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A triangle can be referred to as a structure with three sides and three vertices which are straight in nature. The sum of the three of a triangle always sums upto 180o and the sum of the exterior angles adds upto 360o. The triangles chapter is a very important lesson for the class 10 students of CBSE. It has a high weightage in the board exams. It is a part of unit 4 Geometry and carries a total of 15 periods. The whole unit carries a total of 15 marks.
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Keyterms: Triangle, exterior angles, Geometry, Acute angled triangle, Right angled triangle, Obtuse angled triangle
Types of Triangle
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There are different types of triangles:
Based on the angle
- Acute angled triangle
An acute-angled triangle is a type of triangle in which the sum of all the angles is 180 degrees and each angle of the triangle is acute (each angle is less than 90 degrees).
- Right-angled Triangle
A right-angled triangle is a type of triangle in which the sum of all the angles is 180 degrees and the angle of the triangle is 90 degrees.
- Obtuse Angled Triangle
An obtuse-angled triangle is a type of triangle in which the sum of all the angles is 180 degrees and one of the angles is obtuse( more than 90 degrees).
Based on the length of the side of a Triangle
- Scalene Triangle
A scalene triangle is a type of triangle in which each side of the triangle has a different measure i.e.; no two sides of the triangle have the same length.
- Isosceles Triangle
An isosceles triangle is a type of triangle in which the length of any two sides of the triangle are equivalent (i.e. same).
- Symmetrical Triangle
A symmetrical Triangle is a type of triangle in which each side of the triangle is equivalent to the other i.e.; all the sides of the triangle have the same length.
Important Similarity Concepts
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- Similar figures
Two figures are said to be similar if they have identical properties or shapes but not exact shapes or properties.
- Congruent triangle
Two or more triangles are said to be congruent if their corresponding side or angle is the same. Congruent triangles have the same shape and lengths.
All the triangles that are congruent are necessarily similar but not all similar triangles are necessarily congruent.
There are different types of triangle congruence criteria to determine if the triangles are congruent or similar or not-
- SSS (side-side-side) similarity criterion
Two triangles are said to be sss congruent if three sides of one triangle are congruent to the three sides of the other given triangle.
SSS (side-side-side) similarity criterion
Here, as AB≅DE, BC≅EF, AC≅DF.
Therefore ΔABC≅ΔDEF by sss congruence criterion.
- Side-Angle-Side (SAS)
two triangles are said to be SAS congruent if any two sides of the first triangle and two sides of the second triangle along with one of the angles of each of the triangles are congruent.
Side-Angle-Side (SAS)
Here, as AB≅DE, AC≅DF, and ∠A≅∠D.
Therefore, ΔABC≅ΔDEF by SAS congruence criteria.
- Angle-Side-Angle (ASA)
Two triangles are to be ASA congruent if any two angles of the first triangle and any side are congruent to the two angles and a side of the second triangle.
Angle-Side-Angle (ASA)
Here, as, ∠A≅∠D, ∠B≅∠E, and AB≅DE.
Therefore,ΔABC≅ΔDEF by ASA congruence.
- Angle-Angle-Side (AAS)
Two triangles are said to be AAS congruent if any two angles and a single non included side of one triangle is congruent to the two angles and non included side of the other triangle
Angle-Angle-Side (AAS)
Here, as ∠D≅∠A, ∠E≅∠B, and BC≅EF. Therefore, ΔDEF≅ΔABC congruent by AAS congruence.
- Hypotenuse-Leg congruence
It is said to be Hypotenuse Leg congruence if the hypotenuse side and the leg of one right triangle are congruent to the same of the other given triangle.
Hypotenuse-Leg congruence
Here, as AC≅DF, AB≅DE, ∠B, and ∠E are right angles. Therefore, ΔABC≅ΔDEF is congruent by hypotenuse leg congruence.
- Angle-Side-Side (ASS)
Two triangles are an aid to be ASS congruent If any two sides and the non-included angle of one triangle are congruent to the other triangle's two sides and the non-included angle. The two triangles may not always be congruent.
