SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
Question – 1 : If each angle of the triangle is decreased by 10 degrees then the angles are in the ratio of 3:5:7 . What is the value of the second smallest angle?
Answer – (A) : 60
Answer – (B) : 30
Answer – (C) : 40
Answer – (D) : 60
View Answer
Correct Answer : (D)
Explanation : Let the angles after decreasing be 3x,5x and 7xEach angle in a triangle will be 3x+10,5x+10 and 7x+10Sum of angles in a triangle =1803x+10+5x+10+7x+10=18015x=150x=10Required angles are 40, 60 and 80Second smallest angle =60
Question – 2 : In △ABC,DE‖BC , find the value of x .
Answer – (A) : 3 cm
Answer – (B) : 6 cm
Answer – (C) : 9 cm
Answer – (D) : 12 cm
View Answer
Correct Answer : (A)
Explanation : Given,In △ABC , DE‖BCADBD=AEEC(Thales’ theorem)xx+1=x+3x+5x(x+5)=(x+3)(x+1)x2+5x=x2+3x+x+3x2+5x−x2−3x−x=3∴x=3 cm
Question – 3 : If the three medians of a triangle are same, then the triangle is:
Answer – (A) : equilateral
Answer – (B) : isosceles
Answer – (C) : right-angled
Answer – (D) : obtuse-angled
View Answer
Correct Answer : (A)
Explanation : According to question,AD = BE = CF = median thenAB = BC = CA∴ The triangle is an equilateral triangle.
Question – 4 : If in triangle ABC , if side AB=63,AC=12 and BC=6 , then angle B equals:
Answer – (A) : 90∘
Answer – (B) : 60∘
Answer – (C) : 90∘
Answer – (D) : 45∘
View Answer
Correct Answer : (C)
Explanation : In △ABCGiven:AB=63 cmAC=12 cmBC=6 cmAC2=AB2+BC2∴ By Converse of Pythagoras Theorem, △ABC is an Right Angle Triangle at B .∴ angle B=90∘
Question – 5 : In △ABC , if 3∠A=4∠B=6∠C , then ∠A is equal to:
Answer – (A) : 80∘
Answer – (B) : 60∘
Answer – (C) : 40∘
Answer – (D) : 30∘
View Answer
Correct Answer : (A)
Explanation : As we know,The Sum of all angles in a triangle is 180∘ .LCM of 3,4 and 6 is 12 .Now,⇒3∠A12=4∠B12=6∠C12=k∠A=4k∠B=3k∠C=2k∠A+∠B+∠C=180∘⇒4k+3k+2k=180∘⇒9k=180∘⇒k=20∘∠A=4×20∘=80∘∴∠A=80∘
Question – 6 : In a triangle, if two of the angles are complementary, then the measure of the third angle is:
Answer – (A) : 90∘
Answer – (B) : 45∘
Answer – (C) : 75∘
Answer – (D) : 90∘
View Answer
Correct Answer : (D)
Explanation : Given that, the two angles of a triangle are complementary if their sum is 90∘ .As we know that,Let the two angles be x and y such that (x+y)=90∘ . Let the third angle be z .Sum of the measures of all angles of triangle =180∘Now,x+y+z=180∘⇒90∘+z=180∘⇒z=180∘−90∘⇒z=90∘Thus, the measure of the third angle is 90∘ .
Question – 7 : △ABC and △DEF are equilateral triangles. Circumcircles are drawn around both triangles as shown in the figure. If the radius of the larger circle is R then find out the perimeter of △DEF ?
Answer – (A) : 33R2 cm
Answer – (B) : 32R4 cm
Answer – (C) : 6R3 cm
Answer – (D) : 33R2 cm
View Answer
Correct Answer : (D)
Explanation : let in △ABC the side is ‘ a ‘ and its circumcircle has radius ‘ R ‘ and its incircle has radius ‘ r ‘ soa3=R⇒a=3RThe radius of the incircle :a23=r⇒r=R323⇒R2Let the side of △DEF is ‘ b ‘, sob3=R2⇒b=R32The perimeter of △DEF=3R32=33R2cm
Question – 8 : If one angle of a triangle is equal to sum of the other two angles, then triangle is delinitely:
Answer – (A) : A right triangle
Answer – (B) : A right triangle
Answer – (C) : An equilateral triangle
Answer – (D) : An obtuse triangle
View Answer
Correct Answer : (B)
Explanation : If one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right angled triangle.In △ABC ,∠A=∠B+∠CNow, sum of angles =180∠A+∠B+∠C=180∠A+∠A=180∠A=90∘Thus, △ABC , is a right angled triangle.
