

InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
701. |
State whether the statements are True or False. If in ΔABC and ΔDEF, AB = DE, ∠A = ∠D and BC = EF then the two triangle ABC and DEF are congruent by SAS criterion. |
Answer» If in ΔABC and ΔDEF, AB = DE, ∠A = ∠D and BC = EF then the two triangle ABC and DEF are congruent by SAS criterion. False |
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702. |
In a quadrilateral ABCD, ∠B = ∠90°, \(AD^2=AB^2+BC^2+CD^2\) prove that ∠ACD = 90° |
Answer» We have ∠B = 90° and AD2 = AB2 + BC2 + CD2 (Given) But AB2 + BC2 = AC2 AD2 = AC2 + CD2 By converse of by Pythagoras ∠ACD = 90° |
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703. |
In an equilateral △ABC,AD ⊥BC, prove that \(AD^2=38D^2\), |
Answer» We have ⊿ ABC is an equilateral triangle and AD⊥BC In ⊿ ADB⊿ ADC ∠ADB = ∠ADC = 90° AB = AC (Given) AD = AD (Common) ⊿ ADB ≅⊿ ADC (By RHS condition) ∴ BD = CD = BC/2 ……. (i) In ⊿ ABD BC2 = AD2 + BD2 BC2 = AD2 + BD2 [Given AB = BC] (2BD)2 = AD2 + BD2 [From (i)] 4BD2 - BD2 = AD2 AD2 = 3BD2 |
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704. |
∆ABC is a right triangle right-angled at A and AC ⊥ BD. Show that(i) \(AB^2=BC.BD\)(ii) \(AC^2=BC.DC\)(iii) \(AD^2=BD.CD\)(iv) \(\frac{AB^2}{AC^2}\) = \(\frac{BD}{CD}\) |
Answer» (i) In ⊿ABD and In ⊿CAB ∠DAB=∠ACB=90° ∠ABD=∠CBA [Common] ∠ADB=∠CAB [remaining angle] So, ⊿ADB≅⊿CAB [By AAA Similarity] ∴ AB/CB=BD/AB AB2 = BC x BD (ii) Let ∠CAB = x InΔCBA = 180 - 90° - x ∠CBA = 90° - x Similarly in ΔCAD ∠CAD = 90° - ∠CAD = 90° - x ∠CDA = 90° - ∠CAB = 90° - x ∠CDA = 180° - 90° - (90° - x) ∠CDA = x Now in ΔCBA and ΔCAD we may observe that ∠CBA = ∠CAD ∠CAB = ∠CDA ∠ACB = ∠DCA = 90° Therefore ΔCBA ~ ΔCAD ( by AAA rule) Therefore AC/DC = BC/AC AC2 = DC x BC (iii) In DCA and ΔDAB <DCA = DAB (both angles are equal to 90°) <CDA =, <ADB (common) <DAC = <DBA ΔDCA= ΔDAB (AAA condition) Therefore DC/DA=DA/DB AD2 = BD x CD (iv) From part (I) AB2=CBxBD From part (II) AC2 = DC x BC Hence AB2/AC2 = CB x BD/DC x BC AB2/AC2 = BD/DC Hence proved |
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705. |
In figure, if ∠D = ∠C, then it is true that ∆ADE ∼ ∆ACB? Why? |
Answer» True In ∆ADE and ∆ACB, Given, ∠D = ∠C ∠A = ∠A [common angle] As we know sum of all the angles of a triangle is equals to 180∘. So, by angle sum property of triangle the third angle of both triangles must be equal. ∠E = ∠B ∴ ∆ADE ∼ ∆ACB [by AAA similarity criterion] |
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706. |
In the figure x° = A) 45°B) 55°C) 100°D) 80° |
Answer» Correct option is C) 100° |
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707. |
In the following figure, ABD is a triangle right angled at A and AC ⊥BD. Show that (i) AB2= BC.BD (ii) AC2 = BC.DC (iii) AD2 = BD.CD |
Answer» Data: In ∆ABD, ∠A = 90°, AC ⊥ BD. To Proved: AB2 = BC.BD (ii) AC2 = BC.DC iii) AD2 = BD.CD (i) AB2 = BC.BD ∆ACB ~ ∆BAD (. Theorem7) ∴ \(\frac{AB}{BD} = \frac{BC}{AB}\) ∴ AB2 = BC × BD. (ii) AC2 = BC.DC ∆BCA ~ ∆ACD ∴ \(\frac{AB}{AD} = \frac{AC}{CD} = \frac{BC}{AC}\) ∴ AC × AC = BC × CD ∴ AC2 = BC × CD (iii) AD2 = BD.CD ∆ACD ~ ∆BAD ∴ \(\frac{AD}{BD} = \frac{CD}{AD} = \frac{AC}{AB}\) ∴ AD × AD = BD × DC ∴ AD2 = BD × DC |
