Triangles Class 10 Extra Questions Maths Chapter 6 with Solutions

Triangles Class 10 Extra Questions Maths Chapter 6 with Answers

Extra Questions for Class 10 Maths Chapter 6 Triangles. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also download NCERT Solutions Class 10 Maths to help you to revise complete syllabus and score more marks in your examinations.

Triangles Class 10 Extra Questions Very Short Answer Type

Question 1.
A girl walks 500 m towards east and then 1200 m towards north. Find her distance from the starting point. (CBSE Delhi 2016)
Answer:
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Let A be the starting point. Then, we have
(AC)² = (AB)² + (BC)²
= (500 m)² + (1200 m)²
= 250000 m² + 1440000 m²
= 1690000 m² ⇒ √l690000 m²
∴ AC = 1300 m
Hence, the required distance = 1300 m

Question 2.
In the given figure, ∠CAB = 90° and AD ⊥ BC. If AC = 25 cm, AB = 1 m and BD = 96.08 cm, then find the value of AD.
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Answer:
In ∆BAC and ∆BDA, we have
∠BAC = ∠BDA = 90°
∠B = ∠B [Common]
⇒ ∆BAC ~ ∆BDA [By AA similarity axiom]
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Question 3.
Given ∆ABC ~ ∆PQR, if \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{1}{3}\), then find \(\frac{{ar} \Delta \mathrm{ABC}}{{ar} \Delta \mathrm{PQR}}\).
Answer:
Since areas of two similar triangles is proportional to squares of their corresponding sides
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Question 4.
In the given figure, if ∠PQR = ∠PRX, then find ar (∆PRX) : ar (∆PQR).
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Answer:
In ∆PRX and ∆PQR, we have
∠P = ∠P and ∠PQR = ∠PRX.
∴ ∆PRX ~ PQR (by AA similarity rule)
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Question 5.
∆ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ∆DEF ~ ∆ABC and FE = 4 cm, then find the perimeter of ADEF.
Answer:
∆ABC ~ ∆DEF
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⇒ DE = 6 cm, DF = 5 cm
∴ Perimeter of DEF = DE + EF + DF = 6 cm + 4 cm + 5 cm = 15 cm.

Question 6.
In the figure, ∠M = ∠N = 46°. Express x in terms of a, b and c where a, b, and c are lengths of LM, MN and NK respectively. (CBSE 2009)
Answer:
In ∆ KML and ∆ KNP
∠M = ∠N = 46° [Given]
∠MKL = ∠NKP [Common]
⇒ ∆ KML ~ ∆ KNP [AA similarity rule]
∴ Their corresponding sides are proportional.
⇒ \(\frac{\mathrm{LM}}{\mathrm{PN}}=\frac{\mathrm{MK}}{\mathrm{NK}}\) ⇒ \(\frac{a}{x}=\frac{b+c}{c}\)
⇒ x(b + c) = ac ⇒ x = \(\frac{a c}{b+c}\)

Question 7.
The areas of two similar triangles ABC and DEF are 100 cm² and 81 cm² respectively. If EF = 5 cm, then find BC.
Answer:
∵ ABC ~ ∆ DEF
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Question 8.
In given Fig., ST || RQ, PS = 3 cm and SR = 4 cm. Find the ratio of the area of ∆PST to the area of ∆PRQ. [CBSE Sample Paper, 2017]
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Answer:
PR = PS + SR = 3 + 4 = 7 cm
ST || RQ ⇒ ∆PST ~ ∆PRQ
⇒ \(\frac{{ar}(\Delta \mathrm{PST})}{{ar}(\triangle \mathrm{PRQ})}=\frac{\mathrm{PS}^2}{\mathrm{PR}^2}\)
= \(\frac{3^2}{7^2}=\frac{9}{49}\)\frac{3^2}{7^2}=\frac{9}{49}
Reqd. Ratio = 9 : 49.