Question – 9 : For △ABC∼△PQR , If the length of AB is 10 more than the length of BC and the length of PQ is twice to that of the length of QR , then find the length of BC .
Answer – (A) : 10 cm
Answer – (B) : 25 cm
Answer – (C) : 10 cm
Answer – (D) : 15 cm
View Answer
Correct Answer : (C)
Explanation : Given:For ΔABC∼ΔPQR ,In ΔABC :BC=x cmAB=(x+10) cmIn ΔPQR :PQ=2QRFormula:For ΔABC∼ΔPQR :ABPQ=BCQR=ACPRAccording to the question,⇒ ABPQ=BCQRIt can be written as:⇒ABBC=PQQR⇒x+10x=2xx⇒x+10x=2⇒x+10=2x⇒x=10 cmTherefore, ‘ 10 cm’ is the required answer.
Question – 10 : If ABC is an equilateral triangle of side a , then its altitude is equal to:
Answer – (A) : 32a
Answer – (B) : 32a
Answer – (C) : 3a
Answer – (D) : 35a
View Answer
Correct Answer : (B)
Explanation : Given,ABC is an equilateral triangle side =aAs we know that,So, we haveAB=AC=BC=aIn △ACM ,∴ sin60∘=CMAC⇒32=CMa⇒CM=a×32CM=32aAltitude =CM=32a
Question – 11 : The side AC of a triangle ABC is produced to D such that BC = CD. ∠ACB is 70°, then what is ∠ADB equal to?
Answer – (A) : 35°
Answer – (B) : 45°
Answer – (C) : 70°
Answer – (D) : 110°
View Answer
Correct Answer : (A)
Explanation : Given:BC = CD.∠ACB = 70°Concept:Linear Pair of angles are formed when two lines intersect each other at a single point.The sum of angles of a linear pair is always equal to 180°.We have,⇒ ∠ACB + ∠BCD = 180°⇒ ∠BCD = 180° – 70° = 110°In ΔBCD,⇒ BC = CD⇒ ∠CBD = ∠CDB [angles opposite to equal side]By angle sum property of a triangle∠BCD + ∠CBD + ∠CDB = 180°⇒ 2∠CDB = 180° – ∠BCD⇒ 2∠CDB = 180° – 110° = 70°⇒ ∠CDB = ∠ADB = 35°∴ ∠ADB is a measure of 35°.
Question – 12 : Area of a triangle with vertices at (2,3),(−1,0) and (2,−4) is:
Answer – (A) : 212 sq units
Answer – (B) : 212 sq units
Answer – (C) : 21 sq units
Answer – (D) : None of these
View Answer
Correct Answer : (B)
Explanation : Given (2,3),(−1,0),(2,−4)Let us Assume,A(x1,y1)=(2,3)B(x2,y2)=(−1,0)C(x3,y3)=(2,−4)Area of triangle =12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]Now,Area of given triangle=12[{2(0−(−4}+(−1){(−4−3)+2(3−0)}]=12{8+7+6}=212 sq units
Question – 13 : A 24 cm long wire is bent to form a triangle with one of the angles as 60∘ . What is the altitude of the triangle having the greatest possible area?
Answer – (A) : 43 cm
Answer – (B) : 23 cm
Answer – (C) : 6 cm
Answer – (D) : 3 cm
View Answer
Correct Answer : (A)
Explanation : Given,Length of wire =24 cmOne angle of the triangle =60∘Area is maximum when the triangle is equilateral.As we know equilateral triangle has 3 equal sidesSide of the triangle =243=8 cmAs we know height of triangle =32aAfter putting the value of a=32×8=43 cm
Question – 14 : If area of similar triangles △ABC and △DEF be 64 sq cm and 121 sq cm and EF= 15.4 cm then BC equals:
Answer – (A) : 11.2 cm
Answer – (B) : 18 cm
Answer – (C) : 11.2 cm
Answer – (D) : 8.2 cm
View Answer
Correct Answer : (C)
Explanation : Given,Area of similar triangles △ABC and △DEF be 64 sq cm and 121 sq cm .EF=15.4 cmFormula:If △ABC∼DEF , thenArea of △ABCArea of △DEF =(BCEF)2⇒64121=(BC15.4)2⇒BC15.4=811∴BC=11.2 cm
Question – 15 : In the figure if ∠ACB=∠CDA,AC=8 cm and AD=3 cm, find BD .
