TRIANGLE MCQ LIST

 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:

• Pythagoras Theorem (Triplets)
• Formulas of Areas and Perimeters
• Mass Point geometry
• Similar Triangles
• Special Triangles [Equilateral, Isosceles, Right Triangles, 30-60-90 triangles]

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, $\angle BAC={120}^{\circ }$ and AD is the bisector of $\angle BAC$ . If $\frac{\left(AD\right)\left(AB\right)}{BD}=\frac{AE}{EC}(AE+EC)$ and $\angle EDC=\angle ECD,$ what is the ratio of $\angle B$ and $\angle C$ ?

[1] 1:1

[2] 1:2

[3] 2:3

[4] 5:6

Answer & Solution

Option # 2

AD is the bisector of $\angle BAC$

$\Rightarrow \frac{AB}{AC}=\frac{BD}{DC}\dots -\left(1\right)$

Given

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

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

$\frac{AD}{DC}=\frac{AE}{EC}(becauseDC=\frac{BD\times AC}{AB})$

$\Rightarrow$ DE is the angle bisector of $\angle ADC$

$\Rightarrow \angle EDC=\angle ADE=\angle ECD=y($ say )

$\angle ADB={180}^{\circ }-2y$

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

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

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

$\left(\ln \Delta ABC\right)$

$\Rightarrow y=40$

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

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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$\sqrt{3}m$

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

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

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

Answer & Solution

Option # 4

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

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

(Since area is given and $BC$ is given)

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

Now is right-angled triangle

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

$\Rightarrow BD=8$ and $DC=BC-BD=42$

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

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

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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] $\frac{3}{14}$

Answer & Solution

Option # 3

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

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

Area of $\Delta XYZ=$ area of $\Delta ABC-$ (Area of $\Delta AXZ+$ Area of $\Delta BXY+$ Area of $\Delta CYZ)$

$\frac{\text{Area of}\Delta A\times Z}{\text{Area of}\Delta ABC}=\frac{\frac{1}{2}\left(AX\right)\left(AZ\right)\sin \angle A}{\frac{1}{2}\left(AB\right)\left(AC\right)\sin \angle A}=\left(\frac{AX}{AB}\right)\left(\frac{AZ}{AC}\right)$

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

Similarly, Area of $\Delta BXY=\left(\frac{BX}{BA}\right)\left(\frac{BY}{BC}\right)$ $=\left(\frac{3}{7}\right)\left(\frac{2}{5}\right)=\frac{6}{35}$

$\frac{\text{Area of}\Delta CYZ}{\text{Area of}\Delta ABC}=\left(\frac{CY}{CB}\right)\left(\frac{CZ}{CA}\right)=\left(\frac{3}{5}\right)\left(\frac{2}{3}\right)=\frac{2}{5}$

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

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

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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] $\sqrt{3}:2$

Answer & Solution

Option # 1

Given that DE and FG are both parallel to $B$ .

$\Rightarrow$ DE is parallel to $FG$ .

$\Rightarrow \Delta ADE-\Delta AGF$

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

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Question 5:
The hypotenuse of a right-angled triangle is 20$\sqrt{3}cm$ 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$\pi (4-2\sqrt{3})$

Answer & Solution

Option # 4

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

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

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

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

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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

Answer & Solution

Option # 3

Let $CF=x,AD=AF,CE=CF,BE=BD$

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

$\Rightarrow 7+2x=49$

$x=21$

$AC=4+21=25$

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

$\angle AEB+\angle BFC=$

[1] ${60}^{\circ }$

[2] ${30}^{\circ }$

[3] ${45}^{\circ }$

[4] Cannot be determined

Answer & Solution

Option # 2

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

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

$\Rightarrow$ Triangles ABE and CFB are congruent.

$\Rightarrow \angle BFC=\angle BAE$

In $\Delta ABE,\angle ABE+\angle AEB+\angle BAE={180}^{\circ }$

${150}^{\circ }+\angle AEB+\angle BFC={180}^{\circ }$

$\Rightarrow \angle AEB+\angle BFC={30}^{\circ }$

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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

Answer & Solution

Option # 2

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

$\frac{\text{Area of}\Delta ADC}{\text{Area of}\Delta ABC}=\frac{DQ}{BN}$ and $\frac{\text{Area of}\Delta ABC}{\text{Area of}\Delta ABC}=\frac{DP}{CM}$

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

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

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

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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

Answer & Solution

Option # 2

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

${\left(\text{PM}\right)}^2={\left(\text{PQ}\right)}^2-{\left(\text{QM}\right)}^2\dots \left(1\right)$

${\left(\text{PM}\right)}^2={\left(\text{PS}\right)}^2-{\left(\text{MS}\right)}^2\dots \left(2\right)$

From (1) and (2)

$(PQ{)}^2-(PS{)}^2=(QM{)}^2-(MS{)}^2$

$=(QM+MS)(QM-MS)$

$=\left(QS\right)(RM-MS)$

$=\left(QS\right)\left(SR\right)$

$\Rightarrow \left(QS\right)\left(SR\right)=(11{)}^2-(10{)}^2=21(becausePQ=11cm,PS=10cm)$

$\left(because\$QSandSRareintegersandPQRisatriangle\right)$\Rightarrow\mathrm { QR } = \mathrm { QS } + \mathrm { SR } = 10 \mathrm { cm }$

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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

Answer & Solution

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 = $\sqrt{{10}^2+{10.5}^2}=14.5cm$

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

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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] $\sqrt{2929}$

Answer & Solution

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.

