818 + Interview Questions in TRIANGLES IN GENERAL AWARENESS Page 10 InterviewSolution

451.	State whether the statement are True or False.It is possible to have a triangle in which each angle is equal to 60°
Answer» True Since in equilateral triangle each angle is equal to 60°.

451.

State whether the statement are True or False.It is possible to have a triangle in which each angle is equal to 60°

Answer»

True

Since in equilateral triangle each angle is equal to 60°.

Discussion

452.	State whether the statement are True or False.It is possible to have a triangle in which two angles are acute.
Answer» True. It is possible to have a triangle in which two angles are acute.

452.

State whether the statement are True or False.It is possible to have a triangle in which two angles are acute.

Answer»

True.

It is possible to have a triangle in which two angles are acute.

Discussion

453.	State whether the statement are True or False.It is possible to have a triangle in which each angle is greater than 60°
Answer» False If all angles of a triangle are greater than 60°, then their sum will be greater than 180°. But in a triangle sum of all angles is 180° ∴ Triangle is not possible.

453.

State whether the statement are True or False.It is possible to have a triangle in which each angle is greater than 60°

Answer»

False

If all angles of a triangle are greater than 60°, then their sum will be greater than 180°. But in a triangle sum of all angles is 180°

∴ Triangle is not possible.

Discussion

454.	The bisectors of exterior angles at B and C of Δ ABC meet at O, if ∠A= x°, then ∠BOC =A. 90°+ \(\frac{x°}{2}\)B. 90°- \(\frac{x°}{2}\)C. 180°+ \(\frac{x°}{2}\)D. 180°- \(\frac{x°}{2}\)
Answer» ∠OBC = 180° - ∠B - \(\frac{1}{2}\)(180° - ∠B) ∠OBC = 90° - \(\frac{1}{2}\)∠B And, ∠OCB = 180° - ∠C - \(\frac{1}{2}\)(180° - ∠C) ∠OCB = 90° - \(\frac{1}{2}\)∠C In ΔBOC ∠BOC + ∠OCB + ∠OBC = 180° ∠BOC + 90° - \(\frac{1}{2}\)∠C + 90° - \(\frac{1}{2}\)∠B = 180° ∠BOC = \(\frac{1}{2}\)(∠B + ∠C) ∠BOC = \(\frac{1}{2}\)(180° - ∠A) [From ] ∠BOC = 90° - \(\frac{1}{2}\)∠A ∠BOC = 90° - \(\frac{x}{2}\)

454.

The bisectors of exterior angles at B and C of Δ ABC meet at O, if ∠A= x°, then ∠BOC =A. 90°+ \(\frac{x°}{2}\)B. 90°- \(\frac{x°}{2}\)C. 180°+ \(\frac{x°}{2}\)D. 180°- \(\frac{x°}{2}\)

Answer»

∠OBC = 180° - ∠B - \(\frac{1}{2}\)(180° - ∠B)

∠OBC = 90° - \(\frac{1}{2}\)∠B

And,

∠OCB = 180° - ∠C - \(\frac{1}{2}\)(180° - ∠C)

∠OCB = 90° - \(\frac{1}{2}\)∠C

In ΔBOC

∠BOC + ∠OCB + ∠OBC = 180°

∠BOC + 90° - \(\frac{1}{2}\)∠C + 90° - \(\frac{1}{2}\)∠B = 180°

∠BOC = \(\frac{1}{2}\)(∠B + ∠C)

∠BOC = \(\frac{1}{2}\)(180° - ∠A) [From ]

∠BOC = 90° - \(\frac{1}{2}\)∠A

∠BOC = 90° - \(\frac{x}{2}\)

Discussion

455.	In triangle ABC, AB = 5 cm, BC = 4 cm, CA = 3 cm, which of the following is TRUE? A) ∠A = ∠B = ∠C B) ∠B > ∠A > ∠C C) ∠A < ∠B < ∠C D) ∠B < ∠A < ∠C
Answer» B) ∠B > ∠A > ∠C

Discussion

456.	The angles of a triangle are in the ratio 1 : 2 : 3. Find the angles.
Answer» Given that ratio of the angle = 1: 2: 3 Sum of the terms of the ratio – 1 + 2 + 3 = 6 Sum of the angles of a triangle = 1800 ∴ 1st angle = 1/6 x 180° = 30° (angle sum property) ∴ 2nd angle = 2/6 x 180° = 60° (angle – sum property) ∴ 3rd angle = 3/6 x 180° = 90° (angle – sum property)

456.

The angles of a triangle are in the ratio 1 : 2 : 3. Find the angles.

Answer»

Given that ratio of the angle = 1: 2: 3

Sum of the terms of the ratio – 1 + 2 + 3 = 6

Sum of the angles of a triangle = 1800

∴ 1st angle = 1/6 x 180° = 30° (angle sum property)

∴ 2nd angle = 2/6 x 180° = 60° (angle – sum property)

∴ 3rd angle = 3/6 x 180° = 90° (angle – sum property)

Discussion

457.	Fill in the blanks to make the statements true.If ∆POR and ∆XYZ are congruent under the correspondence QPR ⟷ XYZ, then(i) ∠R = ________(ii) QR = ________(iii) ∠P = ________(iv) OP = ________(v) ∠Q = ________(vi) RP = ________
Answer» If ∆PQR and ∆XYZ are congruent under the correspondence QPR ↔ XYZ, then (i) ∠R = ∠Z (ii) QR = XZ (iii) ∠P = ∠Y (iv) QP = XY (v) ∠Q = ∠X (vi) RP = ZY

457.

Fill in the blanks to make the statements true.If ∆POR and ∆XYZ are congruent under the correspondence QPR ⟷ XYZ, then(i) ∠R = (ii) QR = (iii) ∠P = (iv) OP = (v) ∠Q = (vi) RP =

Answer»

If ∆PQR and ∆XYZ are congruent under the correspondence QPR ↔ XYZ, then

(i) ∠R = ∠Z

(ii) QR = XZ

(iii) ∠P = ∠Y

(iv) QP = XY

(v) ∠Q = ∠X

(vi) RP = ZY

Discussion

458.	Fill in the blanks to make the statements true.In Fig., ∆PQR ≅ ∆ ________
Answer» XZY: In ∆PQR and ∆XZY, PQ = XZ = 3.5 cm ∠PQR = ∠XZY = 45° QR = ZY = 5 cm ∴ ∆PQR ≅ ∆XZY [SAS criterion]

458.

Fill in the blanks to make the statements true.In Fig., ∆PQR ≅ ∆ ________

Answer»

XZY:

In ∆PQR and ∆XZY,

PQ = XZ = 3.5 cm

∠PQR = ∠XZY = 45°

QR = ZY = 5 cm

∴ ∆PQR ≅ ∆XZY [SAS criterion]

Discussion

459.	Fill in the blanks to make the statements true.In the given figure, ∆PQR ≅ ∆ ________
Answer» RSP : In ∆PQR and ∆RSP, QR = SP = 4.1 cm ∠PRQ = ∠RPS = 45° PR = RP [common] ∴ ∆PQR ≅ ∆RSP [SAS criterion]

459.

Fill in the blanks to make the statements true.In the given figure, ∆PQR ≅ ∆ ________

Answer»

RSP :

In ∆PQR and ∆RSP,

QR = SP = 4.1 cm

∠PRQ = ∠RPS = 45°

PR = RP [common]

∴ ∆PQR ≅ ∆RSP [SAS criterion]

Discussion

460.	Fill in the blanks to make the statements true.In the given figure, ∆ ________ ≅ ∆PQR
Answer» DRQ: In ∆PQR and ∆DRQ, ∠PRQ = ∠DQR = 40° ∠PQR = ∠DRQ = 30° + 40° = 70° QR = RQ [common] ∴ ∆PQR ≅ ∆DRO [ASA criterion] ∆DRQ............

460.

Fill in the blanks to make the statements true.In the given figure, ∆ ________ ≅ ∆PQR

Answer»

DRQ:

In ∆PQR and ∆DRQ,

∠PRQ = ∠DQR = 40°

∠PQR = ∠DRQ = 30° + 40° = 70°

QR = RQ [common]

∴ ∆PQR ≅ ∆DRO [ASA criterion]

∆DRQ............

Discussion

461.	Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Answer» For any Q and any R ST. QR>OP and OR>OP `/_OPQ` `angleOPQ=90^0` sum of angles=`angleOQP+angleQOP+90^0=180^0` `angleOQP+angleQOP=90^0` `angleOQP=angleOPQ-angleQOP` `angleOQPThe side opposite to greater angle is longer OP

Discussion

462.	Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Answer» P is a point on straight line l. PR is the segmnent for l drawn such that PQ ⊥ l. Now, in ∆PQR, ∠PQR = 90° (Construction) ∴ ∠QPR + ∠QRP = 90° ∴ ∠QRP < ∠PQR ∴ PQ < PR Any line segment drawn from P to l they form small angle. ∴ PQ ⊥ l is smaller.

462.

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Answer»

P is a point on straight line l.

PR is the segmnent for l drawn such that PQ ⊥ l.

Now, in ∆PQR,

∠PQR = 90° (Construction)

∴ ∠QPR + ∠QRP = 90°

∴ ∠QRP < ∠PQR

∴ PQ < PR

Any line segment drawn from P to l they form small angle.

∴ PQ ⊥ l is smaller.

Discussion

463.	Triangle DEF is right-angled at E, EG is perpendicular to DF and GH is perpendicular to EF. Prove that:1) GH2=EH.HF2) EG2=DG.GF
Answer» Given :- ∠E = 90° . EG ⟂ DF . GH ⟂ EF. To Prove :- (i) GH² = EH * HF (ii) EG² = DG * GF . Solution :- Let us assume that, ∠F = x° . Than, → ∠FEG = (90 - x°) { As, EG ⟂ DF .} Also, in ∆GHF, → ∠HGF = (90 - x°) { As, GH ⟂ EF. } Now, in ∆GHE and ∆FHG, we have, → ∠GHE = ∠FHG (90°) → ∠GEH = ∠FGH (90 - x°) So, → ∆GHE ~ ∆FHG . (By AA similarity) . Therefore, → GH/EH = HF/GH (By CPCT) → *GH² = EH HF (Proved). ________________________________________________________________________ Now, Similarly, → ∠EFD = x° → ∠EDG = (90 - x°) → ∠FEG = (90 - x°) Now, in ∆EGF and ∆DGE, → ∠EGF = ∠DGE. (90°) → ∠FEG = ∠EDG. (90 - x°) So, → ∆EGF ~ ∆DGE .(By AA Similarity.) Therefore, → EG / GF = DG / EG (By CPCT. ) Hence*, → EG² = DG GF. (Proved.)

463.

Triangle DEF is right-angled at E, EG is perpendicular to DF and GH is perpendicular to EF. Prove that:1) GH2=EH.HF2) EG2=DG.GF

Answer»

Given :-

∠E = 90° .
EG ⟂ DF .
GH ⟂ EF.

To Prove :-

(i) GH² = EH * HF
(ii) EG² = DG * GF .

Solution :-

Let us assume that, ∠F = x° .

Than,

→ ∠FEG = (90 - x°) { As, EG ⟂ DF .}

Also, in ∆GHF,

→ ∠HGF = (90 - x°) { As, GH ⟂ EF. }

Now, in ∆GHE and ∆FHG, we have,

→ ∠GHE = ∠FHG (90°)

→ ∠GEH = ∠FGH (90 - x°)

So,

→ ∆GHE ~ ∆FHG . (By AA similarity) .

Therefore,

→ GH/EH = HF/GH (By CPCT)

→ GH² = EH * HF (Proved).

________________________________________________________________________

Now, Similarly,

→ ∠EFD = x°

→ ∠EDG = (90 - x°)

→ ∠FEG = (90 - x°)

Now, in ∆EGF and ∆DGE,

→ ∠EGF = ∠DGE. (90°)

→ ∠FEG = ∠EDG. (90 - x°)

So,

→ ∆EGF ~ ∆DGE .(By AA Similarity.)

Therefore,

→ EG / GF = DG / EG (By CPCT. )

Hence,

→ EG² = DG * GF. (Proved.)

Discussion

464.	From the given figure, what is the additional information needed to prove ΔOBE ≅ ΔOCD ?A) BE = CD B) ΔBOE = ΔCOD C) ΔOBE = ΔOCD D) BD = CE
Answer» B) ΔBOE = ΔCOD

Discussion

465.	□ABCD is a Rhombus prove that the diagonals divide the rhombus into four congruent triangles, (let the point of intersection is O.) For the above problem what is R.T.P. ? A) ΔABC ≅ ΔACD ≅ ΔABD ≅ ΔBCD B) ΔABC ≅ ΔACD C) ΔABD ≅ ΔBCD D) ΔOAB ≅ ΔOBC ≅ ΔOCD ≅ ΔOAD
Answer» D) AOAB ≅ AOBC ≅ AOCD ≅ AOAD

Discussion

466.	ABCD is a trapezium in which AB\|\|DC and its diagonals intersect each other at the point O. Show that `(A O)/(B O)=(C O)/(D O)`.
Answer» `In/_ACD` EO\|\|DC `(AE)/(ED)=(AO)/(OC)`-(1) `In/_BCD` OF\|\|DC `(BO)/(OD)=(BF)/(FC)` `In/_ADB` EO\|\|AB `(EA)/(DE)=(OB)/(OD)` putting this value in equation 1 `(OC)/(OB)((AO)/(OC)=(OB)/(OD))(OC)/(OB)` `(AO)/(OB)=(OC)/(OD)`

Discussion

467.	In the given figure. ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ∆ABC ~ ∆AMP
Answer» Data: ∆ABC and ∆AMP are two right triangles, right angled at B and M respectvely. To Prove: i) ∆ABC ~ ∆AMP (ii) \(\frac{CA}{PA} = \frac{BC}{MP}\) (i) In ∆ABC and ∆AMP, ∠ABC = ∠AMP = 90° (data) ∠CAB = ∠MAP (common) ∴ ∠ACB = ∠MPA ∴ These are equiangular triangles. Similarity criterion for ∆ is A.A.A. ∴ ