30^\circ$ and $n = 12$.

Diagonals $= \dfrac{12 \times 9}{2} = 54$.

Solution

AB is perpendicular to both parallel sides, so AB is the height $h$.

$CD^2 = h^2 + (8 - 3)^2$. Perimeter: $h + 3 + 8 + CD = 36 \Rightarrow CD = 25 - h$.

$(25 - h)^2 = h^2 + 25 \Rightarrow 625 - 50h = 25 \Rightarrow h = 12$.

Area $= \dfrac{1}{2}(3 + 8) \times 12 = 66$.

Solution

Let the sides be $l$ and $b$: $l + b = \dfrac{P}{2}$, and the diagonal is a diameter: $l^2 + b^2 = 4R^2$.

Area $= lb = \dfrac{(l + b)^2 - (l^2 + b^2)}{2} = \dfrac{P^2/4 - 4R^2}{2} = \dfrac{P^2}{8} - 2R^2$.

Solution

$AD = AB\sin B = AB\sin\theta$ and $BE = AB\sin A = AB\sin 45^\circ$.

$\dfrac{AD}{BE} = \dfrac{\sin\theta}{1/\sqrt{2}} = \sqrt{2}\sin\theta$.

Solution

Let A have $n$ sides and B have $2n$.

$\dfrac{180 - 360/n}{180 - 180/n} = \dfrac{3}{4} \Rightarrow 720 - \dfrac{1440}{n} = 540 - \dfrac{540}{n} \Rightarrow \dfrac{900}{n} = 180$.

$n = 5$, so B has $10$ sides.

Solution

Triangles ADC and ABD share the height from A, so $DC = \dfrac{1}{2}BD$: $BD = 2$, $DC = 1$.

$AD^2 = 3^2 + 2^2 - 2 \times 3 \times 2\cos 60^\circ = 7$.

$AD = \sqrt{7}$.

Solution

O is the centroid, and E, D are midpoints of AB and AC, so $ED \parallel BC$ and $ED = \dfrac{BC}{2}$.

Triangle BOC has area $\dfrac{1}{3} \times 108 = 36$, and triangle EOD is similar to it with ratio $\dfrac{1}{2}$.

Area of EOD $= \dfrac{36}{4} = 9$.

Solution

Each side must be shorter than the sum of the other three.

$x < 1 + 2 + 4 = 7$ and $4 < 1 + 2 + x \Rightarrow x > 1$.

$x = 2, 3, 4, 5, 6$: $5$ possible lengths.

Solution

A lies on the circle with diameter BC, so $\angle BAC = 90^\circ$ and $BC = 8\sqrt{2}$; circle 1 has radius $4\sqrt{2}$, circle 2 has radius 8.

On A's side of BC, the half of circle 1 lies entirely within circle 2: area $= \dfrac{1}{2}\pi(4\sqrt{2})^2 = 16\pi$.

On the other side, the overlap is circle 2's minor segment over chord BC: $\dfrac{90}{360}\pi(64) - \dfrac{1}{2}(8)(8) = 16\pi - 32$.

Total $= 32\pi - 32 = 32(\pi - 1)$.

Solution

Triangle area $= \dfrac{\sqrt{3}}{4} \times 144 = 36\sqrt{3}$.

Hexagon area $= \dfrac{3\sqrt{3}}{2}s^2 = 36\sqrt{3} \Rightarrow s^2 = 24 \Rightarrow s = 2\sqrt{6}$.

Solution

In a regular hexagon of side 2: $AC = 2\sqrt{3}$ and $\angle ACD = 90^\circ$.

$CT = 1$, so $AT = \sqrt{12 + 1} = \sqrt{13}$.

Solution

Inradius $= 4$. For a right triangle, $r = \dfrac{AB + BC - AC}{2}$: $8 = AB + 10 - AC \Rightarrow AC = AB + 2$.

$AC^2 = AB^2 + 100 \Rightarrow 4AB + 4 = 100 \Rightarrow AB = 24$.

Area $= \dfrac{1}{2} \times 24 \times 10 = 120$.

Solution

$d_1 d_2 = 24$ and $d_1^2 + d_2^2 = 4 \times 25 = 100$.

$(d_1 + d_2)^2 = 148 \Rightarrow d_1 + d_2 = 2\sqrt{37}$; $(d_1 - d_2)^2 = 52 \Rightarrow d_1 - d_2 = 2\sqrt{13}$.

Longer diagonal $= \sqrt{37} + \sqrt{13}$.

Solution

ABPD is a parallelogram, so $DP = AB = a$, and P is the midpoint, so $PC = a$, $CD = 2a$. Let the height be $h$.

Parallelogram $= ah$; triangle BPC $= \dfrac{1}{2}ah$. Difference $= \dfrac{1}{2}ah = 10 \Rightarrow ah = 20$.

Trapezium $= \dfrac{1}{2}(a + 2a)h = 1.5ah = 30$.