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708. |
Two ……………. are always similar (A) Triangles (B) Rectangles (C) Squares(D) Trapeziums |
Answer» Correct option is (C) Squares Two squares are always similar but two triangles, rectangles and trapeziums need not be always similar. Correct option is: (C) Squares |
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709. |
Rashmi claims that no triangle can have more than one right angle. Do you agree with her. Why ? |
Answer» Yes, Rashmi is correct. A triangle can’t have more than one right angle. As the sum of two right angles is 180° there will be no scope for 3rd angle. |
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710. |
In ΔABC, ⌊C = 90° then its hypotenuse is ……………….?A) ABB) BCC) ACD) None |
Answer» Correct option is A) AB |
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711. |
In ΔMAN, \(\overline{MA}=\overline{AN}\) then it is a ………………. type of triangle.A) ScaleneB) IsoscelesC) EquilateralD) Right angled triangle |
Answer» Correct option is B) Isosceles |
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712. |
In the given figure BC || DE and AD = DB = 3.4 and AC = 14.So find AE and EC. |
Answer» In the given ΔABC, DE || BC is given. So from Thales theorem, \(\frac{AD}{DB}=\frac{AE}{EC}\) In this problem, AD = DB = 3.4 is given. And also AC = 14 is given. So \(\frac{AD}{DB}=\frac{AE}{EC}\) ⇒ \(\frac{3.4}{3.4}=\frac{AE}{EC}\) ⇒ AE = EC But AC = 14 is given. AC = AE + EC = 2AE = 14 ⇒ AE = EC = 7 cm |
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713. |
In ∆ABC, DE ║ BC such that AD/DB = 3/5 . If AC = 5.6cm, then AE = ? (a) 4.2cm (b) 3.1cm (c) 2.8cm (d) 2.1cm |
Answer» (d) 2.1 cm It is given that DE || BC. Applying Thales’ theorem, we get: AD/DB = AE/EC Let AE be x cm. Therefore, EC = (5.6 – x) cm ⇒ 3/5 = x/5.6 − x ⇒ 3(5.6 − x) = 5x ⇒ 16.8 − 3x = 5x ⇒ 8x = 16.8 ⇒ x = 2.1 cm |
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714. |
In ∆ABC, DE║BC so that AD = 2.4cm, AE = 3.2cm and EC = 4.8cm. Then, AB = ? (a) 3.6cm (b) 6cm (c) 6.4cm (d) 7.2cm |
Answer» (b) 6 cm It is given that DE || BC. Applying basic proportionality theorem, we have: AD/BD = AE/ EC ⇒ 2.4/BD = 3.2/4.8 ⇒ BD = 2.4×4.8/3.2 = 3.6 cm Therefore, AB = AD + BD = 2.4 + 3.6 = 6 cm |
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715. |
The lengths of two sides of a triangle are 6 cm and 9 cm. Write all the possible lengths of the third side. |
Answer» The possible length of the third side must greater than (9— 6) and less than (9 + 6). ∴ The third side may be 4 cm, 5 cm, 6 cm, 7 cm, 8 cm, 9 cm, 10 cm, 11 cm, 12 cm, 13 cm, 14 cm. |
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716. |
Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer. |
Answer» No, it is not possible to construct a triangle with lengths of sides 4 cm, 3 cm and 7 cm. Justification: We know that, Sum of any two sides of a triangle is always greater than the third side. Here, the sum of two sides whose lengths are 4 cm and 3 cm = 4 cm + 3 cm = 7 cm, Which is equal to the length of third side, i.e., 7 cm. Hence, it is not possible to construct a triangle with lengths of sides 4 cm, 3 cm and 7 cm. |