Triangles Class 10 Extra Questions Short Answer Type-1

Question 1.
In the given figure, DEFG is a square and ∠BAC = 90°. Show that FG² = BG × FC.
Or
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. [CBSE SQP 2019 Standard]
Answer:
In ∆ADE and ∆GBD,
∠ADE = ∠GBD (Corresponding ∠s)
∠EAD = ∠DGB (Each 90° given)
⇒ ∆ADE ~ ∆GBD ……. (i) (AA similarity)
In ∆ADE and ∆FEC,
∠AED = ∠ECF (Corresponding ∠s)
∠DAE = ∠CFE (Each 90° given)
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∆ADE ~ ∆FEC ………. (ii) (AA similarity)
From equation (i) and (ii) ⇒ ∆GBD ~ ∆FEC
⇒ = \(\frac{G D}{F C}=\frac{G B}{F E}\)
⇒ GD × FE = GB × FC
⇒ FG × FG = GB × FC [∵ GD = FG = FE]
∴ FG² = BG × FC

Or

Given: ∆ABC with AB = BC = CA. AD ⊥ BC
To prove: 3AB² = 4 AD²
Proof: In right ∠d ∆ADB, by Pythagoras theorem
AB² = AD² + BD²
⇒ AB² = AD² + \(\left(\frac{B C}{2}\right)^2\)
⇒ AB² = AD² + \(\frac{\mathrm{BC}^2}{4}\)
(∵ BD = \(\frac{1}{2}\) BC altitude of equilateral A is also its medianj
⇒ 4AB² = 4AD² + AB² [∵ BC = AB]
⇒ 3AB² = 4AD² Q.E.D.

Question 2.
Find the length of an altitude of an equilateral triangle of side a.
Answer:
Let ∆ABC be equilateral.
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Draw AM ⊥ BC
Now, BM = MC = a/2
In ∆BMA
AB² = AM² + BM²
[By Pythagoras theorem]
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Question 3.
Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P.
Prove that AP × PC = BP × PD.
Answer:
Given: Two ∆s BAC and BDC are drawn on the same hypotenuse BC and on same side of BC. AC and BD intersect at P.
To prove: AP × PC = BP × PD
Proof: In ∆BPA and ∆CPD
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∠1 = ∠2 (Vertically opp.angles)
∠BAP = ∠CDP = 90°
∴ ∆BPA ~ ∆CPD
∴ \(\frac{\mathrm{AP}}{\mathrm{DP}}=\frac{\mathrm{BP}}{\mathrm{CP}}\)
⇒ AP × CP = BP × DP
⇒ AP × PC = BP × PD

Question 4.
In fig., ∠1 = ∠2 and ∠3 = ∠4. Show that PT. QR = PR. ST.
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Answer:
∠1 = ∠2 (Given)
⇒ ∠1 + ∠5 = ∠2 + ∠5
or ∠SPT = ∠QPR
and ∠3 = ∠4 (Given)
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Question 5.
In ∆ABC, D and E are the points on the sides AB and AC respectively such that DE || BC.
If AD = 6x – 7, DB = 4x – 3, AE = 3x – 3, and EC = 2x – 1 then find. the value of ‘x’.
Answer:
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Here AD = 6x – 7, DB = 4x – 3,
AE = 3x – 3, EC = 2x – 1
∵ DE || BC
By BPT \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
\(\frac{6 x-7}{4 x-3}=\frac{3 x-3}{2 x-1}\)
On solving we get x = 2.

Question 6.
In given fig. S and T trisect QR of right ∆PQR right angled at Q. Prove that 8PT² = 3PR² + 5PS²
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Answer:
In given fig.,
Let,
x = RT = TS = SQ, ∠Q = 90°
PR² = PQ² + QR² ……. (i)
PT² = PQ² + QT² ……. (ii)
PS² = PQ² + QS² ……. (iii)
Now, 8PT² – 3PR²
= 8 (PQ² + QT²) – 3(PQ² + QR²)
= 8(PQ² + 4x²) – 3(PQ² + 9x²)
= 5PQ² + 5x²
= 5(PQ² + x²) = 5PS²
⇒ 8 PT² = 3PR² + 5 PS².
Hence proved]