Angle-Side-Side (ASS)
Here, though AC≅DF, BC≅EF, ∠A≅∠D, but ΔABC is not congruent to ΔDEF.
- Angle-Angle-Angle (AAA)
It is said to be AAA congruence if the three angles of one triangle are congruent to three angles of the other triangle, the two triangles may not always be congruent
Angle-Angle-Angle (AAA)
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Important Questions
Question: If the area of two triangles that are similar is equal, demonstrate that they are congruent. (4 marks)
Answer: Given - ΔABC and ΔPQR, 2 similar triangles with equal area.
To prove - ΔABC ≅ ΔPQR.
Proof:
ΔABC ~ ΔPQR (Given)
∴ Area of (ΔABC)/Area of (ΔPQR) = BC2/QR2 = AB2/PQ2 = AC2/PR2
(from Similarity rule)
⇒ BC2/QR2 = AB2/PQ2 = AC2/PR2 = 1 [Since, ar(ΔABC) = ar(ΔPQR)]
⇒ BC2/QR2 = 1
⇒ AB2/PQ2 = 1
⇒ AC2/PR2 = 1
Therefore, BC = QR;
AB = PQ; and AC = PR.
Therefore, ΔABC ≅ ΔPQR by SSS criterion of congruence.
Question: State and prove Thales’s theorem. (5 marks)
Answer: Statement: A line which is drawn parallel to any side of a triangle will intercross the other two sides of the triangle at different points. The other two remaining sides are then said to be divided in an equivalent ratio.
Given - In a ΔABC , DE is parallel to BC.
To prove - ADDB = AEEC.
Construction -
In the triangle, draw a perpendicular EM to AD AND DN to AE. Also, connect point B to E and point C to point D.
Proof:
IN ΔADE and ΔBDE,
ar(ΔADE)ar(ΔBDE) = (½ X AD X EM).(½ X DB X EM)= ADDB
[As, area of a triangle = ½ X Base X Altitude)
IN ΔADE and ΔCDE
ar (ΔADE)ar(ΔCDE) = (½ x AE X DN) x(½ X EC X DN) = AEEC
Since, it is given that DE || BC
ar(ΔBDE) = ar(ΔCDE).
Therefore, ADDB = AEEC
Question: State Pythagoras theorem. (3 marks)
Answer. The Pythagorean Theorem shows that for a right triangle, the sum of the areas of the squares formed by the legs of the triangle is equal to the area of the square formed by the triangle's hypotenuse.
a2 + b2 = c2
where a and b are the sides of the right triangle and c is the hypotenuse of the triangle.
Therefore,
The length of the hypotenuse,
\(c = \sqrt{a^2 + b^2}\)The lengths of side \(a = \sqrt{c^2 - b^2}\) and \(b = \sqrt{c^2 - a^2}\)
Question: For a right triangle, hypotenuse c = 10 and side a = 6. Find the length of side b. (2 marks)
Answer. Given- Hypotenuse, c = 10, a = 6 is
From above we can find the other by using the formula
\(b = \sqrt{c^2 - a^2}\) and substituting values of c and a. \(b = \sqrt{c^2 - a^2}\)= 100 – 36
= 64
= 8
Question: Prove that the ratios of the areas of any two similar triangles is equivalent to the square of the ratio of their similar sides. (5 marks)
Answer.
Given- ΔABC and ΔDEF are similar
To prove- ar(ΔABC)ar(ΔDEF) = AB2DE2 = BC2EF2 = AC2DF2.
Construction
Draw a line AM perpendicular to BC and DN perpendicular to EF.
According to the stated theorem,
area of ΔABC. area of ΔPQR = (ABPQ)2 =(BCQR)2 = (CARP)2
As Area of triangle = ½ × Base × Height
Draw perpendicular to vertex A and P To find the area of ΔABC and ΔPQR.