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717. |
In the given figure, AD = AE and `AD^2 = BD xx EC` prove that: triangles ABD and CAE are similar. |
Answer» AD=AE `/_ADE=/_AED` `/_ADB=/_AEC` `AD^2=BD*EC` `(AD)/(BD)=(EC)/(AD)` `(AD)/(BD)=(EC)/(AE)` `In/_ABD and /_CAE` `(AD)/(BD)=(EC)/(AE)` `/_ADB=/_AEC` `/_ABD cong /_CAE`. |
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718. |
In the given figure, side BC of a ∆ABC is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F respectively, and AD is produced to X so that D is the midpoint of OX. Prove that AO : AX = AF : AB and show that EF║BC. |
Answer» It is give that BC is bisected at D. ∴ BD = DC It is also given that OD =OX The diagonals OX and BC of quadrilateral BOCX bisect each other. Therefore, BOCX is a parallelogram. ∴ BO || CX and BX || CO BX || CF and CX || BE BX || OF and CX || OE Applying Thales’ theorem in ∆ ABX, we get: AO/ AX = AF/ AB …(1) Also, in ∆ ACX, CX || OE. Therefore by Thales’ theorem, we get: AO/AX = AE/ AC …(2) From (1) and (2), we have: AO/ AX = AE/AC Applying the converse of Theorem in ∆ ABC, EF || CB. This completes the proof. |
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719. |
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle? |
Answer» Assume ABC and PQR to be 2 triangle. We, have ΔABC ~ΔPQR Perimeter of ΔABC = 25 cm Perimeter of ΔPQR = 15 cm AB = 9 cm PQ = ? Since, ΔABC ~ΔPQR Then, ratio of perimeter of triangles = ratio of corresponding sides So \(\frac{25}{15}\) = \(\frac{AB}{PQ}\), (Corresponding parts of similar triangle area proportion) Or \(\frac{25}{15}\) = \(\frac{9}{PQ}\) Or PQ = 135/25 Or PQ = 5.4 cm |
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720. |
Is it possible to construct a triangle with length of its sides as 8 cm ,7 cm and 4 cm ? Give reason for your answer |
Answer» Yes, because in each case the sum of two a sides is greater than the third side `7+4gt8,` `8+4gt7,` `7+8gt4,` Hence,it is possible to construct a triangle with given sides. |
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721. |
In figure ,D and E are Points on side BC of a `Delta ABC` such that BD=CE and AD=AE.Show that `Delta ABD cong Delta ACE.` |
Answer» Given D and E are the point on side BC of a `Delta ABC `Such that BD =CE and AD=AE. To show `Delta ABD cong Delta ACE` Proof We have,`" " AD=AE " "[given]` `rArr " "angleADE=angleAED " "…(i)` `" "["since angle opposite to equal sides are equal "]` We have `" "angleADB+angleADE=180 ^(@) " " ["linear Pair axiom"] ` `rArr " "angleADB=180^(@)-angleAED` `" "=180^(@)-angleAED " " ["from Eq.(i)"]` `In Delta ABD and Delta ACE,` `" "angleADB=angleAEC " "[because angleAEC + angleAED=180^(@),"Linear Pair axiom"]` `" "BD=CE " "["given"]` `therefore " "DeltaABD cong Delta ACE " "["by SAS congruence rule"]` |
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722. |
D and E are points on the sides AB and AC respectively of a `Delta ABC` such that `DE||BC`. Find the value of x, when (i) `AD =x cm, DB=(x-2) cm`, `AE=(x+2) cm and EC =(x-1) cm`. (ii) `AD=4 cm, DB =(x-4) cm, AE =8 cm` and `EC=(3x-19) cm`. `(iii) AD =(7x-4) cm, AE =(5x-2) cm, DB (3x+4) cm and EC= 3x cm`. |