Answer – (A) : 553 cm
Answer – (B) : 553 cm
Answer – (C) : 643 cm
Answer – (D) : 357 cm
View Answer
Correct Answer : (B)
Explanation : In triangle ACB and ADC∠A=∠A∠ACB=∠CDATherefore triangle ACB and ADC are similar,ACAD=ABACAC2=AD×AB82=3×AB⇒AB=643This implies,BD=643−AD⇒BD=553
Question – 16 : If the sides of a right-angled triangle are three consecutive integers, then the length of the smallest side is:
Answer – (A) : 3 units
Answer – (B) : 2 units
Answer – (C) : 4 units
Answer – (D) : 5 units
View Answer
Correct Answer : (A)
Explanation : Let the length of the shortest side be x – 1.So, the length of the other two sides will be x and x + 1.Since, the three sides form a right angle triangle,(x+1)2 = (x-1)2 + (x)2 ⇒ x2 – 4x = 0The roots of a quadratic equation ax2 + bx + c = 0 is given by x=−b±b2−4ac2aSo, the roots are x=−(−4)±(−4)2−4×1×02×1 ⇒ x = 4 or 0Since, the length of a side cannot be 0, So, x = 4 unitsSo, x – 1 = 4 – 1 = 3 units
Question – 17 : A tangent MN touches a circle at point A, ΔABC is inscribed in a circle such that the length CB is the diameter of the circle. If ∠ABC = 30°, then find the ∠BAN.
Answer – (A) : 60°
Answer – (B) : 80°
Answer – (C) : 20°
Answer – (D) : 60°
View Answer
Correct Answer : (D)
Explanation : Given:A tangent MN touches a circle at point A,ΔABC is inscribed in a circle such that the length BC is the diameter of the circle.∠ABC = 30°,Formula:(1) The angle subtended by the diameter of the circle at any point on the circumference is 90°.(2) The angle between a tangent and a chord is equal to the angle in the alternate segment.∠CAB = ∠CBDAccording to the question,The angle subtended by the diameter is 90°.Therefore,∠BAC = 90°In ΔABC,⇒ ∠ABC + ∠BAC + ∠ACB = 180°⇒ 30° + 90° + ∠ACB = 180⇒ 120° + ∠ACB = 180°⇒ ∠ACB = 60°By applying the alternate segment theorem,∠BAN = ∠ACB = 60°Therefore, 60° is the required answer.
Question – 18 : What is the area of a right-angled triangle, if the radius of the circumcircle is 5 cm and altitude drawn to the hypotenuse is 4 cm ?
Answer – (A) : 20 cm2
Answer – (B) : 18 cm2
Answer – (C) : 16 cm2
Answer – (D) : 10 cm2
View Answer
Correct Answer : (A)
Explanation : Given:Radius of circumcircle =5 cmAltitude to the hypotenuse =4 cmAs we know,The hypotenuse is the diameter of a circle if an angle of a triangle is 90∘ which is inscribed in a circle, that triangle formed by the diameter of the circle.Area of triangle =(B×H)2WhereB is the base of the triangleH is the height of the triangleThe hypotenuse of tringle is BC=2× of the radius of circles(OC)BC=2×5=10 cmThe altitude drawn to hypoteneus is AD=4 cmThe area of △ABC=(10×4)2=20 cm2∴ The area of the right-angle triangle is 20 cm2
Question – 19 : A triangle has vertices (1,6),(3,0) and (−3,−7) . Its area in square units is:
Answer – (A) : 25
Answer – (B) : 25
Answer – (C) : 30
Answer – (D) : 40
View Answer
Correct Answer : (B)
Explanation : Given:A triangle has vertices (1,6),(3,0) and (−3,−7) .We know that,Area of triangle =(12)|(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|According to the question,=(12)|(1(0−(−7+3(−7−6)+(−3)(6−0|=(12)|(7−39−18)|=(12)×|−50|=25∴ Its area in square units is 25 .