Now, the area of ΔABC = ½ × BC × AD
area of ΔPQR = ½ × QR × PE
The ratio of areas of the triangles are:
area of ΔABC/area of ΔPQR = ½ × BC × AD / ½ × QR × PE
⇒ area of ΔABC/area of ΔPQR = BC × AD / QR × PE ……………. (1)
Now in ΔABD and ΔPQE, we get
∠ABC = ∠PQR (Since ΔABC ~ ΔPQR)
∠ADB = ∠PEQ ( both angles are 90°)
From AA criterion of similarity ΔADB ~ ΔPEQ
⇒ AD/PE = AB/PQ …………….(2)
Since it is known that ΔABC~ ΔPQR,
AB/PQ = BC/QR = AC/PR …………….(3)
Substituting this value in equation (1), we would get
area of ΔABC/area of ΔPQR = AB/PQ × AD/PE
Using equation (2), we would get
area of ΔABC/area of ΔPQR = AB/PQ × AB/PQ
⇒area of ΔABC/area of ΔPQR =(ABPQ)2
From the above equation (3),
area of ΔABC/area of ΔPQR = (ABPQ)2 =(BCQR)2 = (CARP)2
Hence proved.
Question: The Two diagonals of a rhombus are 32 cm and 24 cm. Calculate the height of the rhombus or the altitude’s length. (2013) (4 marks)
Answer: Diagonals of a rhombus bisect (cut each other at the middle)
Therefore , AC ⊥ BD,
OA = OC = AC/2 ⇒24/2 = 12 cm
OB = OD = BD/2 ⇒32/2 = 16 cm
In right ΔBOC (by pythagoras theorem);
(BC) = \(\sqrt{(OC)^2 + (OB)^2}\)
= \(\sqrt{(12)^2 + (16)^2}\)
= \(\sqrt{144 + 256}\)
∴ Side of rhombus, BC = \(\sqrt{400}\) = 20 cm
ar. (Δ ABC) = ar. (Δ ABC)
(Taking BC as base) = (Taking BC as base)
\(\frac{1}{2}\) x BC x AM = \(\frac{1}{2}\) x AC x OB
…..[ \(\because\) area of Δ = \(\frac{1}{2}\) x base x corr. altitude
20 x AM = 24 x 16
∴ AM = \(\frac{24 \times 16}{20} = \frac{96}{5} = 19.2 cm\)
Therefore, the altitude or height of the rhombus is 19.2 cm in length.
Question: Let ΔABC ~ ΔPQR, if the perimeter of ΔABC is 32 cm, the perimeter of ΔPQR is 48 cm, and the length of PR is 6 cm, calculate the length of AC from the data given. (2012) (2 marks)
Answer:
We know that ΔABC ~ ΔPQR - [Given]
Now, by similarities of the triangles we get,
\(\frac{perimeter of \Delta ABC}{perimeter of \Delta PQR}\) = \(\frac{AC}{PR}\)
⇒ \(\frac{32}{48} = \frac{AC}{6}\)
⇒ AC = 4 cm
Question: Let ΔABC ~ ΔDEF. Now, if the length of AB is 4 cm, BC is 3.5 cm, CA is 2.5 cm and DF is 7.5 cm, calculate the perimeter of the ΔDEF with the given data. (2012, 2017) (3 marks)
Ans: ΔABC – ΔDEF …[Given]
Now by the congruency of the triangles we get :
Perimeter of ΔABC / Perimeter of ΔDEF = AC / DF
AB + BC + CA / Perimeter of ΔDEF = AC / DF
Therefore, the perimeter of the ΔDEF is given by
10/x = 2.5/7.5
x=30.
The perimeter of ΔDEF is 30 cm.
Question: In the given ΔABC, DE is parallel to BC, then calculate the value of x. (2015)
Answer:
In the given ΔABC we know that , DE || BC …[Given ]
\(\frac{AD}{BD} = \frac{AE}{EC}\) – (By Thale’s Theorem)
\(\frac{X}{(X-1)} = \frac{(X+3)}{(X+5)}\)
x(x+5) = (x+3)(x+1)
x2 + 5x = x2 + 3x + x + 3
x2 + 5x - x2 - 3x - x = 3
By solving the equation we get x = 3 cm.