Question 7.
In fig., ∠CAB = 90° and AD ⊥ BC If AC = 75 cm, AB = 1 m, and BD = 1.25 m, find AD.
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Answer:
In given fig.,
∠CAB = 90°, AD ⊥ BC
AC = 75 cm, AB = 1 m
and BD = 1.25 m
In ∆ABC = ∆DBA
∠CAB = ∠ADB = 90°
∠CBA = ∠ABD
∴ ∆ABC ~ ∆DBA
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Question 8.
In an isosceles triangle ABC, if AC = BC and AB² = 2AC², prove that ∠C is right angle.
Answer:
It is given that:
AB² = 2AC² = AC² + AC²
AB² = AC² + BC² (∵ AC = BC)
By the converse of Pythagoras Theorem, we have that the triangle ABC is right angled at C.
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⇒ ∠ACB = 90°
⇒ ∠C is right angle.
Hence proved

Question 9.
PB and QA are the perpendiculars to segment AB. If PO = 5 cm, QO = 7 cm and ar (∆BOP) = 150 cm², find the ar (∆QOA).
Answer:
It is ∆BOP and ∆QOA, we have
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Question 10.
M and N are points on the sides PQ and PR respectively of a ∆PQR. State whether MN || QR if PM = 4 cm, QM = 4.5 cm, PN = 4 cm, NR = Answer:
In given figure,
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PN = 4 cm, QM = 4.5 cm
PN = 4 cm, NR = 4.5 cm
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Triangles Class 10 Extra Questions Short Answer Type-2

Question 1.
The perpendicular from A on the side BC of a ∆ABC intersect BC at D, such that DB = 3 CD. Prove that 2 AB² = 2 AC² + BC². [CBSE 2019(C)]
Answer:
Given: ∆ABC with AD ⊥ BC.
Also, DB = 3 CD.
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To prove: 2AB² = 2AC2² + BC²
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⇒ AC² – AB² = – \(\frac{x^2}{2}\)
⇒ 2AC² – 2AB²= – x²
⇒ 2AC² – 2AB² = – BC² [Replacing x by BC]
⇒ 2AC² + BC² = 2AB²
∴ 2AB² = 2AC² + BC²

Question 2.
In a ∆ABC, if ∠A = 90° and AD ⊥ BC, prove that AD² = BD × DC. [CBSE 2019]
Answer:
Given: Right ∆ABC and AD ⊥ BC
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To prove: AD² = BD × DC
Proof: In ∆BDA, ∠1 + ∠B = 90° (∵ ∠D = 90° )
⇒ ∠1 = 90° – ∠B …………. (i)
Also in right ∆BAC
∠C = 90° – ∠B …………….. (ii)
(i) and (ii)
⇒ ∠1 = ∠C
Similarly, ∠2 = ∠B
In ∆s ADC and ABD
∠1 = ∠C
∠2 = ∠B
⇒ ∠ADC ~ ∠BDA
∴ \(\frac{\mathrm{AD}}{\mathrm{DC}}=\frac{\mathrm{BD}}{\mathrm{AD}}\)
⇒ AD² = BD × DC
Hence proved

Question 3.
In the fig., ∠D = ∠E and \(\frac{A D}{D B}=\frac{A E}{E C}\). Prove that ∆BAC is an isosceles triangle.
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Answer:
Here, \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) [Given]
⇒ DE || BC
[By converse of Basic Proportionality Theorem]
Now, ∠D = ∠B [Corresponding angle]
∠E = ∠C
But ∠D = ∠E [Given]
Hence ∠B = ∠C
∴ AB = AC
[Sides opp. to equal angles of a ∆ are equal]
∴ ∆BAC is an isosceles ∆.

