Online Mock Tests
Chapter 2: Polynomials
Chapter 3: Pair of Liner Equation in Two Variable
Chapter 4: Quadatric Euation
Chapter 5: Arithematic Progressions
Chapter 6: Triangles
Chapter 7: Coordinate Geometry
Chapter 8: Introduction to Trignometry & its Equation
Chapter 9: Circles
Chapter 10: Construction
Chapter 11: Area Related To Circles
Chapter 12: Surface Areas and Volumes
Chapter 13: Statistics and Probability
Chapter 6: Triangles
NCERT solutions for Mathematics Exemplar Class 10 Chapter 6 Triangles Exercise 6.1 [Pages 60 - 62]
Choose the correct alternative:
In Figure, ∠BAC = 90° and AD ⊥ BC. Then, ______.
BD . CD = BC2
AB . AC = BC2
BD . CD = AD2
AB . AC = AD2
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is ______.
If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF, then which of the following is not true?
BC . EF = A C. FD
AB . EF = AC . DE
BC . DE = AB . EF
BC . DE = AB . FD
If in two triangles ABC and PQR, `(AB)/(QR) = (BC)/(PR) = (CA)/(PQ)`, then ______.
ΔPQR ~ ΔCAB
ΔPQR ~ ΔABC
ΔCBA ~ ΔPQR
ΔBCA ~ ΔPQR
In figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to ______.
If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?
`(EF)/(PR) = (DF)/(PQ)`
`(DE)/(PQ) = (EF)/(RP)`
`(DE)/(QR) = (DF)/(PQ)`
`(EF)/(RP) = (DE)/(QR)`
In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles are ______.
congruent but not similar
similar but not congruent
neither congruent nor similar
congruent as well as similar
It is given that ΔABC ~ ΔPQR, with `(BC)/(QR) = 1/3`. Then, `(ar(PRQ))/(ar(BCA))` is equal to ______.
It is given that ΔABC ~ ΔDFE, ∠A =30°, ∠C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, the following is true ______.
DE = 12 cm, ∠F = 50°
DE = 12 cm, ∠F = 100°
EF = 12 cm, ∠D = 100°
EF = 12 cm, ∠D = 30°
If in triangles ABC and DEF, `(AB)/(DE) = (BC)/(FD)`, then they will be similar, when ______.
∠B = ∠E
∠A = ∠D
∠B = ∠D
∠A = ∠F
If ∆ABC ~ ∆QRP, `(ar(ABC))/(ar(PQR)) = 9/4`, AB = 18 cm and BC = 15 cm, then PR is equal to ______.
If S is a point on side PQ of a ΔQR such that PS = QS = RS, then ______.
PR . QR = RS2
QS2 + RS2 = QR2
PR2 + QR2 = PQ2
PS2 + RS2 = PR2
NCERT solutions for Mathematics Exemplar Class 10 Chapter 6 Triangles Exercise 6.2 [Pages 63 - 65]
Is the triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Give reasons for your answer.
It is given that ΔDEF ~ ΔRPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P? Why?
A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm. Is AB || QR? Give reasons for your answer.
In figure, BD and CE intersect each other at the point P. Is ΔPBC ~ ΔPDE? Why?
In ∆PQR and ∆MST, ∠P = 55°, ∠Q = 25°, ∠M = 100° and ∠S = 25°. Is ∆QPR ~ ∆TSM? Why?
Is the following statement true? Why? “Two quadrilaterals are similar, if their corresponding angles are equal”.
Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?
Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
The ratio of the corresponding altitudes of two similar triangles is `3/5`. Is it correct to say that ratio of their areas is `6/5`? Why?
D is a point on side QR of ΔPQR such that PD ⊥ QR. Will it be correct to say that ΔPQD ~ ΔRPD? Why?
In figure, if ∠D = ∠C, then it is true that ΔADE ~ ΔACB? Why?
Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.
NCERT solutions for Mathematics Exemplar Class 10 Chapter 6 Triangles Exercise 6.3 [Pages 66 - 69]
In a ∆PQR, PR2 – PQ2 = QR2 and M is a point on side PR such that QM ⊥ PR. Prove that QM2 = PM × MR.
Find the value of x for which DE || AB in figure.
In figure, if ∠1 =∠2 and ΔNSQ ≅ ΔMTR, then prove that ΔPTS ~ ΔPRQ.
In the figure `(QR)/(QS) = (QT)/(PR)` and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.
In figure, if AB || DC and AC, PQ intersect each other at the point O. Prove that OA . CQ = OC . AP
Find the altitude of an equilateral triangle of side 8 cm.
If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of ∆ABC.
In figure, if DE || BC, find the ratio of ar(ΔADE) and ar (DECB).
ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.
Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm2, find the area of the larger triangle.
In a triangle ∆PQR, N is a point on PR such that QN ⊥ PR. If PN. NR = QN2, prove that ∠PQR = 90°.
Areas of two similar triangles are 36 cm2 and 100 cm2. If the length of a side of the larger triangle is 20 cm, find the length of the corresponding side of the smaller triangle.
In the figure, if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, find BD.
A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.
Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.
NCERT solutions for Mathematics Exemplar Class 10 Chapter 6 Triangles Exercise 6.4 [Pages 73 - 76]
In figure, if ∠A = ∠C, AB = 6 cm, BP = 15 cm, AP = 12 cm and CP = 4 cm, then find the lengths of PD and CD.
It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.
Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.
In figure, if PQRS is a parallelogram and AB || PS, then prove that OC || SR.
A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
For going to a city B from city A, there is a route via city C such that AC ⊥ CB, AC = 2x km and CB = 2(x + 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.
A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.
Tick the correct answer and justify:
A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3 m, find how far she is away from the base of the pole.
In figure, ABC is a triangle right angled at B and BD ⊥ AC. If AD = 4 cm, and CD = 5 cm, find BD and AB.
In figure, PQR is a right triangle right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.
In ∆PQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b)(a – b) = (c + d)(c – d).
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2
In figure, l || m and line segments AB, CD and EF are concurrent at point P. Prove that `(AE)/(BF) = (AC)/(BD) = (CE)/(FD)`.
In figure, PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.
In figure, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE. Prove that `(BD)/(CD) = (BF)/(CE)`.
Prove that the area of the semicircle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.
Prove that the area of the equilateral triangle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.
Chapter 6: Triangles
NCERT solutions for Mathematics Exemplar Class 10 chapter 6 - Triangles
NCERT solutions for Mathematics Exemplar Class 10 chapter 6 (Triangles) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics Exemplar Class 10 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Mathematics Exemplar Class 10 chapter 6 Triangles are Similar Figures, Similarity of Triangles, Basic Proportionality Theorem (Thales Theorem), Criteria for Similarity of Triangles, Areas of Similar Triangles, Application of Pythagoras Theorem in Acute Angle and Obtuse Angle, Triangles Examples and Solutions, Angle Bisector, Ratio of Sides of Triangle, Right-angled Triangles and Pythagoras Property, Similarity of Triangles, Similarity of Triangles.
Using NCERT Class 10 solutions Triangles exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10 prefer NCERT Textbook Solutions to score more in exam.
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