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# Geometry: Triangle Questions for CAT with Solutions

Questions on Triangles from Geometry regularly appear in CAT exam. Most of these questions are based on the simple concepts and theorem on triangles. It has been observed that the questions on triangles frequently tests the following concepts:

It is strongly suggested that the aspirant should be good at applying the above concepts. Below we are providing 20 questions on triangles for practice.

## Triangles Practice Problems

Question 1:
In the given figure, $\angle B A C = 120 ^ { \circ }$ and AD is the bisector of $\angle B A C$ . If $\frac { ( A D ) ( A B ) } { B D } = \frac { A E } { E C } ( A E + E C )$ and $\angle E D C = \angle E C D ,$ what is the ratio of $\angle B$ and $\angle C$ ?

[1] 1:1

[2] 1:2

[3] 2:3

[4] 5:6

Option # 2

AD is the bisector of $\angle \mathrm { BAC }$

$\Rightarrow \frac { \mathrm { AB } } { \mathrm { AC } } = \frac { \mathrm { BD } } { \mathrm { DC } } \ldots - ( 1 )$

Given

$\frac { \mathrm { AD } ( \mathrm { AB } ) } { \mathrm { BD } } = \frac { \mathrm { AE } ( \mathrm { AC } ) } { \mathrm { EC } }$

$\frac { \mathrm { AD } ( \mathrm { A } \mathrm { B } ) } { \mathrm { BD } ( \mathrm { AC } ) } = \frac { \mathrm { AE } } { \mathrm { EC } }$

$\frac { \mathrm { AD } } { \mathrm { DC } } = \frac { \mathrm { AE } } { \mathrm { EC } } \left( \mathrm { because }\ \mathrm { DC } = \frac { \mathrm { BD } \times \mathrm { AC } } { \mathrm { AB } } \right)$

$\Rightarrow$ DE is the angle bisector of $\angle A D C$

$\Rightarrow \angle E D C = \angle A D E = \angle E C D = y ($ say )

$\angle A D B = 180 ^ { \circ } - 2 y$

$\Rightarrow \angle B = 180 ^ { \circ } - \left( 60 ^ { \circ } + 180 ^ { \circ } - 2 y \right)$

$\angle B = 2 y - 60 ^ { \circ }$

$2 y - 60 ^ { \circ } + 120 ^ { \circ } + y = 180 ^ { \circ }$

$( \ln \Delta A B C )$

$\Rightarrow y = 40$

$\frac { \angle B } { \angle C } = \frac { 80 ^ { \circ } - 60 ^ { \circ } } { 40 ^ { \circ } } = \frac { 1 } { 2 }$

Question 2:
If the two sides of a triangle are 50 and 20 respectively, and the area of the triangle is 150. Find the length of the third side.

[1] 40$\sqrt { 3 } \mathrm { m }$

[2] 40$\sqrt { 2} \mathrm { m }$

[3] 30$\sqrt { 3 } \mathrm { m }$

[4] 40$\sqrt { 2 } \mathrm { m }$

Option # 4

Area of Triangle ABC = (1/2) (BC) (AD) (where AD is the altitude)

Now, AD has to be equal to $\frac { \text { Area } \times 2 } { \mathrm { BC } }$

(Since area is given and $\mathrm { BC }$ is given)

$\Rightarrow \mathrm { AD } = \frac { ( 150 ) ( 2 ) } { 50 }= 6$

Now is right-angled triangle

$\Rightarrow \mathrm { BD } ^ { 2 } = \mathrm { AB } ^ { 2 } - \mathrm { AD } ^ { 2 } = 10 ^ { 2 } - 6 ^ { 2 } = 8 ^ { 2 }$

$\Rightarrow \mathrm { BD } = 8$ and $\mathrm { DC } = \mathrm { BC } - \mathrm { BD } = 42$

$\Rightarrow \mathrm { AC } ^ { 2 } = \mathrm { AD } ^ { 2 } + \mathrm { DC } ^ { 2 } = ( 6 ) ^ { 2 } + ( 42 ) ^ { 2 }$

$\Rightarrow \mathrm { AC } = \sqrt { 1800 } = 30 \sqrt { 2 }$

Question 3:
X, Y and Z are points on the sides AB, BC and AC of the triangle ABC, such that AX : XB = 4:3 BY : YC = 2: 3 and CZ : ZA = 2:1. Find the ratio of the area of the triangle XYZ to that of the triangle ABC.

[1] $\frac { 4 } { 35 }$

[2] $\frac { 4 } { 21 }$

[3] $\frac { 5 } { 21 }$

[4] $\frac { 3 } { 14 }$

Option # 3

Area of the triangle $ABC = \frac { 1 } { 2 } ( AB ) ( BC ) \sin \angle B$

$= \frac { 1 } { 2 } ( \mathrm { AB } ) ( \mathrm { AC } ) \sin \angle \mathrm { A } = \frac { 1 } { 2 } ( \mathrm { AC } ) ( \mathrm { BC } ) \sin \angle \mathrm { C }$

Area of $\Delta X Y Z =$ area of $\Delta A B C -$ (Area of $\Delta A X Z +$ Area of $\Delta B X Y +$ Area of $\Delta C Y Z )$

$\frac { \text { Area of } \Delta A \times Z } { \text { Area of } \Delta A B C } = \frac { \frac { 1 } { 2 } ( A X ) ( A Z ) \sin \angle A } { \frac { 1 } { 2 } ( A B ) ( A C ) \sin \angle A } = \left( \frac { A X } { A B } \right) \left( \frac { A Z } { A C } \right)$

$= \left( \frac { 4 } { 7 } \right) \left( \frac { 1 } { 3 } \right) = \frac { 4 } { 21 }$

Similarly, Area of $\Delta B X Y = \left( \frac { B X } { B A } \right) \left( \frac { B Y } { B C } \right)$ $= \left( \frac { 3 } { 7 } \right) \left( \frac { 2 } { 5 } \right) = \frac { 6 } { 35 }$

$\frac { \text { Area of } \Delta C Y Z } { \text { Area of } \Delta A B C } = \left( \frac { C Y } { C B } \right) \left( \frac { C Z } { C A } \right) = \left( \frac { 3 } { 5 } \right) \left( \frac { 2 } { 3 } \right) = \frac { 2 } { 5 }$

Therefore Area of $\Delta \mathrm { ABC } = 1 - \left( \frac { 4 } { 21 } + \frac { 6 } { 35 } + \frac { 2 } { 5 } \right)$

$= - \left( \frac { 20 + 18 + 42 } { 5 \times 7 \times 3 } \right) = 1 - \frac { 16 } { 21 } = \frac { 5 } { 21 }$

Question 4:
In the triangle ABC given, DE and F6 are drawn parallel to BC, such that the areas of triangle ADE, quadrilateral EGFD and quadrilateral GCBF are all equal. What is the ratio of the lengths of DE and FG?

[1] $1 : \sqrt { 2 }$

[2] $\sqrt { 2 } : 3$

[3] 2:3

[4] $\sqrt { 3 } : 2$

Option # 1

Given that DE and FG are both parallel to $\mathrm { B }$ .

$\Rightarrow$ DE is parallel to $\mathrm { FG }$ .

$\Rightarrow \Delta \mathrm { ADE } - \Delta \mathrm { AGF }$

$\begin{array}{*{35}{l}}\Rightarrow {{\left( \frac{\text{DE}}{FG} \right)}^{2}}=\frac{1}{2}\Rightarrow \frac{DE}{FG}=\frac{1}{\sqrt{2}} & \text{ } \\ \end{array}$

Question 5:
The hypotenuse of a right-angled triangle is 20$\sqrt { 3 } \mathrm { cm }$ and one of its angles is $30 ^ { \circ } .$ Find the area (in sq.cm) of the largest circle that can be cut out from the triangle.