Question 4.
In given figure ∠1 = ∠2; ∠2 = ∠4
DE = 4; CE = x + 1, AE = 2x + 4; BE = 4x – 2, find x.
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Answer:
In ∆DCE and ∆ABE, we have
∠2 = ∠4
∠DEC = ∠AEB
[Vertically opposite angles]
Hence, ∆DCE ~ ∆BEA
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⇒ (x + 1)(4x – 2) = 4(2x + 4)
⇒ 4x² + 2x – 2 = 8x + 16
⇒ 4x² – 6x – 18 = 0
⇒ 2x² – 3x – 9 = 0
By splitting method
⇒ 2x² – 6x + 3x – 9 = 0
⇒ (x – 3) (2x + 3) = 0
x = 3 and x = – \(\frac{3}{2}\)

Question 5.
In the given fig. PA, QB and RC each is perpendicular to AC such that PA = x, RC = y, QB = z, AB = a and BC = b.
Prove that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\).
Answer:
Here PA ⊥ AC
and QB ⊥ AC
⇒ QB || PA
Thus, in ∆PAC QB || PA.
⇒ ∆QBC ~ ∆PAC
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Question 6.
If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
Answer:
We have a quadrilateral ABCD such that its diagonals intersect at O and
\(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\) …………… (1)
∵ ∠AOB = ∠COD [Vertically opposite angles]
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∴ Using SAS similarity, we have
∆AOB ~ ∆ COD
⇒ Their corresponding angles are equal.
i.e., ∠1 = ∠2
But they form a pair of int. alt. angles.
⇒ AB || DC
⇒ ABCD is a trapezium.
Hence proved

Question 7.
The area of two similar triangles are 81 cm2 and 49 cm2 respectively. If one of the sides of the first triangle is 6.3 cm, find the corresponding side of the other triangle.
Answer:
∆ABC ~ ∆PQR
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Question 8.
In figure, M is mid-point of side CD of a parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced at E. Prove that EL = 2BL.
Answer:
In ∆DEM and ∆CBM,
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∠DEM = ∠MBC [alt.angles]
∠EDM = ∠MCB [alt.angles]
DM = MC [given]
∠DEM = ∠CBM [AAS congruence]
⇒ DE = BC
but AE = AD + DE
= AD + BC
⇒ AE = 2BC
[∵ AD = BCas ABCD is a || gm] … (i)
In ∆AEL and ∆CBL,
∠1 = ∠2 [vertically opp. angles]
∠AEL = ∠LBC
[Alt. angles as DE || BC, BE acting as transversal]
∴ ∆AEL ~ ∆CBL [By AA similarity]
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Question 9.
Any point X is taken on the side BC of a triangle ABC and XM, XN are drawn parallel to BA, CA meeting CA, BA at M and N respectively. MN meets BC produced in T.
Prove that: TX² = TB × TC.
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Answer:
Here BA || XM ⇒ BN || XM
and CA || XN ⇒ CM||XN
Now in ATMX, BN||XM
∴ By Corollary to B.P.T., we have
\(\frac{\mathrm{TB}}{\mathrm{TX}}=\frac{\mathrm{TN}}{\mathrm{TM}}\) ……….. (i)
Again, in ATMC, XN || CM
By using corollary to B.P.T., we have
\(\frac{\mathrm{TX}}{\mathrm{TC}}=\frac{\mathrm{TN}}{\mathrm{TM}}\)
From (i) and (ii), we get
\(\frac{\mathrm{TX}}{\mathrm{TC}}=\frac{\mathrm{TB}}{\mathrm{TX}}\)
⇒ TX² = TB × TC

Triangles Class 10 Extra Questions Long Answer Type 1

Question 1.
A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake, the Peacock pouncies on it. If their speeds are equal, at what distance from the whole is the snake caught?
Answer:
Let the position of hole is at ‘H’ and snake be at ‘S’. If the point ‘P’ represents the peacock and let the distance between hole to the place where they meet = x metres.
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∴ SH = 27 m
∵ Their speeds are equal
∴ SC = (27 – x) m
PC = (27 – x) m
In right ∆ PHC, we have:
CH² + PH² = PC² [By Pythagoras Theorem]
⇒ (9)² + (x)² = (27 – x)²
⇒ 81 + x² = 729 – 54x + x²
⇒ 54x = 729 – 81 = 648
⇒ x = \(\frac{648}{54}\)= 12 m
Hence, the required distance = 12 m.

Question 2.
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. [CBSE Sample Paper 2017, CBSE 2018, 2019, 2019(C) 2019 SQP (Basic)]
Using the above, solve the following:
A ladder reaches a window which is 12 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 9 m high. Find the width of the street if the length of the ladder is 15 m.
Answer:
Let ∆ ABC is right angled at B.
Construction: Draw BD ⊥ AC.
In ∆ ADB and ∆ ABC
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∠ADB = ∠ABC = 90°
∠A = ∠A (Common)
∴ ∆ ADB ~ ∆ ABC
⇒ \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
⇒ AB² = AD × AC ………….. (i)
Similarly, ∆CDB ~ ∆CBA
⇒ \(\frac{\mathrm{CD}}{\mathrm{BC}}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
⇒ BC² = CD × AC …………… (ii)
Add (i) and (ii)
AB² + BC² = AD × AC + CD × AC
= (AD + CD) × AC
= AC × AC = AC²
⇒ AB² + BC² = AC²
Hence proved

Second part:
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Let AB and CD be two parallel walls and BE and ED be two positions of the same ladder of length 15 m.
In rt.∠d ∆EAB
AE² + AB² = BE²
⇒ AE² = BE² – AB² = 15² – 12² = 225 -144 = 81
⇒ AE = √81 = 9
In ∆ECD,
EC² = DE² – DC²
= 15² – 9²
= 225 – 81
= 144
∴ EC = √144 = 12
∴ Width of the street = AE + EC = 9 + 12 = 21 m.

Question 3.
ABC is a triangle right angled atC. If pis length of the perpendicular from C to AB and AB = c, BC = a and CA = b, then prove that:
(i) pc = ab
(ii) \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\)
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Answer:
Given: ∆ABC in which ∠C = 90°, CD ⊥ AB, CD = p, BC = a, AB = c and AC = b.
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Question 4.
In a trapezium ABCD, AC and BD are intersecting at O, AB || DC and AB = 2 CD. If the area of ∆AOB = 84 cm², find the area of ∆COD. [CBSE 2009]
Answer:
In ∆AOB and ∆COD
∠1 = ∠2
[Alternate interior angle ∵ AB || DC]
∠3 = ∠4
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Question 5.
In the figure, O is any point inside a rectangle ABCD such that OB = 6 cm, OD = 8 cm and OA = 5 cm. Find the length of OC. [CBSE 2009 C]
Answer:
Let us draw EOF || AB ⇒ OE ⊥ AD and OF⊥BC
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In ∆ OFB
∵ ∠F = 90°
∴ Using Pythagoras theorem, we have
OB² = OF² + BF²
In ∆ OED ∠E = 90°
∴ Using Pythagoras theorem, we have
OD² = OE² + DE²
Adding (1) and (2), we get OB² + OD²
= OF² + BF² + OE² + DE²
= OF² + AE² + OE2² + CF² [ ∵ BF = AE and CF = DE]
= (OF² + CF²) + (OE² + AE²)
= OC² + OA² = OC² + 5²
⇒ 6² + 8² = OC² + 5²
⇒ 36 + 64 = OC² + 25
OC² = 36 + 64 – 25 = 75
⇒ OC = √75 = √25 × 3 = √3 × 5
Thus, OC = 5√3 cm.

Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. [CBSE 2019(C)]
Answer:
Given: ∆ABC ~ ∆PQR
To prove:
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Triangles Class 10 Extra Questions HOTS

Question 1.
A quadrilateral OABC in which OA = OC. The bisector of ∠AOB meet AC at D and AB at F and the bisector of ∠COB meets AC at E and BC at G. Prove that ∆ODE ~ ∆OFG.
Answer:
We know that bisector of an angle of a triangle divides the opposite side in the ratio of sides containing the angle.
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⇒ FG || DE …(iv)
[by converse of Basic Proportionality Theorem]
In ∆ODE and ∆OFG, we have
⇒ AODE ~ AOFG [SAS similarity]
[Hence proved]