$AC=2\times 25=50cm$

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

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

$AB\times BC=672$

Solving ( 1 ) and ( 2 )

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

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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

Answer & Solution

Option # 3

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

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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

Answer & Solution

Option # 3

Let AD be the median to the largest side BC.

By Apollonius Theorem,

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

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

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

$AD=\sqrt{12}=2\sqrt{3}cm$

The median to the longest side = 3.5 cm nearly

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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] $2\sqrt{2}$

Answer & Solution

Option # 1

Let the Triangle ABC have the centroid as G.

$AB=AC,$ so $GB=GC=\sqrt{2},$ so $GE=GF=\frac{1}{\sqrt{2}},$

so $EB=\sqrt{\frac{5}{2}},$ so $AB=AC=\sqrt{10}$

$AD=ht=\sqrt{10-1}=3\cdot$ 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.

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Question 15:
$\Delta ABC$ and $\Delta DBC$ are right $\Delta$ with common hypotenuse BC. The side Ac and BD are extended to intersect at P , then $\frac{AP\times PC}{DP\times PB}=?$

[1] 1

[2] 2

[3] 1/3

[4] 4

Answer & Solution

Option # 1

$\Delta CDP\sim \Delta ABP$

as $\angle CDP=\angle BAP$

$\angle DPC=\angle BPA$

$\Rightarrow \frac{DP}{AP}=\frac{CP}{BP}\Rightarrow \frac{DP\times BP}{AP\times PC}=1$

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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

Answer & Solution

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

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

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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] $\frac{ab}{a+b}$

[3] $\frac{a+b}{ab}$

[4] ${\left(\frac{ab}{a+b}\right)}^2$

Answer & Solution

Option # 4

Let the side of square CEDF is $x$ .

$\Delta$ AFD $\sim \Delta DEB$

$\frac{AF}{FD}=\frac{DE}{EB}$

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

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

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

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

[1] $\frac{1}{\sqrt{2}}$

[2] $\frac{\sqrt{2}+2}{\sqrt{2}}$

[3] $\frac{1}{2}$

[4] $\frac{\sqrt{2}-1}{\sqrt{2}}$

Answer & Solution

Option # 4

$ar\left(\Delta ABC\right)=2$ ar $\left(\Delta XBY\right)$

$\frac{ar(\Delta \times BY)}{ar\left(\Delta ABC\right)}=\frac{1}{2}$

But $\Delta XBY\sim \Delta ABC(becauseXY\parallel AC)$

$\Rightarrow \frac{ar\left(\Delta XBY\right)}{ar\left(\Delta ABC\right)}=\frac{{XB}^2}{{AB}^2}$ (Area Thm.)

$\frac{XB}{AB}=\frac{1}{\sqrt{2}}$

$\frac{AB-AX}{AB}=\frac{1}{\sqrt{2}}\Rightarrow \frac{AX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}$

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

[1] 9$cm$

[2] 10$cm$

[3] 11$cm$

[4] Cannot be determined

Answer & Solution

Option # 2

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

${\left(\text{PM}\right)}^2={\left(\text{PQ}\right)}^2-{\left(\text{OM}\right)}^2\dots \left(1\right)$

${\left(\text{PM}\right)}^2={\left(\text{PS}\right)}^2-{\left(\text{MS}\right)}^2\dots \left(2\right)$

From (1) and (2)

$(PQ{)}^2-(PS{)}^2=(QM{)}^2-(MS{)}^2$

$=(QM+MS)(QM-MS)$

$=\left(QS\right)(RM-MS)$

$=\left(QS\right)\left(SR\right)$

$\Rightarrow \left(QS\right)\left(SR\right)=(11{)}^2-(10{)}^2=21(becausePQ=11cm,PS=10cm)$

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

since$QS$ and $SR$ are integers and $PQR$ is a triangle

$\Rightarrow QR=QS+SR=10cm$

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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

Answer & Solution

Option # 2

In triangle PSR, UT and RS are parallel.

Therefore, PT/PS = PU/PR.

In triangle PQR. US and RQ are parallel.

$\Rightarrow PSPQ=PUPR=2/3\left(given\right)$

$\Rightarrow (PT+TS):SQ=2:3=10:15$

and $PT:TS=2:3=4:6$

$\Rightarrow (PT:TS):SQ=4:6:15$ and $TS:SQ=2:5$