[1] 180$\pi$

[2] 75$\pi ( 4 + 2 \sqrt { 3 } )$

[3] 300$\pi$

[4] 75$\pi ( 4 - 2 \sqrt { 3 } )$

Option # 4

In a right angled triangle (of perpendicular sides a and b ,and hypotenuse c) the inradius $= ( a + b - c ) / 2$

Hence $r = ( 10 \sqrt { 3 } + 30 - 20 \sqrt { 3 } ) / 2 = \quad 15 - 5 \sqrt { 3 }$

$= 5 \sqrt { 3 } ( \sqrt { 3 } - 1 )$

Therefore area of the incircle $= \pi 75 ( 4 - 2 \sqrt { 3 } ) \mathrm { cm } ^ { 2 }$

Question 6:
ABC is a triangle right-angled at B. The circle inscribed in the triangle touches AB, BC and CA at D, E and F respectively. If BE = 3 cm and AD = 4 cm, f‌ind AC.

[1] 13 cm

[2] 17 cm

[3] 25 cm

[4] 29 cm

Option # 3

Let $C F = x , A D = A F , C E = C F , B E = B D$

$( 4 + x ) ^ { 2 } - ( 3 + x ) ^ { 2 } = ( 4 + 3 ) ^ { 2 }$ are tangents.

$\Rightarrow 7 + 2 x = 49$

$x = 21$

$A C = 4 + 21 = 25$

Question 7:
In the figure given below, A B C D is a rectangle and BCE and CDF are equilateral triangles.

$\angle A E B + \angle B F C =$

[1] $60 ^ { \circ }$

[2] $30 ^ { \circ }$

[3] $45 ^ { \circ }$

[4] Cannot be determined

Option # 2

Since triangles $\mathrm { BEC }$ and CDF are equilateral triangles, we get $\angle \mathrm { ABE } = \angle \mathrm { BCF } = 90 ^ { \circ } + 60 ^ { \circ } = 150 ^ { \circ }$

$\mathrm { AB } = \mathrm { CF } , \mathrm { BC } = \mathrm { BE }$ and $\angle \mathrm { ABE } = \angle \mathrm { BCF }$

$\Rightarrow$ Triangles ABE and CFB are congruent.

$\Rightarrow \angle B F C = \angle B A E$

In $\Delta A B E , \angle A B E + \angle A E B + \angle B A E = 180 ^ { \circ }$

$150 ^ { \circ } + \angle A E B + \angle B F C = 180 ^ { \circ }$

$\Rightarrow \angle A E B + \angle B F C = 30 ^ { \circ }$

Question 8:
In triangle ABC, D is a point on BC. P and Q are points on AB and AC respectively such that DP is perpendicular to AB and DQ is perpendicular to AC. If the altitudes from B to AC and C to AB are 30 cm and 40 cm respectively and DO = 6, f‌ind DP.

[1] 24cm

[2] 32 cm

[3] 36 cm

[4] 480m

Option # 2

Let the altitudes from $B$ and $C$ be BN and CM

$\frac { \text { Area of } \Delta A D C } { \text { Area of } \Delta A B C } = \frac { D Q } { B N }$ and $\frac { \text { Area of } \Delta A B C } { \text { Area of } \Delta A B C } = \frac { D P } { C M }$

$\frac { \text { Area of } \Delta \mathrm { ADC } } { \text { Area of } \Delta \mathrm { ADC } } + \frac { \text { Area of } \Delta \mathrm { ABD } } { \text { Area of } \Delta \mathrm { ABC } } = 1$

$\frac { \mathrm { DP } } { \mathrm { CM } } = 1 - \frac { 1 } { 5 } = \frac { 4 } { 5 }$

$\Rightarrow \mathrm { DP } = \frac { 4 } { 5 } ( \mathrm { CM } ) = \frac { 4 } { 5 } ( 40 ) = 32$

Question 9:
In a APQR, P0 = PR = 11 cm and S is a point on QR such that PS = 10 cm. If the lengths of QB and SR, when expressed in cm, are integers, then find the length of OR.