Answer» Correct Answer - `(i) x=4 (ii) x=11 (ii) x=4` | |
723. |
D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC. If AD/DB = 4/7 and AC = 6.6 cm, find AE. |
Answer» From given triangle, points D and E are on the sides AB and AC respectively such that DE || BC. AD/DB = 4/7 or AD = 4 cm, DB = 7 cm, and AC = 6.6 By Thale’s Theorem: AD/DB = AE/EC Add 1 on both sides AD/DB + 1 = AE/EC + 1 (AD + DB)/DB = (AE + EC)/EC (4+7)/7 = AC/EC = 6.6/EC EC = (6.6 x 7)/11 = 4.2 And, AE = AC – EC AE = 6.6 – 4.2 AE = 2.4 cm |
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724. |
In ΔABC and ΔDEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥BC and DM ⊥ EF, find AL : DM. |
Answer» Since \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = \(\frac{1}{2}\) Then, ΔABC ~ ΔDEF (By SS similarity) Now, In ΔABL ~ ΔDEM ∠B = ∠E (ΔABC ~ΔDEF) ∠ALB = ∠DME (Each 90°) Then, ΔABL ~ ΔDEM (By SS similarity) So, \(\frac{AB}{DE}\) = \(\frac{AL}{DM}\) (Corresponding parts of similar triangle area proportion) or \(\frac{5}{10}\) = \(\frac{AL}{DM}\) or, \(\frac{1}{2}\) = \(\frac{AL}{DM}\) |
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725. |
D and E are the points on the sides AB and AC respectively of a ΔABC such that AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = 5/2 DE |
Answer» We have, \(\frac{AD}{DB}\) = \(\frac{8}{12}\) = \(\frac{2}{3}\) And, \(\frac{AD}{EC}\) = \(\frac{6}{9}\) = \(\frac{2}{3}\) Since, \(\frac{AD}{DB}\) = \(\frac{AD}{EC}\) Then , by converse of basic proportionality theorem. DE||BC In Δ ADE and Δ ABC ∠A = ∠A (common) ∠ADE = ∠B (Corresponding angles) Then, Δ ADE ~ Δ ABC (By AA similarity) \(\frac{AD}{AB}\) = \(\frac{DE}{BC}\) (Corresponding parts of similar triangle are proportion) \(\frac{8}{20}\) = \(\frac{DE}{BC}\) \(\frac{2}{5}\) = \(\frac{DE}{BC}\) BC = \(\frac{5}2\) DE |
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726. |
If D and E are points on sides AB and AC respectively of a Δ ABC such that DE II BC and BD = CE. Prove that Δ ABC is isosceles. |
Answer» We have DE∥BC by the converse of proportionality theorem AD/DB=AE/EC AD/DB=AE/DB [BD=CE] AD=AE Adding D both sides AD+BD=AE+DB AD+BD=AE+EC [BD=CE] AB=AC ⊿ABC is isosceles |
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727. |
In a Δ ABC , P and Q are points on sides AB and AC respectively, such that PQ II BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find AB and PQ.. |
Answer» WE have, PQ∥BC We have AP/PB = AQ/QC 2.4/PB = 2/3 PB = 3x2.4/2 PB = 3x1.2 PB = 3.6 cm Now AB = AP+PB AB = 2.4+3.6 AB = 6 cm Now IN ⊿ APQ and ⊿ ABC ∠A = ∠A [Common] ∠APQ = ∠ABC [PQ∥BC] ⊿ APQ ~ ⊿ ABC [By AA criteria] AB/AP = BC/PQ PQ = 6x2.4/6 PQ = 2.4 cm |
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728. |
In a Δ ABC , AD is the bisector of ∠A, meeting side BC at D. (i) If BD = 2.5 cm, AB = 5 cm and AV = 4.2 cm, find DC. (ii) If BD = 2 cm, AB = 5 cm and DC = 3 cm, find AC. (iii) If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD. (iv) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC. (v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm, find AB. (vi) If AB = 5.6 cm, AC = 6 cm and DC = 6 cm, find BC. (vii) If AD = 5.6 cm, BC = 6 cm and BD = 3.2 cm, find AC. (viii) If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC. |