Question – 20 : What is the area of the triangle ABC with sides a=10cm and c=4 cm angle B=30∘?
Answer – (A) : 10 cm2
Answer – (B) : 12 cm2
Answer – (C) : 10 cm2
Answer – (D) : 8 cm2
View Answer
Correct Answer : (C)
Explanation : Given,a=10cm , c=4cm , ∠B=30∘Area of a triangle =12× base × altitude=12×c×asin∠CBA=12×10×4sin30∘=5×4×12 [∵sin30∘=12]=10 cm2
Question – 21 : If the exterior angle of a triangle is 60∘ and the interior opposite angles are in the ratio 1:3 , then the angles of the triangle are:
Answer – (A) : 120∘,45∘,15∘
Answer – (B) : 120∘,10∘,50∘
Answer – (C) : 120∘,45∘,15∘
Answer – (D) : 60∘,90∘,30∘
View Answer
Correct Answer : (C)
Explanation : Given that, the exterior angle of a triangle is 60∘ and the interior opposite angles are in the ratio 1:3 .Let the interior angles be x and 3x . We know that, the exterior angle of triangle is equal to sum of interior opposite angles. So,x+3x=60∘⇒4x=60∘⇒x=60∘4⇒x=15∘So, the angles are:x=15∘3x=3×15∘=45∘As we know that, the sum of all the angles of a triangle is 180∘ . So, the remaining angle be y .∴ x+3x+y=180∘⇒15∘+45∘+y=180∘⇒60∘+y=180∘⇒y=180∘−60∘⇒y=120∘Thus, the angles of the triangle are 120∘,45∘,15∘ .
Question – 22 : The sum of two angles of a triangle is 115° and their difference is 15° . Find the smallest angle of the triangle.
Answer – (A) : 50°
Answer – (B) : 50°
Answer – (C) : 52°
Answer – (D) : None of these
View Answer
Correct Answer : (B)
Explanation : Let three angles of the triangle be ∠A, ∠B, and ∠C∠A+∠B=115° …..(i)∠A-∠B=15° …..(ii)∠A+∠B+∠C=180° …..(iii)From equation (i) and equation (iii)∠C=180°-115°=65°From equation (i) and equation (ii)∠A=65° and ∠B=50°
Question – 23 : The angles of a triangle are in the ratio of 4 :1 :1 . Then the ratio of the longest side to the perimeter is
Answer – (A) : 32+3
Answer – (B) : 23
Answer – (C) : 21+3
Answer – (D) : 32+3
View Answer
Correct Answer : (D)
Explanation : Let Angles be 4x, x and xAccording to the question,4x+x+x=180x=30BD is perpendicular to side ACIn ∆ABD ,longest side =ACThe perimeter of △ABC=AB+BC+CARequired ratio =232+2+3+3=32+3
Question – 24 : The internal bisector of ∠ A and the external bisector of ∠ B of triangle ABC meet at D . If ∠ACB=40° , find ∠ADB
Answer – (A) : 20∘
Answer – (B) : 50∘
Answer – (C) : 80∘
Answer – (D) : 15∘
View Answer
Correct Answer : (A)
Explanation : ∠CBE=∠ACB+∠CAB∠CBD=20°+∠DAB∠CBD=∠ADB+∠CBD-20°∠ADB=20°
Question – 25 : ABC is a triangle and D is a point on the side BC. If BC = 12 cm, BD = 9 cm and ∠ADC = ∠BAC, then the length of AC is equal to
Answer – (A) : 6 cm
Answer – (B) : 6 cm
Answer – (C) : 8 cm
Answer – (D) : 9 cm
View Answer
Correct Answer : (B)
Explanation : Given: BC = 12 cm, BD = 9 cmAs D is a point on the side BCBC = BD + CD⇒12 = 9 + CD⇒CD = 3 cmIn ΔBAC and ΔADC∠ADC = ∠BAC (Given)∠C = ∠C (Common angle)∴ ΔBAC ~ ΔADC (AA similarity criteria)The ratio of sides is also equal.BC : AC = AC : CDAC × AC = BC × CD⇒AC2 = 12 × 3 = 36⇒AC = 6 cm
Question – 26 : Which of the following is not a criterion for congruence of triangles?
