CBSE Class 10 Mathematics Triangles Worksheet Set 02

Key Notes

Triangles

• Congruence of Triangles
• Criteria for Congruence of Triangles
• Some Properties of a Triangle
• Inequalities in a Triangle

 

• Triangle- A closed figure formed by three intersecting lines is called a triangle. A triangle has three sides, three angles and three vertices.
• Congruent figures- Congruent means equal in all respects or figures whose shapes and sizes are both the same for example, two circles of the same radii are congruent. Also two squares of the same sides are congruent.
• Congruent Triangles- two triangles are congruent if and only if one of them can be made to superpose on the other, so as to cover it exactly.
• If two triangles \( ABC \) and \( PQR \) are congruent under the correspondence \( A \leftrightarrow P, B \leftrightarrow Q \) and \( C \leftrightarrow R \) then symbolically, it is expressed as \( \Delta ABC \cong \Delta PQR \).
• In congruent triangles corresponding parts are equal and we write 'CPCT' for corresponding parts of congruent triangles.
• SAS congruency rule - Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle. For example: \( \Delta ABC \) and \( \Delta PQR \) as satisfy SAS congruent criterion.
• ASA Congruence Rule- Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. For examples \( \Delta ABC \) and \( \Delta DEF \) shown below satisfy ASA congruence criterion.
• AAS Congruence Rule- Two triangle are congruent if any two pairs of angles and one pair of corresponding sides are equal for example \( \Delta ABC \) and \( \Delta DEF \) shown below satisfy AAS congruence criterion.
• AAS criterion for congruence of triangles is a particular case of ASA criterion.
• Isosceles Triangle- A triangle in which two sides are equal is called an isosceles triangle. For example: \( \Delta ABC \) shown below is an isosceles triangle with \( AB=AC \).
• Angle opposite to equal sides of a triangle are equal.
• Sides opposite to equal angles of a triangle are equal.
• Each angle of an equilateral triangle is \( 60^\circ \).
• SSS congruence Rule - If three sides of one triangle are equal to the three sides of another triangle then the two triangles are congruent for example \( \Delta ABC \) and \( \Delta DEF \) as satisfy SSS congruence criterion.
• RHS Congruence Rule- If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle then the two triangle are congruent. For example: \( \Delta ABC \) and \( \Delta PQR \) shown below satisfy RHS congruence criterion.
• RHS stands for right angle - Hypotenuse side.
• A point equidistant from two given points lies on the perpendicular bisector of the line segment joining the two points and its converse.
• A point equidistant from two intersecting lines lies on the bisectors of the angles formed by the two lines.
• In a triangle, angle opposite to the longer side is larger (greater).
• In a triangle, side opposite to the large (greater) angle is longer.
• Sum of any two sides of a triangle is greater than the third side.

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 :

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.

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

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.

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

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.

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. 

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

Ans. 

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. 

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

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.

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. 

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°.

Ans. In right angled ΔABC, AC² = AB² + BC²                ...(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)

 

 

 

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