Answer» (i) we have Angle BAD=CAD Here AD bisects ∠A BD/DC=AB/AC 2.5/DC=5/4.2 DC=2.5 x 4.2/5 DC=2.1 cm (ii) Here AD bisects ∠A AB/DC=AB/AC 2/3=5/AC AC=15/2 AC=7.5 cm (iii) in △ ABC A bisects ∠A BD/DC = AB/BC BD/2.8 = 3.5/4.2 BD =3.5 x 2.8/4.2 BD = 7/3 BD = 2.33 cm (iv) In△ABC, AD bisects ∠A BD/DC = AB/AC X/6 - x = 10/14 14x = 60 - 10x 14x + 10x = 60 24x = 60 x = 60/24 x = 5/2 x = 2.5 BD = 2.5 DC = 6-2.5 DC = 3.5 (v) AB/AC=BD/DC AB/4.2 = BC - DC/DC AB/4.2 = 10-6/6 AB/4.2 = 4/6 AB = 4 x 4.2/6 AB = 2.8 cm (vi) BD/DC=AB/AC BD/6 = 5.6/6 BD = 5.6 BC = BD + DC BC = 5.6 + 6 BC = 11.6 cm (viii) In△ABC, AD bisects ∠A AB/AC = BD/DC 5.6/AC = 3.2/BC - BD 5.6/AC = 3.2/6 - 3.2 5.6/AC = 3.2/2.8 AC x 3.2 = 2.8 x 5.6 AC = 2.8 x 5.6/3.2 AC = 7 x 0.7 AC = 4.9 cm (ix) let BD =x,then DC=12 - X BD/DC = AB/BC x/12-x = 10/6 6x =120 - 10x 6x + 10x = 120 16x = 120 x = 120/16 x = 7.5 BD = 7.5 cm DC = 12 - x DC =12 - 7.5 DC = 4.5 cm |
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729. |
From the given figure which corresponding parts are need to equal when ΔADC ≅ ΔABC by S.A.S rule?A) CD = CBB) ⌊A CB = ⌊D CAC) AD = ABD) ⌊D = ⌊B |
Answer» Correct option is C) AD = AB |
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730. |
ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.i) State the three pairs of equal parts in ΔADB and ΔADC.ii) Is ΔADB = ΔADC ? Why or why not ?iii) Is ∠B = ∠C ? Why or why not ? |
Answer» i)) ∠ADB = ∠ADC Right angle AB = AC hypotenuse; AD = AD common side ii) Yes, by R.H.S congruence iii) Yes by c.p.c.t |
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731. |
The line segments joining the midpoints of the adjacent sides of a quadrilateral form (a) parallelogram (b) trapezium (c) rectangle (d) square |
Answer» (a) A parallelogram The line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram. |
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732. |
The areas of two similar triangles `Delta ABC` and `Delta PQR ` are `25 cm^(2)` and `49 cm^(2)` respectively. If `QR=9.8` cm, find `BC`. |
Answer» We know that the ratio or the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. `:. (ar (Delta ABC))/(ar (Delta PQR))=(BC^(2))/(QR^(2))` ltbr `rArr ((BC)/(QR)^(2))= (ar (Delta ABC))/(ar (Delta PQR))=(25)/(49)=((5)/(7))^(2)` ltbr `rArr (BC)/(QR)=(5)/(7)rArr (BC)/(9.8)=(5)/(7)` `rArr BC(5xx9.8)/(7) cm= 5xx1.4 m= 7 cm` Hence, BC=7cm |
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733. |
The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR. |
Answer» We have, ΔABC ~ ΔPQR Area(ΔABC) = BC2 Area (ΔPQR) QR2 (4.5)2/QR2 = 9/16 4.5/QR = 3/4 QR = 4 x 4.5/3 QR = 6cm |
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734. |
Thecorresponding altitudes of two similar triangles are 6 cm and 9 cmrespectively. Find the ratio of their areas. |
Answer» Correct Answer - `4:9` Require ratio `=((6)^(2))/((9)^(2))=(36)/(81)=(4)/(9)=4:9` |
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735. |
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? |
