Triangles
Mathematics · Class 10 · All
Right-angled isosceles triangle at B, AP perp bisector of BC. AC² =
All 110 questions in this quiz
Which statement best summarizes the chapter "Triangles"?
- Similar triangles always have equal sides.
- Only right triangles can be similar.
- The chapter explains similarity criteria, proportionality theorems, relationships between corresponding sides and areas, and the Pythagoras Theorem, which are essential for solving geometric problems.
- Every triangle is similar to every other triangle.
Which of the following is NOT a similarity criterion?
- AA
- SAS
- SSS
- ASA
If a line divides two sides of a triangle proportionally, then according to the converse of BPT, the line is:
- Perpendicular to the third side
- Parallel to the third side
- Equal to the third side
- Bisects the third side
Which statement about similar triangles is true?
- Corresponding angles are unequal.
- Corresponding sides are proportional.
- All sides are equal.
- Areas are always equal.
If AB/DE = BC/EF = AC/DF, then the triangles are:
- Congruent
- Similar by SSS criterion
- Right triangles only
- Isosceles triangles
Which theorem is used in a right triangle?
- Midpoint Theorem
- Basic Proportionality Theorem
- Pythagoras Theorem
- Angle Bisector Theorem
In two similar triangles, if one side is doubled, then every corresponding side is:
- Halved
- Doubled
- Tripled
- Unchanged
If two similar triangles have corresponding sides in the ratio 3:5, then the ratio of their areas is:
- 3:5
- 6:10
- 9:25
- 15:25
Which similarity criterion requires one angle to be equal and the including sides proportional?
- AA
- SAS
- SSS
- RHS
If the areas of two similar triangles are in the ratio 25:49, then the ratio of their corresponding sides is:
- 25:49
- 5:7
- 7:5
- 10:14
Which theorem states that a line drawn parallel to one side of a triangle divides the other two sides proportionally?
- Pythagoras Theorem
- Basic Proportionality Theorem (BPT)
- Midpoint Theorem
- Angle Bisector Theorem
If two triangles are similar, the ratio of their corresponding sides is:
- Equal
- Zero
- Infinite
- Always 1
If two triangles are similar, then their corresponding angles are:
- Supplementary
- Equal
- Complementary
- Unequal
Which criterion is known as the Side-Side-Side (SSS) similarity criterion?
- Corresponding angles are equal.
- Corresponding sides are proportional.
- One side is equal.
- Only one angle is equal.
The corresponding sides of two similar triangles are in the ratio 7 : 9. If the perimeter of the smaller triangle is 63 cm, what is the perimeter of the larger triangle?
- 72 cm
- 81 cm
- 84 cm
- 90 cm
In triangle ABC, D and E are points on AB and AC respectively such that DE is parallel to BC. If AD = 5 cm, AB = 15 cm and AC = 18 cm, then find AE.
- 4 cm
- 5 cm
- 6 cm
- 8 cm
Triangles ABC and PQR are similar. If the ratio of their corresponding sides is 3 : 5 and the area of triangle PQR is 200 cm², then what is the area of triangle ABC?
- 72 cm²
- 80 cm²
- 90 cm²
- 120 cm²
In triangle ABC, point D lies on BC such that BD : DC = 2 : 3. If the area of triangle ABD is 48 cm², find the area of triangle ADC.
- 60 cm²
- 72 cm²
- 75 cm²
- 80 cm²
Two similar triangles have corresponding sides in the ratio 4 : 7. If the difference between their perimeters is 39 cm, find the perimeter of the larger triangle.
- 91 cm
- 84 cm
- 78 cm
- 98 cm
In triangle ABC, points D and E lie on sides AB and AC respectively such that DE is parallel to BC. If AD = 6 cm, DB = 3 cm and AE = 8 cm, then find the length of EC.
- 2 cm
- 3 cm
- 4 cm
- 6 cm
If Σ(3/4)ˣ from x=0 to ∞ is finite, then
- x<4
- x>4
- x<1/4
- x>1/4
If 1st, 2nd and middle terms of AP are a, b, c, sum of series is
- 2c(c–a)/(b–a)
- 2c(b–a)/(c–a)
- 2c(c–a)/(b–a)
- c(c–a)/(b–a)
Ratio of sum of m and n terms of AP is m²:n². Ratio of mth and nth terms is
- m–1:n–1
- 2m+1:2n+1
- 2m–1:2n–1
- m:n
If a₁,a₂,...,aₙ are in AP, sum of 1/√(a₁+√a₂)+... equals
- (n–1)/√(a₁+aₙ)
- n/√(a₁+aₙ)
- (n–1)/√(a₁·aₙ)
- none
If AM of roots of quadratic is 8/5 and AM of reciprocals is 8/7, the quadratic is
- 5x²–8x+7=0
- 5x²–16x+7=0
- 7x²–16x+5=0
- 7x²+16x+5=0
If a,b,c are in AP, which is NOT true?
- 1/a,1/b,1/c are in HP
- a+k,b+k,c+k are in AP
- a²,b²,c² are in AP
- a+b,c+a,b+c are in AP
7th term of AP is 40. Sum of first 13 terms is
- 520
- 53
- 2080
- 1040
5th term of GP is 2; product of first 9 terms is
- 256
- 128
- 18
- 512
Sum of infinite GP is 20 and sum of squares is 100. Common ratio is
- 5
- 1/2
- 3/4
- 2/3
If nth terms of APs 3,10,17,... and 63,65,67,... are equal, then n =
- 13
- 15
- 9
- 8
If AM of roots of quadratic is 8 and GM is 5, the equation is
- x²+16x–25=0
- x²–8x+5=0
- x²–16x+25=0
- x²–16x–25=0
In AP of 81 terms, 41st term is 10. Sum of series is
- 41×41
- 10×81
- 10×41
- 81×10/2
Value of (7+77+777+...)/7 is
- 3
- 6
- 9
- 1
If pth, qth, rth terms of GP are in GP, then p,q,r are in
- AP
- GP
- HP
- p=q=r
If aˣ=bʸ=c^z and a,b,c are in GP, then x,y,z are in
- AP
- GP
- HP
- AGP
If p,q,r are in AP and x,y,z are in GP, then x^(q–r)·y^(r–p)·z^(p–q) =
- 1
- xyz
- x+y+z
- p+q+r
If p,q,r are in AP, then pth, qth, rth terms of any GP are in
- AP
- GP
- HP
- AGP
If Sᵣ denotes sum of first r terms of GP, then Sₙ, S₂ₙ–Sₙ, S₃ₙ–S₂ₙ are in
- AP
- GP
- HP
- AGP
If a>0,b>0,c>0 are in GP, logₐx, logᵦx, log_cx are in
- AP
- GP
- HP
- AGP
If A is single AM and S is sum of n AMs between a and b, then S/A depends on
- n,a,b
- n,a
- n,b
- n
If a,b,c are in AP, then 10^(ax+10), 10^(bx+10), 10^(cx+10) are in
- AP
- GP only when x>0
- GP for all x
- GP only when x<0
Third term of GP is 4. Product of first five terms is
- 4³
- 4⁵
- 4⁴
- 4⁶
20th term of series 2×4+4×6+6×8+... is
- 1600
- 1680
- 420
- 840
Sum of n terms of AP is Pn+Qn². Common difference is
- P+Q
- P–Q
- 2P
- 2Q
First term of AP is p²+1, sum of (2p+1) terms can be expressed as
- (p+1)²
- (2p+1)(p+1)²
- (p+1)³
- p³+(p+1)³
AP with n (odd) terms and middle term m. Sum of AP is
- 2mn
- mn/2
- mn
- mn²
If a,b,c,d,e form AP, then a–4b+6c–4d+e =
- 1
- 2
- 0
- 3
If a,b,c,d,e,f are in AP, then e–c equals
- 2(c–a)
- 2(f–d)
- 2(d–c)
- d–c
If 3rd term of AP is 12 and 7th term is 24, then 10th term is
- 36
- 39
- 30
- 33
If a, b, c are in AP, then 3ᵃ, 3ᵇ, 3ᶜ are in
- AP
- GP
- HP
- AGP
For AP with first term a and common difference d, the nth term is
- aₙ=a+nd
- aₙ=a–(n+1)d
- aₙ=2a+(n–1)d
- aₙ=a+(n–1)d
If a=–1, d=1/2, the AP terms are
- –1, 4, ...
- –1, –1/2, 0, 1/2,...
- –1, –1/2, 0, 1/2,...
- –1, 4,...
