CBSE Class 10 Mathematics Triangles Worksheet Set B

Read and download free pdf of CBSE Class 10 Mathematics Triangles Worksheet Set B. Download printable Mathematics Class 10 Worksheets in pdf format, CBSE Class 10 Mathematics Chapter 6 Triangles Worksheet has been prepared as per the latest syllabus and exam pattern issued by CBSE, NCERT and KVS. Also download free pdf Mathematics Class 10 Assignments and practice them daily to get better marks in tests and exams for Class 10. Free chapter wise worksheets with answers have been designed by Class 10 teachers as per latest examination pattern

Chapter 6 Triangles Mathematics Worksheet for Class 10

Class 10 Mathematics students should refer to the following printable worksheet in Pdf in Class 10. This test paper with questions and solutions for Class 10 Mathematics will be very useful for tests and exams and help you to score better marks

Class 10 Mathematics Chapter 6 Triangles Worksheet Pdf

 

Question. All equilateral triangles are __________ .
Ans. Similar

Question. In __________ triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Ans. Right

Question. Let ΔABC ~ ΔDEF and their areas be respectively 81 cm2 and 144 cm2. If EF = 24 cm, then length of side BC is _________ cm.
Ans. 18 cm

Question. Pythagoras Theorem is valid for right angled triangle.
Ans. True

Question. Match the following :

CBSE Class 10 Mathematics Triangles_17

Ans. (a) (iii) AAA similarity criterion.
(b) (iv) SSS similarity criterion.
(c) (i) SAS similarity criterion.

Question. If ΔABC ~ ΔDEF, ar(ΔDEF) = 100 cm2 and AB/DE = 1/2, then ar(ΔABC) is
(a) 50 cm2
(b) 25 cm2
(c) 4 cm2
(d) 200 cm2
Ans. (b) 25 cm2

Question. If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF, then which of the following is not true?"
(a) BC.EF = AC.FD
(b) AB.EF = AC.DE
(c) BC.DE = AB.EF
(d) BC.DE = AB.FD
Ans. (c) BC.DE = AB.EF

Question. A vertical pole of length 3 m casts a shadow of 7 m and a tower casts a shadow of 28 m at a time. The height of tower is
(a) 10 m
(b) 12 m
(c) 14 m
(d) 16 m
Ans. (b) 12 m

Question. In the given Fig. ΔAHK ~ ΔABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC.

CBSE Class 10 Mathematics Triangles_18

Ans. AK/AC = HK/BC ⇒ 10/AC = 7/3.5 ⇒ AC = 5 cm

Question. It is given that ΔDEF ~ ΔRPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P?
Ans. ∠D = ∠R (True)
∠F = ∠P (False)

Question. Write the statement of Basic Proportionality Theorem.
Ans. If a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio.

Question. If the corresponding Medians of two similar triangles are in the ratio 5 : 7. Then find the ratio of their sides.
Ans. 5 : 7

Question. If ΔABC ~ ΔQRP, [Area(ΔABC)] / [Area(ΔPQR)] = 9/4 , AB = 18 cm, BC = 15 cm, then find the length of PR.
Ans. 10 cm

Question. The areas of two similar triangles ΔABC and ΔDEF are 225 cm2 and 81 cm2 respectively. If the longest side of the larger triangle ΔABC be 30 cm, find the longest side of the smaller triangle DEF.
Ans. Let longest side of the ΔDEF be x cm.
225/81 = (30/x)2
x = 18 cm

Question. In the given Fig., DE || AC and DC || AP Prove that BE/BC = EC/CP

CBSE Class 10 Mathematics Triangles_19

Ans. DE || AC, AD/DB = EC/BE     ...(1) [∵BPT]
DC || AP, AD/DB = CP/BC             ...(2) [∵ BPT]
From (1) and (2), we get
BE/EC = BC/CP

Question. In the given Fig. PQR is a triangle, right angled at Q. If XY || QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1 : 2. Calculate the lengths of PR and QR.