Answer – (A) : SSA
Answer – (B) : SAS
Answer – (C) : SSA
Answer – (D) : ASA
View Answer
Correct Answer : (C)
Explanation : SSA is not a criterion for Congruency. If two triangles seem to be congruent by SSA rule, they cannot be said congruent.
Question – 27 : △ABC is an isosceles right-angled triangle. If ‘ r ‘ is it’s in radius and ‘ R ‘ is its circumradius, then Rr equals to :
Answer – (A) : 2+1
Answer – (B) : 3+2
Answer – (C) : 52
Answer – (D) : 3+2
View Answer
Correct Answer : (A)
Explanation : △ABC is an isosceles right-angled triangle.‘ r ‘ is it’s in radius and ‘ R ‘ is its circumradiusSides of right angled isoceles triangle will be ,a,a and 2a .∴Δ=12⋅a⋅a=a22s=a+a+a22=a(2+2)2∴ Circumradius (R)=abc4Δ=a⋅a⋅2a4(a22)=a2Inradius (r)=Δs=a22a(2+2)2=a2+2∴R:r=a2:a2+2=2+2:2=2+1:1
Question – 28 : In a triangle, values of all the angles are integers (in degree measure). Which one of the following cannot be the proportion of their measures?
Answer – (A) : ∶∶ 6∶7∶8
Answer – (B) : ∶∶ 3∶4∶5
Answer – (C) : ∶∶ 5∶6∶7
Answer – (D) : ∶∶ 6∶7∶8
View Answer
Correct Answer : (D)
Explanation : We know that,Sum of interior angle of triangle =180∘1) 1:2:3So, x+2x+3x=6xThen 6x=180∘,x=30∘Angles 30∘,60∘ and 90∘2) 3:4:5So, 3x+4x+5x=180∘Then 12x=180∘,x=15Angles 45∘,60∘ and 75∘3) 5:6:7So, 5x+6x+7x=180∘Then 18x=180∘,x=10∘Angles 50∘,60∘ and 70∘4) 6:7:8So, 6x+7x+8x=180∘Then 21x=180∘,x=8.57∘Angles 51.42∘,59.99∘ and 68.57∘∴ Option (D) is not integar angle in triangle.
Question – 29 : The angle of elevation of the top of a tower from two-point at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. What is the height of the tower?
Answer – (A) : 6 m
Answer – (B) : 6 m
Answer – (C) : 10 m
Answer – (D) : 8 m
View Answer
Correct Answer : (B)
Explanation : Given:AB is the tower.P and Q are the points at distance of 4 m and 9 m respectively.From figure, PB=4 m, QB=9 mLet the angle of elevation from P be α and the angle of elevation from Q be β .Given that α and β are supplementary.Thus, α+β=90In △ABP ,tanα=ABBP …..(i)In △ABQ ,tanβ=ABBQtan(90−α)=ABBQ (Since, α+β=90 )cotα=ABBQ1tanα=ABBQSo, tanα=BQAB …..(ii)From equation (i) and equation (ii)ABBP=BQABAB2=BQ×BPAB2=4×9AB2=36Therefore, AB=6So, the height of the tower is 6 m.
Question – 30 : The sides of a triangle are m,n and m2+n2+mn . What is the sum of the acute angles of the triangle?
Answer – (A) : 60∘
Answer – (B) : 60∘
Answer – (C) : 75∘
Answer – (D) : 90∘
View Answer
Correct Answer : (B)
Explanation : Let m=n=1 unitThen m2+n2+mn=3 unitUsing cosine rule;cosθ=a2+b2−c22abAfter putting the value,cosθ=12+12−322×1×1⇒cosθ=1+1−32×1×1⇒cosθ=−12∴θ=120∘ [∵cos120∘=−12]Now, the sum of the acute angles of the triangle =180∘−120∘=60∘
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
SSC TRIANGLE MCQ ENGLISH
v