Question: In the below-given figure, the angle A is 90°, AD ⊥ BC. If the length of BD is 2 cm and CD-length is 8 cm, find the length of AD in the figure. (2012, 2017) (3 marks)
Answer: We know that,
ΔADB ~ ΔCDA …[If a perpendicular is drawn from the vertex of the right angle of a right Δ to the hypotenuse then the length on both sides of the ⊥ are similar to each other and D]
∴ BD/AD=AD/CD …[as Sides are proportional]
(AD)2 = BD/DC
(AD)2 = 2 x 8 = 16cm
⇒ AD = 4 cm
Therefore, the length of AD is 4cm.
Question: A ladder of 6.5m in length is placed against a wall in a way that its foot is 2.5m away from the foot of the wall. Calculate the height of the wall in a case where the top of the ladder meets the top of the wall at a point. (2015) (3 marks)
Answer: Let us suppose AC is the length of the ladder and AB is the length of the wall. A is the point of contact of both wall and ladder.
We know that AC= 6.5M = 13/2m
BC = 2.5M = 5/2m
In the right ΔABC,
AB2+BC2=AC2 ( By using Pythagoras theorem)
AB2+(5/2 )2= (13/2)2
AB2= 144 / 4 m
AB2 = 36 m
AB = 6m
Therefore, the height of the required wall is 6 m tall.
Question: As shown in the below-given figure, QA ⊥ AB and PB ⊥ AB. If the length of AO is 20 cm, the length of BO is 12 cm, and the length of PB is 18 cm, find the length of AQ. (4 marks)
Answer: In the given figure of ΔOAQ and ΔOBP,
∠OAQ = ∠OBP … [angle of Each is 90°]
∠AOQ = ∠BOP … [as they are vertically opposite angles]
Therefore , ΔOAQ ~ ΔOBP [BY AA]
AO/BO = AQ/PB [As the sides are proportional ]
20/12 = AQ/18
AQ = 30 cm.
Therefore, the length of side AQ is 30 cm.
Question: In the below given figure, AD⊥BC and BD is 3 times CD. Prove that 2(AC)2 = 2(AB)2 + (BC)2 (2012) (3 marks)
Answer: In the given figure;
BC = BD + DC = BD + 3BD = 4BD [from given ]
∴ BC4 = BD.
In right. ?ADB, AD2 = AB2 – BD2 ….(ii) [Pythagoras theorem]
In right. ?ADC, AD2 = AC2 – CD2 …(iii) [Pythagoras Theorem]
From (ii) and (iii), we get
(AC)2 – (CD)2 = (AB)2 –( BD)2
(AC)2 = (AB)2 – (BD)2 + (CD)2
= (AB)2 - (CD3)2 + (CD)2
= (AB)2 - (8/9) (CD)2
= (AB)2 - (8/9) (3BD)2
= (AC)2 = (AB)2 +
⇒ 2AC2 = 2AB2 + BC2
Hence proved
Question: In ΔABC, AP ⊥ BC at P and it is given that (AC)2 = (BC)2 – (AB)2, then prove that (PA)2 = PB × CP. (2015) (4 marks)
Answer: AC2 = BC2 – AB2 …[Given]
AC2 + AB2 = BC2
∴ ∠BAC = 90° … [By the converse of the Pythagoras’ theorem ]
also ΔAPB ~ ΔCPA
[If a perpendicular is drawn from the vertex of the right angle of a triangle to the hypotenuse then the lengths on both sides of the perpendicular are similar to each other and the whole triangle]
∴ AP/CP = PB/PA [In similar Δ's, corresponding sides are similar]
∴ PA2 = PB.CP
Hence, the required equation is proved.
Question: Let ABCD be a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = + BD2+ AC2 (3 marks)
Answer. Given: In rhombus ABCD, AC and BD diagonals intersect each other at a point O.
To prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Proof:
We know that ,
AC ⊥ BD [since rhombus’ diagonals bisect each other at 90 degrees]
∴ OA = OC
also , OB = OD
In right. ΔAOB,
AB2 = OA2 + OB2 [by using Pythagoras’ theorem]
AB2 = (AC/2)2+(BD/2)2
4AB2 = AC2 + BD2
AB2 + AB2 + AB2 + AB2= BD2 + AC2
∴ AB2 + BC2 + CD2 + DA2 =BD2+AC2
Hence proved.
Question: In a given rectangle ABCD, let E be the middle point of AD. If length AD is 40 m and AB is 48 m, then find the length of EB. (2014) (4 marks)
Answer. We know that E is the midpoint of AD [Given]
AE = 40/2m = 20 m
∠A = 90° [as Angle of a rectangle is 90]
In right. ΔBAE,
(EB)2 = (AB)2 + (AE)2 [by Pythagoras’ theorem]
(EB)2 = (48)2 + (20)2
(EB)2 = 2304 + 400 = 2704
⇒ EB = √2704 m
⇒ EB= 52 m
Therefore, the required length of side EB is 52m.
Question: State the converse of Pythagoras theorem and also prove the converse of Pythagoras theorem. Using that, solve the following:
In ΔABC, if the length of AB is √63 cm, BC is 6 cm and AC is 12 cm, find the ∠B. (2015) (5 marks)
Answer: CONVERSE OF PYTHAGORAS THEOREM
Statement: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the opposite angle to the first side is a right angle(90).
To prove: ∠ABC = 90°
Construct: Draw a right angle ΔDEF in which DE = BC and EF = AB.
Proof: In right. ΔABC,
(AB)2 + (BC)2 = (AC)2 (i) [Pythagoras theorem]
In right. ΔDEF;
(DE)2 + (EF)2 = (DF)2 [Pythagoras theorem]
(BC)2 + (AB)2 = (DF)2 (ii)…[since DE = BC; EF = AB]
From (i) and (ii),
AC2 = DF2 = AC = DF
we get, DE = BC …[from the construction]
EF = AB …[from construction]
DF = AC … [Proved]
∴ ΔDEF = ΔABC … (by SSS congruence)
∴ ∠DEF = ∠ABC.
∴ ∠DEF = 90°
Also , ∠ABC = 90°
Given: In the right. ?ABC,
(AB)2 + (BC)2 = (AC)2
(AB)2 + (BC)2 = (√63)2 + (6)2
= 108 + 36
= 144
= (12)2
we know that , AB2 + BC2 = AC2
∴ ∠B = 90° [from above theorem]
Hence proved.
Question: In a given ΔABC, the perpendicular from A which is on the side BC of a ΔABC intersects BC at D such that DB = 3 CD. show that 2 AB2 = 2 AC2 + BC2 (2013, 2017) (5 marks)
Solution: In right. ΔADB,
(AD)2 = (AB)2 – (BD)2 …(i) [by Pythagoras theorem]
In right. ΔADC,
(AD)2 = (AC)2 – (DC)2 (ii) [by Pythagoras theorem]
From above (i) and (ii), we will get
(AB)2 – (BD)2 = (AC)2 – (DC)2
(AB)2 = (AC)2 + (BD)2 – (DC)2
⇒ BC = BD + DC
= 3CD + CD = 4 CD …[∴ BD = 3CD (from Given)]
⇒ (BC)2 = 16 (CD)2 …(iv) [Squaring on both sides ] ⇒(AB)2 = (AC)2 + (BD)2 – (DC)2
= (AC)2 + 9(DC)2 – (DC)2 [∴ BD = 3 CD ⇒ (BD)2 = 9 (CD)2 (from given)]
= (AC)2 + 8 (DC)2
= (AC)2 + 16(DC)2
= (AC)2 + (BC)2
∴ 2(AB)2 = 2(AC)2 + (BC)2
Hence Proved
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