Question 2.
Bisector of ∠B and ∠C in ∆ABC meet each other at X. Line AX cuts the side BC in Y. Prove that
\(\frac{\mathbf{A X}}{\mathbf{X Y}}=\frac{\mathbf{A B}+\mathbf{A C}}{\mathbf{B C}}\)
Answer:
In ∆ABY, BX is bisector of ∠B.
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[Bisector of an angle of a triangle divides the opp. sides in the ratio of sides containing the angles.] ….. (i)
Similarly, in ∆ACY, CX is bisector of ∠C.
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Question 3.
CE and DE are equal chords of a circle with centre O. If ∠AOB = 90°, then find ar(∆CED): ar(∆AOB).
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Answer:
In ∆CED, we have ∠CED = 90°
[∵ Angle in semicircle is a right angle]
∠1 = ∠2 = 45°
[∵ CE = DE (given) and ∠1 + ∠2 = 90°]
In ∆AOB, ∠AOB = 90° [given]
⇒ OA = OB [Radii of same circle]
⇒ ∠3 = ∠4 = 45°
Now in ∆CED and ∆AOB
∠CED = ∠AOB [each 90°]
∴ ∠1 = ∠3 [each 45°]
∴ ∠2 = ∠4 [each 45°]
∆CED ~ ∆AOB [AAA similarity]
⇒ \(\frac{{ar}(\Delta \mathrm{CED})}{{ar}(\Delta \mathrm{AOB})}=\frac{\mathrm{CD}^2}{\mathrm{AB}^2}\) [Using Theorem] …….. (i)
Now CD = 2OD = 2OB
⇒ CD² = 4(OB²)
In right ∆AOB
AB² = OA² + OB² = OB² + OB² = 2(OB)²
[∵ OA = OB (Radii of same circle)]
From (i) and (ii), we get
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Thus, ar (∆CED): ar (∆AOB) = 2:1

Question 4.
The internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.
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Answer:
Given: A ∆ ABC such that ,X bisector of ∠A meets BC in D.
To prove: \(\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}\)
Construction: BT parallel to AD, meeting CA produced at point T.
Proof: Now BT || AD and CT is a transversal.
⇒ ∠1 = ∠3 ……. (i)
[Corresponding angles]
Further BT || AD and AB is a transversal.
⇒ ∠2 = ∠4 [Alternate angles] ……. (ii)
But ∠1 = ∠2 ……. (iii)
[Given as AD is bisector of ∠BAC]
From (i), (ii) and (iii), we get
∠3= ∠4
⇒ AB = AT …(to)
[∵ Sides opposite to equal angles are also equal]
Also in ABCT, DA || BT
∴ By B.P.T.
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Multiple Choice Questions

Choose the correct option out of four given in each of the following:

Question 1.
In adjoining fig, O is the point of intersection of two chords AB and CD such that OB = OD, then triangles OAC and ODB are:
(a) isosceles and similar
(b) equilateral and similar
(c) isosceles but not similar
(d) equilateral but not similar
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Answer:
(a) isosceles and similar

Question 2.
In the adjoining fig. XY || QR and \(\frac{P X}{X Q}=\frac{P Y}{Y R}\) = \(\frac{1}{2}\), then
(a) XY = QR
(b) XY = \(\frac{1}{3}\) QR
(c) XY² = QR²
(d) XY = \(\frac{1}{2}\) QR
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Answer:
(b) XY = \(\frac{1}{3}\) QR

Question 3.
If ∆ABC ~ ∆EDF and Image may be NSFW.
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(a) BC.EF = AC.FD
(b) AB.EF=AC.DE
(c) BC.DE = AB.EF
(d) BC.DE = AB.FD
Answer:
(c) BC.DE = AB.EF

Question 4.
In fig. ∠PQR = 90° and QM ⊥ PR. Then,
(a) PM.MR = PR²
(b) QP.QR = PR²
(c) PM.MR = QM²
(d) QP.QR = QM²
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Answer:
(c) PM.MR = QM²

Question 5.
The ratio of the areas of two similar triangles is equal to the:
(a) ratio of their corresponding sides
(b) ratio of their corresponding attitudes
(c) ratio of the squares of their corresponding sides
(d) ratio of the squares of their perimeter
Answer:
(c) ratio of the squares of their corresponding sides