Answer» False By the property of area of two similar triangles `(("Area"_(1))/("Area"_(2)))=(("Atitude"_(1))/("Atitude"_(2)))^(2)` `rArr (("Area"_(1))/("Area"_(2)))=(3/5)^(2)` [`because("altitude"_(1))/("altitude"_(2))=3/5`] `=9/25ne6/5` So, given statement is not correct. |
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736. |
The ratio of areas of two similar triangles is 1: 2, then the ratio of their altitudes is (A) 1 : √2 (B) 1 : 2 (C) 2 : 1 (D) √2 : 1 |
Answer» Correct option is (A) 1 : √2 The ratio of altitudes of similar triangle = (The ratio of areas of similar triangle\()^\frac12\) \(=(\frac12)^\frac12=\sqrt{\frac12}\) \(=\frac1{\sqrt2}=1:\sqrt2\) Correct option is: (A) 1 : √2 |
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737. |
The ratio of altitudes of two similar triangles is 4 : 9, then ratio of their areas is (A) 2 : 3 (B) 4 : 9 (C) 81 : 16(D) 16 : 81 |
Answer» Correct option is (D) 16 : 81 The ratio of areas of similar triangle is square of the ratio of corresponding sides (or altitude) of similar triangle. \(\therefore\) The ratio of areas of given similar triangle \(=(\frac49)^2\) \(=\frac{16}{81}=16:81\) Correct option is: (D) 16 : 81 |
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738. |
The areas of two similar triangles are 16 cm2 and 25 cm2 respectively, then the ratio of their corresponding sides is ………(A) 5 : 4 (B) 4 : 5 (C) 16 : 25 (D) 25 : 16 |
Answer» Correct option is (B) 4 : 5 Let similar triangles are \(\triangle ABC\;\&\;\triangle DEF\) Also let \(ar(\triangle ABC)=16\,cm^2\;\&\) \(ar(\triangle DEF)=25\,cm^2\) \(\because\) \(\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=(\frac{AB}{DE})^2\) \(\therefore\) \((\frac{AB}{DE})^2=\frac{16}{25}\) \(\Rightarrow\) \(\frac{AB}{DE}=\sqrt{\frac{16}{25}}=\frac45\) \(\therefore AB:DE=4:5\) Hence, the ratio of corresponding sides of given triangle is 4:5. Correct option is: (B) 4 : 5 |
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739. |
Using Theorem 2.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. |
Answer» In ∆ABC, D and E are mid-points of AB and AC. ∴ AD = DB AE = EC AB = 2AD AC = 2AE \(\frac{AD}{AB} = \frac{AE}{AC}\) = \(\frac{1}{2}\) As per S.S.S. Postulate. ∆ADE ~ ∆ABC ∴ They are equiangular triangles. ∴ ∠A is common. ∠ADE = ∠ABC ∠ADE = ∠ACE These are pair of corresponding angles ∴ DE || BC |
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740. |
The areasof two similar triangles `A B C`and `P Q R`are in theratio `9: 16`. If `B C=4. 5 c m`, find thelength of `Q R`. |
Answer» Correct Answer - 6 cm | |
741. |
Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX). |
Answer» Let P be the mid point of AB =`(AP)/(PB)=1/1` Draw a line PQ||BC =`(AQ)/(QC)=(AP)/(PB)` =`(AQ)/(QC)=1/1` AQ=QC PQ line bisects the side AC |
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742. |
The areas of two similar `triangleABC and trianglePQR` are 64 sq. cm and 121 sq. cm. repsectively. If QR= 15.4 cm, find BC. |
Answer» Correct Answer - `11.2 cm` | |
743. |
A vertical pole 6 m log casts a shadow of length 3.6 m on the ground. What is the height of tower which casts a shadow of length 18 m at the same time?A. `10.8 m`B. `28.8m`C. `32.4m`D. 30m |