First five terms of the sequence aₙ = (n–1)/(n+1) are
- 0,1/3,1/2,3/5,2/3
- 0,1/3,2/4,3/5,4/6
- 1/2,2/3,3/4,4/5,5/6
- 1,2,3,4,5
If sums of n, 2n, 3n terms of AP are S₁, S₂, S₃, then S₃/(S₂–S₁) =
- 0
- 1/2
- 2
- 3
Sum of all two-digit natural numbers is
- 4900
- 4500
- 4905
- 4599
If x, a₁, a₂, y and x, b₁, b₂, y are each in AP, then (a₁–a₂)/(b₁–b₂) =
- 0
- –1
- 2
- 1
AM between 10½ and 25½ is
- 17.9
- 18
- 19.5
- 18.5
Sum of series 45²–43²+44²–42²+43²–41²+... to 30 terms is
- 2000
- 1790
- 2210
- 2220
If 'a' is AM of 3 numbers and 'b' is AM of their squares, the AM of pairwise products in terms of a and b is
- (9a²–b)/3
- (9a²–b)/6
- (b–3a²)/2
- (9a²–b)/2
Sum of first 3000 positive integers is
- 450150
- 4051500
- 4501500
- 4508542
Sum of first 14 terms of an AP is 1050 and first term is 10. The 20th term is
- 100
- 300
- 200
- 700
Sum of first 50 terms of AP: 8, 3, –2,...
- –2225
- 2225
- 2220
- –5725
Flower bed rows: 23, 19, 15,..., 7 rose plants. How many rows?
- 4
- 3
- 5
- 7
How many two-digit numbers are divisible by 5?
- 15
- 17
- 18
- 20
What is the sum of first 54 terms of 5, 11, 17, 23,...?
- 287
- 8856
- 5871
- 4856
In an AP, 3rd term is 11 and 7th term is 29. The AP starting values are
- 5, 9, 13,...
- 2, 6.5, 11,...
- 2, 6, 10,...
- 3, 7, 11,...
Which term of the AP: 21, 18, 15,... is –81?
- 36
- 33
- 37
- 35
What is the 10th term of the AP: 3, 10, 17,...?
- 56
- 46
- 17
- 66
Which of the following lists of numbers form an AP? (i) 4,10,16,22,... (ii) 1,–1,–3,–5,... (iii) –2,2,–2,2,... (iv) 1,1,1,2,2,2,...
- i, ii & iv
- ii, iii & iv
- i & ii
- iii & iv
For the AP: 3/2, 1/2, –1/2, –3/2,..., the common difference d is
- d=1
- d=–1
- d=2
- d=5
Right triangle ABC at B, medians AD and CE from A and C. AC=5cm, AD=3√5/2. CE =
- √5 cm
- 2√5 cm
- 2 cm
- 2 cm
Similar triangles areas 121 cm² and 64 cm². Median of first is 12.1 cm. Corresponding median of other is
- 8 cm
- 8.5 cm
- 8.9 cm
- 8.7 cm
Right-angled isosceles triangle at B, AP perp bisector of BC. AC² =
- AP²+2BP²
- AP²–2BP²
- AP²+2BP·CP
- AP²–2BP·CP
AB||MN, PA=x–2, PM=x, PB=x–1, PN=x+2. Value of x is
- 2
- 3
- 4
- 5
Ladder foot is 5.5m from wall, top reaches 9m high. Length of ladder is
- 10.52 m
- 10.54 m
- 11 m
- 10.9 m
Girl 120cm tall walks from lamp (3.6m) at 1.2m/s. Shadow after 4 seconds is
- 1.6 m
- 2.2 m
- 2.4 m
- 2.6 m
Triangle ABC~PQR, angle A=80°, angle B=70°. Find angle R.