Question 6.
The ratio of the corresponding sides of two similar triangles is 2:3, then ratio of the corresponding heights is:
(a) 2:3
(b) 4:9
(c) 3:2
(d) 9:4.
Answer:
(a) 2:3

Question 7.
The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. The ratio of their cor-responding sides is:
(a) 81:49
(b) 9:7
(c) (81)² : (49)²
(d) none of these.
Answer:
(b) 9:7

Question 8.
The areas of two similar triangles are 144cm2 and 81cm2. If one median of the first triangle is 16 cm, length of corresponding median of the second triangle is:
(a) 9 cm
(b) 27 cm
(c) 12 cm
(d) 16 cm
Answer:
(c) 12 cm

Question 9.
The areas of two similar triangles are 12cm2 and 48cm². If the height of the smaller one is 2.1cm. The corresponding height of the bigger triangle is:
(a) 4.41cm
(b) 8.4 cm
(c) 4.2 cm
(d) 6.3 cm
Answer:
(c) 4.2 cm

Question 10.
If ∆ABC is similar to ∆DEF and ∠A = 47°, ∠E = 83°, then ∠C is:
(a) 80°
(b) 83°
(c) 47°
(d) 50°
Answer:
(d) 50°

Question 11.
In a right triangle ABC, in which ∠C = 90° and CD ⊥ AB.
If BC = a, CA = b, AB = c and CD = p.
(a) \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\)
(b) \(\frac{1}{p^2} \neq \frac{1}{a^2}+\frac{1}{b^2}\)
(c) \(\frac{1}{p^2}<\frac{1}{a^2}+\frac{1}{b^2}\) (d) \(\frac{1}{p^2}>\frac{1}{a^2}+\frac{1}{b^2}\)
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Answer:
(a) \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\)

Question 12.
If ∆ABC and ∆DEF are similar and 2AB = DE, BC = 8 cm, then EF is :
(a) 4 cm
(b) 16 cm
(c) 8 cm
(d) 12 cm
Answer:
(b) 16 cm

Question 13.
∆ABC ~ ∆DEF, BC 3cm, EF = 4 cm and area of ∆ABC = 54 cm2 then the area of ∆DEF is:
(a) 72 cm
(b) 40.5 cm
(c) 96 cm²
(d) none of these.
Answer:
(c) 96 cm²

Question 14.
The hypotenuse ‘ d and one side V of a right triangle are consecutive numbers. The square of the other side of the triangle is:
(a) ac
(b) \(\frac{c}{a}\)
(c) c + a
(d) c – a
Answer:
(c) c + a

Question 15.
In a right triangle, the square on the hypotenuse equals twice the product of other two sides. One of the acute angles of the triangle is :
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Answer:
(b) 45°

Fill in the Blanks

Question 1.
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are _____________ to the whole triangle
Answer:
similar

Question 2.
The ratio of the areas of two similar triangles is equal to the ratio of the _____________ on their corresponding sides.
Answer:
square

Question 3.
In _____________ triangle the square on the hypotenuse is equal to the sum of the squares on the other two sides.
Answer:
right angled

Question 4.
In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angle opposite to the first side is a _____________ angle.
Answer:
right

Question 5.
The ratio of the areas of two similar triangles is equal to the _____________ of their corresponding medians.
Answer:
square

Question 6.
The ratio of the areas of two similar triangles is equal to the _____________ of their corresponding altitudes.
Answer:
square

Question 7.
The ratio of corresponding sides of two similar triangles is _____________ to their corresponding perimeters.
Answer:
proportional

Question 8.
A triangle whose sides measures 12 cm, 5 cm and 13 cm is a _____________ triangle.
Answer:
Right angled

Question 9.
If corresponding sides of a triangle are proportional then triangles are similar and this criteria of similarity is briefly written as _____________ .
Answer:
SSS

Question 10.
If two corresponding angles of two triangles are equal then triangles are _____________ (similar/not similar).
Answer:
similar

Extra Questions for Class 10 Maths

NCERT Solutions for Class 10 Maths

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