Solution

$\dfrac{[ADE]}{[ABC]} = \dfrac{AD}{AB} \times \dfrac{AE}{AC} = \dfrac{2}{3} \times \dfrac{2}{5} = \dfrac{4}{15}$.

$[ABC] = 8 \times \dfrac{15}{4} = 30$.

Solution

Let the side fenced at ₹200 be $x$ and the other side $y$: cost $= 200x + 100x + 200y = 300x + 200y$, with $xy = 60000$.

By AM–GM: $300x + 200y \geq 2\sqrt{300 \times 200 \times 60000} = 2 \times 60000 = 120000$.

Lowest cost = ₹120000 (at $x = 200$, $y = 300$).

Solution

Side $= 10$ m. $d_1 d_2 = 192$ and $d_1^2 + d_2^2 = 400$.

$(d_1 + d_2)^2 = 784 \Rightarrow d_1 + d_2 = 28$ m.

Cost $= 28 \times 125 = 3500$.

Solution

$AD = DE \Rightarrow \angle ADE = 180^\circ - 2A$. $BD = DF \Rightarrow \angle BDF = 180^\circ - 2B$.

$\angle FDE = 180^\circ - (180^\circ - 2A) - (180^\circ - 2B) = 2(A + B) - 180^\circ$.

$A + B = 130^\circ$, so $\angle FDE = 80^\circ$.

Solution

In triangle ADC, let $DC = x$: $20^2 = 10^2 + x^2 - 2(10)x\cos 30^\circ \Rightarrow x^2 - 10\sqrt{3}x - 300 = 0$.

$x = \dfrac{10\sqrt{3} + \sqrt{1500}}{2} = 5\sqrt{3} + 5\sqrt{15}$.

Area $= AD \times DC \times \sin 30^\circ = 5x = 25(\sqrt{3} + \sqrt{15})$.

Solution

The top piece is a cone of height 9, similar with ratio $\dfrac{1}{3}$: volume $= \dfrac{V}{27}$. The bottom piece is $\dfrac{26V}{27}$.

Difference $= \dfrac{25V}{27} = 225 \Rightarrow V = 243$ cc.

Solution

Painted (circle) $= A$, unpainted $= \dfrac{2}{3}A$, total $= \dfrac{5}{3}A = 135 \Rightarrow A = 81$, so $r = \dfrac{9}{\sqrt{\pi}}$.

The circle touches two opposite sides, so one side $= 2r = \dfrac{18}{\sqrt{\pi}}$; the other $= \dfrac{135\sqrt{\pi}}{18} = 7.5\sqrt{\pi}$.

Perimeter $= \dfrac{36}{\sqrt{\pi}} + 15\sqrt{\pi} = 3\sqrt{\pi}\left(5 + \dfrac{12}{\pi}\right)$.

Solution

Side $= \sqrt{6^2 + 8^2} = 10$; area $= \dfrac{12 \times 16}{2} = 96$.

Inradius $= \dfrac{\text{height}}{2} = \dfrac{96/10}{2} = 4.8$.

Ratio $= \dfrac{\pi(4.8)^2}{96} = \dfrac{23.04\pi}{96} = \dfrac{6\pi}{25}$.

Solution

Take O as the origin: $A(0, 4)$, $B(3, 0)$. Line AB: $4x + 3y = 12$.

Distance from O $= \dfrac{12}{5} = 2.4$.

$PQ = 2\sqrt{25 - 5.76} = 2\sqrt{19.24} \approx 8.8$ m.

Solution

Rectangle sides $x$ and $3x$: $R = 3x^2$. Triangle side $t$: $T = \dfrac{\sqrt{3}}{4}t^2$.

$R = T^2 \Rightarrow 3x^2 = \dfrac{3t^4}{16} \Rightarrow x = \dfrac{t^2}{4}$.

Perimeters: $3t + 8x = 90 \Rightarrow 2t^2 + 3t - 90 = 0 \Rightarrow t = 6$, $x = 9$.

Longer side $= 27$.

Solution

Let the radii be $R$ and $R - 1$. The chord touches the inner circle, so it is $R - 1$ from the centre.

Half-chord: $3^2 = R^2 - (R - 1)^2 = 2R - 1 \Rightarrow R = 5$.

Diameter of C1 $= 10$ cm.

Solution

In an equilateral triangle, the sum of perpendiculars from any interior point equals the altitude (Viviani's theorem), so the altitude $= s$.

Side $= \dfrac{2s}{\sqrt{3}}$, so area $= \dfrac{\sqrt{3}}{4} \times \dfrac{4s^2}{3} = \dfrac{s^2}{\sqrt{3}}$.

Solution

BC is perpendicular to DC (and AB), so the height is 4.

Drop a perpendicular from D to AB: with $\angle BAD = 45^\circ$, the horizontal part is also 4.

$AB = DC + 4 = 9$. Area $= \dfrac{1}{2}(9 + 5) \times 4 = 28$.