- 20°
- 30°
- 35°
- 37°
In triangle ABC, AB>AC and AD⊥BC. AB²–AC² =
- BD²+AD²
- BD²+CD²
- BD²+AC²
- BD²–CD²
Rhombus ABCD. AB²+BC²+CD²+DA² =
- OA²+OB²
- OB²+OC²
- OC²+OD²
- AC²+BD²
In right triangle ABC right angled at C, AD is median. AB² =
- AC²–AD²
- AD²–AC²
- 3AC²–4AD²
- 4AD²–3AC²
CD⊥AB, CD=p. Then 1/p² =
- 1/a²+1/b²
- 1/a²–1/b²
- 1/a·b
- 1/(a+b)
ABC obtuse. AC² =
- AB²+BC²–2BC·BD
- AB²+BC²
- AB²+BC²+2BC·BD
- AD²+BD²
Areas of similar triangles 81:49. Ratio of angle-bisector segments is
- 5:4
- 9:7
- 4:5
- 625:256
P, Q on AB, AC of triangle ABC: AP=3.5, PB=7, AQ=3, QC=6, PQ=4.5. BC=
- 11.5 cm
- 13.5 cm
- 14.5 cm
- 15.5 cm
Similar triangles areas 64 cm² and 49 cm². Altitude of first is 6cm. Altitude of second is
- 5.3
- 3.5
- 27.56
- 8.76
ABC and DEF similar, BC=3.5cm, EF=2.5cm, area ABC=7 sq cm. Area DEF is
- 4.59
- 5.49
- 9.54
- 3.57
Boy 90cm tall walks from base of 3.6m building at 1.2m/s. Shadow after 4 seconds is
- 1.6 m
- 1.7 m
- 1.8 m
- 2 m
Two angles of triangle are congruent; third is 30°. Measure of one congruent angle is
- 45°
- 60°
- 75°
- 90°
DE||BC, AE=AC/4, AB=6cm. Value of AD is
- 5 cm
- 1 cm
- 1.5 cm
- 2 cm
In figure CAB~CED with given measurements, value of x is
- 3 m
- 6 m
- 9 m
- 12 cm
Similar triangles have perimeters 25 cm and 15 cm. One side of first is 9 cm. Corresponding side of other is
- 1.4 cm
- 1.6 cm
- 3.4 cm
- 5.4 cm
Student 90cm tall walks from lamp (3.6m high) at 1.2m/s. Shadow length after 4 seconds is
- 1 m
- 1.6 m
- 3.6 m
- 4.8 m
Areas of similar triangles are 169 cm² and 121 cm². Longest side of larger is 26 cm. Longest side of smaller is
- 16 cm
- 22 cm
- 26 cm
- 42 cm
Areas of two similar triangles are 81 cm² and 49 cm². Ratio of corresponding heights is
- 3:7
- 5:7
- 7:9
- 9:7
25m ladder reaches window 20m above ground. Distance of foot from building is
- 5 m
- 10 m
- 15 m
- 20 m
Hypotenuse of right triangle is 6m more than twice shortest side. Third side is 2m less than hypotenuse. Sides are
- 10, 24 and 26
- 20, 24 and 26
- 20, 25 and 26
- 10, 25 and 26
In figure showing a right triangle with altitude to hypotenuse, the ratio involving the segments is
- AB²/AD
- AD²/AB
- AB/AD²
- AD/AB²
In triangle DEF, DE=6√3 cm, DF=12 cm, EF=6 cm. Angle E is
- 120°
- 90°
- 60°
- 45°
In figure, ST||QR. If PS=?, given values from figure
- 2 cm
- 4.5 cm
- 4 cm
- 3 cm
Trapezium ABCD, AB||DC, diagonals meet at O, AB=2CD. Ratio of areas of triangles AOB and COD is
- 1:2
- 2:1
- 1:4
- 4:1
If one pair of corresponding sides are proportional and included angles equal, then the criterion is called
- AAA similarity
- SSS similarity
- SAS similarity
- ASA similarity
P, Q on AB, AC of triangle ABC; PQ||BC divides ABC into 2 equal areas. AP:AB =
- 1:1
- 1:√2
- 1:√2
- (√2–1):1
Using similar triangles in a figure, determine x
- (5√11)/4
- (3√11)/4
- (5√11)/3
- (3√11)/2
Equilateral triangle ABC each side 4. AD⊥AB, AD=1. Area of BCED is
- 8
- 4
- 4.5
- 3.5
Area of right triangle is 40 cm², perimeter 40 cm. Length of hypotenuse is
- 16 cm
- 18 cm
- 17 cm
- 19 cm
x is length of median of equilateral triangle. Area is
- x²
- (x²/3)√3
- (x²√3)/3
- 2x²/3
Two sides of triangle produced beyond base, exterior angles bisected; bisectors include angle equal to
- half sum of base angles
- sum of base angles
- half difference of base angles
- difference of base angles
PSR right-angled at S, D mid-point of SR. Bisector of angle PSR and perp bisector of SR meet at O. Triangle OSD is
- scalene
- equilateral
- isosceles right angled
- acute-angled
Equilateral triangle area X and square area Y have same perimeter. Then
- X>Y
- X<Y
- X=Y
- X≥Y
In an isosceles triangle 'a' is the base and 'b' is the equal side. Its area is
- (a/4)√(4b²–a²)
- (b/4)√(4b²–a²)
- (a/4)√(4a²–b²)
- (b/4)√(4a²–b²)