CBSE Class 10 Mathematics Triangles_20

Ans. PX/XQ = PY/YR ⇒ 1/2 = 4/YR ⇒ YR = 8cm
∴ PR = 8 + 4 = 12cm
QR = √((12)- (6)2) = 6√3 cm

Question. In the given figure, ΔODC ~ ΔOBA, ∠BOC = 115° and ∠CDO = 70°. Find,
(i) ∠DOC,
(ii) ∠DCO,
(iii) ∠OAB,
(iv) ∠OBA.

CBSE Class 10 Mathematics Triangles_21

Ans. (i) 65°
(ii) 45°
(iii) 45°
(iv) 70°

Question. In the given figure, QR/QS = QT/PR and ∠1 = ∠2 then prove that ΔPQS ~ ΔTQR.

CBSE Class 10 Mathematics Triangles_22

Ans. In ΔPQR, ∠1 = ∠2
PR = PQ                                             [Opposite sides of equal angles]
∴ QR/QS = QT/PQ and ∠1 = ∠1            (Common)
∴ ΔPQS ~ ΔTQR                                 (SAS Similarity criterion)

Question. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Ans. 

CBSE Class 10 Mathematics Triangles_23

CBSE Class 10 Mathematics Triangles_24

Question. In the given figure, DE || BC, DE = 3 cm, BC = 9 cm and ar (ΔADE) = 30 cm2. Find ar (BCED).

CBSE Class 10 Mathematics Triangles_25

Ans. 

CBSE Class 10 Mathematics Triangles_26

Question. Two poles of height a metrs and b metres are p metres apart. Prove that the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is given by ab/(a+b) metres.
Ans. 

CBSE Class 10 Mathematics Triangles_27

CBSE Class 10 Mathematics Triangles_28

Question. In the given figure ∠D = ∠E and AD/DB = AE/EC. Prove that ΔBAC is an isoscles triangle.

CBSE Class 10 Mathematics Triangles_29

Ans. AD/DB = AE/EC
By converse of BPT, DE || BC
∴ ∠D = ∠B and ∠E = ∠C                (Corresponding Angles)
But ∠D = ∠E
So, ∠B = ∠C
∴ AB = AC
So, ΔABC is an isosceles triangle.

Question. Two triangles ΔBAC and ΔBDC, right angled at A and D respectively are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P. PRove that AP × PC = DP × PB.

CBSE Class 10 Mathematics Triangles_30

Ans. ΔAPB ~ ΔDPC             (AA Similarity criterion)
AP/DP = PB/PC                  (∵ C.P.S.T.)
AP.PC = DP.PB

Question. 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.
Ans. 

CBSE Class 10 Mathematics Triangles_31

ΔABE ~ ΔCDE
AB/CD = BE/DE
6/1.5 = (3+BD) / 3
BD = 9m

Question. In a quadrilateral ABCD, ∠B = 90°, AD2 = AB2 + BC2 + CD2. Prove that ∠ACD = 90°.

CBSE Class 10 Mathematics Triangles_32

Ans. In right angled ΔABC, AC2 = AB2 + BC               ...(1)
Given, AD2 = (AB2 + BC2) + CD2
⇒ AD2 = AC2 + CD2                               [From (1)]
By converse of Pythagoras theorem, ∠ACD = 90°.

Question. In ΔPQR, PD ⊥ QR such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d and a, b, c, d are positive units. Prove that (a + b) (a – b) = (c + d) (c – d).
Ans. In right angled ΔPDQ,
PD2 = a2 – c2                                ...(1)
In right angled ΔPDR
PD2 = b2 – d2                                ...(2)
From (1) and (2), we have
a2 – c2 = b2 – d2
a2 – b2 = c2 – d2
(a – b) (a + b) = (c + d) (c – d)

 

 

CBSE Class 10 Mathematics Triangles Worksheet Set B 1

CBSE Class 10 Mathematics Triangles Worksheet Set B 2

 

Please click on below link to download CBSE Class 10 Mathematics Triangles Worksheet Set B

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