Answer» Correct Answer - D | |
744. |
In the adjoining figure, show that ∠A + ∠B +∠C + ∠D + ∠E + ∠F = 360°. |
Answer» We know that the sum of all the angles in triangle ACE is 180o. ∠A + ∠C + ∠E = 180o ….. (1) We know that the sum of all the angles in triangle BDF is 180o. ∠B + ∠D + ∠F = 180o ….. (2) Now by adding both equations (1) and (2) we get ∠A + ∠C + ∠E + ∠B + ∠D + ∠F = 180o + 180o On further calculation ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360o Therefore, it is proved that ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360o. |
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745. |
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. |
Answer» We can create two triangles for both pole and tower with their shadows. Please refer to video for the figure.As angle made by sun at the same time are same for both tower and vertical pole. Also, both are perpendicular to ground. So, these two triangles are similar. Thus, Length of pole/Length of tower =Shadow of pole/Shadow of tower `=>6/h = 4/28 =>h = 7**6 = 42` So, height of tower is `42` m. |
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746. |
A vertical pole of length `7.5 m ` casts a shadow 5 long on the ground and at the same time a tower casts a shadow 24 m long. Find the height of the tower. |
Answer» Correct Answer - 36 m | |
747. |
In a `Delta ABC` it is given AD is internal bisector of `angleA` . If `BD=4 cm, DC=5 cm and AB=6 cm`, then AC =? |
Answer» Correct Answer - B::D | |
748. |
ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the midpoints of AB, AC, CD and BD respectively, show that PQRS is a rhombus. |
Answer» In ∆ ABC, P and Q are mid points of AB and AC respectively. So, PQ || BC, and PQ = 1/2 …(1) Similarly, in ∆ADC, …(2) Now, in ∆BCD, SR = 1/2 …(3) Similarly, in ∆ABD, PS = 1/2 = 1/2 …(4) Using (1), (2), (3), and (4). PQ = QR = SR = PS Since, all sides are equal Hence, PQRS is a rhombus. |
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749. |
In Fig., if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH. |
Answer» Given, AB ‖ CD and, EF ‖ BC ∠BAC = 65° and ∠DHF = 35° ∠BAC = ∠ACD (Alternate angles) ∠ACD = 65° ∠DHF = ∠GHC (Vertically opposite angles) ∠GHC = 35° In ΔGHC ∠GCH + ∠GHC + ∠HGC = 180° 65° + 35° + ∠HGC = 180° ∠HGC = 80° ∠AGH + ∠HGC = 180°(Linear pair) ∠AGH + 80° = 180° ∠AGH = 100° |
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750. |
In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2. |
Answer» Given, Quadrilateral ABCD, In which ∠A + ∠D = 90° Construct produce AB and CD to meet at E. Also, join AC and BD. in ∆AED, ∠A + ∠D = 90° [given] ∴∠E = 180°- (∠A + ∠D) = 90° [∵ sum of angles of a triangle = 180°] Then, By Pythagoras theorem, AD2 = AE2 + DE2 In ∆BEC, by Pythagoras theorem, BC2 = BE2 + EF2 On adding both equations, we get, AD2 + BC2 = AE2 + DE2 + BE2 + CE2 In ∆AEC, by Pythagoras theorem, AC2 = AE2 + CE2 And in ∆BED, by Pythagoras theorem, BD2 = BE2 + DE2 On adding both equations, we get, AC2 + BD2 = AE2 + CE2 + BE2 + DE2 ….(i) From Equation (i) and (ii), AC2 + BD2 = AD2 + BC2 Hence proved. |
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