[1] 9 cm

[2] 10 cm

[3] 11cm

[4] Cannot be determined

Option # 2

Let $\overline { \mathrm { PM } } \perp \overline { \mathrm { QR } }$ and $\mathrm { QS } > \mathrm { SR }$

${{(\text{PM})}^{2}}={{(\text{PQ})}^{2}}-{{(\text{QM})}^{2}}\ldots (1)$

${{(\text{PM})}^{2}}={{(\text{PS})}^{2}}-{{(\text{MS})}^{2}}\ldots (2)$

From (1) and (2)

$( P Q ) ^ { 2 } - ( P S ) ^ { 2 } = ( Q M ) ^ { 2 } - ( M S ) ^ { 2 }$

$= ( Q M + M S ) ( Q M - M S )$

$= ( Q S ) ( R M - M S )$

$= ( Q S ) ( S R )$

$\Rightarrow ( \mathrm { QS } ) ( \mathrm { SR } ) = ( 11 ) ^ { 2 } - ( 10 ) ^ { 2 } = 21 ( \mathrm { because }\ P Q = 11 \mathrm { cm } , \mathrm { PS } = 10 \mathrm { cm } )$

$( \mathrm { because }\$ QS and SR are integers and PQR is a triangle)$\Rightarrow$ $\mathrm { QR } = \mathrm { QS } + \mathrm { SR } = 10 \mathrm { cm }$

Question 10:
If two of the sides of a right triangle are 10 cm and 10.5 cm and its inradius is 3 cm, what is its circumradius?

[1] 14.5 cm

[2] 50m

[3] 5.25 cm

[4] 7.25 cm

Option # 4

The hypotenuse is the longest side of a right-angled triangle. Given that two of the sides of a right triangle are 10 cm and 10.5 cm.

If hypotenuse = 10.5 cm, then the sides containing the right angle are 10 cm $\sqrt { ( 10.5 ) ^ { 2 } - 10 ^ { 2 } } = \sqrt { 10.25 } - 3.2$ and But the inradius of the triangle is given as 3 cm. The smallest of the sides is more than 6 cm long. Therefore, the 10.5 cm side is not the hypotenuse. Hence the lengths of the sides containing the right angle are 10 cm and 10.5 cm. 80, hypotenuse = $\sqrt { 10 ^ { 2 } + 10.5 ^ { 2 } } = 14.5 \mathrm { cm }$

The circumradius of the right triangle = $\frac { 14.5 } { 2 } = 7.25 \mathrm { cm }$

Question 11:
The shortest median of a right-angled triangle is 25 units. If the area of the triangle is 336 sq.units, what is the length (in units) of the longest median of the triangle?

[1] $\sqrt { 772 }$

[2] $\sqrt { 821 }$

[3] $\sqrt { 2353 }$

[4] $\sqrt { 2929 }$

Option # 3

The shortest median is the median drawn on to the largest side.

Let AC be the hypotenuse and AB be the shortest side.

$A C = 2 \times 25 = 50 \mathrm { cm }$

Area $= \frac { 1 } { 2 } \times \mathrm { AB } \times \mathrm { BC } = 336$

$\mathrm { AB } ^ { 2 } + \mathrm { BC } ^ { 2 } = 2500$

$\mathrm { AB } \times \mathrm { BC } = 672$

Solving ( 1 ) and ( 2 )

$\begin{array}{*{35}{l}}\text{ We get, AB }=14\text{cm and BC}=48\text{cm} & {} \\ \text{ Length of the longest median }=\sqrt{{{(\text{AB}/2)}^{2}}+\text{B}{{\text{C}}^{2}}} & {} \\ =\sqrt{49+2304}=\sqrt{2353} & \text{ } \\ \end{array}$

Question 12:
Side AB of a triangle ABC is 80 cm long, whose perimeter is 170 cm. If angle ABC = 60 degrees, the shortest side of triangle ABC measures (cm).

[1] 40

[2] 36

[3] 17

[4] 14

Option # 3

$a + b = 90 \mathrm { cm } .$ cosine rule gives $\cos 60 ^ { \circ } = \frac { a ^ { 2 } + 80 ^ { 2 } - b ^ { 2 } } { 2 \times a \times 80 } \cdot$ Solving $a = 17$ and $b = 73$

Question 13:
The sides of a triangle are 5 cm, 7 cm and 10 cm. Find the length of the median to the longest side.

[1] 4.5 cm

[2] 1 cm

[3] 3.5 cm

[4] 4.2 cm

Option # 3

Let AD be the median to the largest side BC.

By Apollonius Theorem,

$A B ^ { 2 } + A C ^ { 2 } = 2 \left( A D ^ { 2 } + B D ^ { 2 } \right)$

$49 + 25 = 2 \left( A D ^ { 2 } + 25 \right)$

$2 A D ^ { 2 } = 24 \Rightarrow A D ^ { 2 } = 12$

$A D = \sqrt { 12 } = 2 \sqrt { 3 } \mathrm { cm }$

The median to the longest side = 3.5 cm nearly

Question 14:
The unequal side of an isosceles triangle is 2 cm. The medians drawn to the equal sides are perpendicular. The area of the triangle is

[1] 3

[2] $\sqrt { 10 }$

[3] $2\sqrt { 3 }$

[4] $2\sqrt { 2 }$

Option # 1

Let the Triangle ABC have the centroid as G.

$A B = A C ,$ so $G B = G C = \sqrt { 2 } ,$ so $G E = G F = \frac { 1 } { \sqrt { 2 } } ,$

so $E B = \sqrt { \frac { 5 } { 2 } } ,$ so $A B = A C = \sqrt { 10 }$

$A D = h t = \sqrt { 10 - 1 } = 3 \cdot$ Area $= \frac { 1 } { 2 } \times 3 \times 2 = 3$

Where D, E and F are the points where the medians from vertex A, B and C respectively meet the opposite sides.

Question 15:
$\Delta A B C$ and $\Delta D B C$ are right $\Delta$ with common hypotenuse BC. The side Ac and BD are extended to intersect at P , then $\frac { A P \times P C } { D P \times P B } = ?$

[1] 1

[2] 2

[3] 1/3

[4] 4

Option # 1

$\Delta C D P \sim \Delta A B P$

as $\angle C D P = \angle B A P$

$\quad \angle D P C = \angle B P A$

$\Rightarrow \quad \frac { D P } { A P } = \frac { C P } { B P } \Rightarrow \frac { D P \times B P } { A P \times P C } = 1$

Question 16:
A triangle has sides 20, 48 and 52 cm. This triangle is cut into 2 pieces of equal areas by a single straight cut. What is the maximum possible sum of the perimeters of the two pieces?

[1] 178

[2] 190

[3] 198

[4] 218

Option # 4

Maximum perimeter will be there when the median is to the shortest side.

So

$x = \sqrt { 48 ^ { 2 } + 10 ^ { 2 } } = \sqrt { 2404 } \approx 49 .$ So the total perimeter

$= 49 \times 2 + 20 + 52 + 48 = 218$

Question 17:
In a right angle triangle ABC, what is the maximum possible area of a square that can be inscribed when one of its vertices coincide with the vertex of right angle of the triangle?

[1] $\frac { a } { b }$

[2] $\frac { a b } { a + b }$

[3] $\frac { a + b } { a b }$

[4] $\left( \frac { a b } { a + b } \right) ^ { 2 }$

Option # 4

Let the side of square CEDF is $x$ .

$\Delta$ AFD $\sim \Delta D E B$

$\frac { A F } { F D } = \frac { D E } { E B }$

$\frac { b - x } { x } = \frac { x } { a - x }$

$x = \frac { a b } { a + b }$

Area of square BDEF $= \left( \frac { a b } { a + b } \right) ^ { 2 }$ . Ans. (4)

Question 18:
In the figure XY || AC and XY divides triangular region ABC into two part equal in area. Then $\frac { \mathbf { A } \mathbf { X } } { \mathbf { A } \mathbf { B } }$ is equal to

[1] $\frac { 1 } { \sqrt { 2 } }$

[2] $\frac { \sqrt { 2 } + 2 } { \sqrt { 2 } }$

[3] $\frac { 1 } { 2 }$

[4] $\frac { \sqrt { 2 } - 1 } { \sqrt { 2 } }$

Option # 4

$\mathrm { ar } ( \Delta \mathrm { ABC } ) = 2$ ar $( \Delta \mathrm { XBY } )$

$\frac { \mathrm { ar } ( \Delta \times B Y ) } { \mathrm { ar } ( \Delta A B C ) } = \frac { 1 } { 2 }$

But $\Delta \mathrm { XBY } \sim \Delta \mathrm { ABC } ( \mathrm { because }\ \mathrm { XY } \| \mathrm { AC } )$

$\Rightarrow \frac { \mathrm { ar } ( \Delta \mathrm { XB } \mathrm { Y } ) } { \mathrm { ar } ( \Delta \mathrm { ABC } ) } = \frac { \mathrm { XB } ^ { 2 } } { \mathrm { AB } ^ { 2 } }$ (Area Thm.)

$\frac { X B } { A B } = \frac { 1 } { \sqrt { 2 } }$

$\frac { A B - A X } { A B } = \frac { 1 } { \sqrt { 2 } } \quad \Rightarrow \frac { A X } { A B } = \frac { \sqrt { 2 } - 1 } { \sqrt { 2 } }$

Question 19:
In a triangle $\mathrm { PQR } , \mathrm { PQ } = \mathrm { PR } = 11 \mathrm { cm }$ and $\mathrm { S }$ is a point on QR such that $\mathrm { PS } = 10 \mathrm { cm }$ . If the lengths of QS and $\mathrm { SR }$ , when expressed in $\mathrm { cm }$ , are both integers, then find the length of $\mathrm { QR }$

[1] 9$\mathrm { cm }$

[2] 10$\mathrm { cm }$

[3] 11$\mathrm { cm }$

[4] Cannot be determined

Option # 2

Let $\overline { \mathrm { PM } } \perp \overline { \mathrm { QR } }$ and $\mathrm { QS } > \mathrm { SR }$

${{(\text{PM})}^{2}}={{(\text{PQ})}^{2}}-{{(\text{OM})}^{2}}\ldots (1)$

${{(\text{PM})}^{2}}={{(\text{PS})}^{2}}-{{(\text{MS})}^{2}}\ldots (2)$

From (1) and (2)

$( P Q ) ^ { 2 } - ( P S ) ^ { 2 } = ( Q M ) ^ { 2 } - ( M S ) ^ { 2 }$

$= ( \mathrm { QM } + \mathrm { MS } ) ( \mathrm { QM } - \mathrm { MS } )$

$= ( \mathrm { QS } ) ( \mathrm { RM } - \mathrm { MS } )$

$= ( \mathrm { QS } ) ( \mathrm { SR } )$

$\Rightarrow ( \mathrm { QS } ) ( \mathrm { SR } ) = ( 11 ) ^ { 2 } - ( 10 ) ^ { 2 } = 21 ( \mathrm { because }\ \mathrm { PQ } = 11 \mathrm { cm } , \mathrm { PS } = 10 \mathrm { cm } )$

$\Rightarrow \mathrm { QS } = 7 \mathrm { cm }$ and $\mathrm { SR } = 3 \mathrm { cm }$ is the only integer solution.

since$QS$ and $\mathrm { SR }$ are integers and $\mathrm { PQR }$ is a triangle

$\Rightarrow Q R = Q S + S R = 10 \mathrm { cm }$

Question 20:
In triangle PQR, T and S are points on PQ and U is a point in PR such that UT and RS are parallel and US and RQ are parallel. If PS : SQ = 2 : 3, f‌ind TS : SQ.

[1] 3:5

[2] 2:5

[3] 4:5

[4] 2:3

Option # 2

In triangle PSR, UT and RS are parallel.

Therefore, PT/PS = PU/PR.

In triangle PQR. US and RQ are parallel.

$\Rightarrow P S P Q = P U P R = 2 / 3 ( g i v e n )$

$\Rightarrow ( P T + T S ) : S Q = 2 : 3 = 10 : 15$

and $P T : T S = 2 : 3 = 4 : 6$

$\Rightarrow ( P T : T S ) : S Q = 4 : 6 : 15$ and $T S : S Q = 2 : 5$

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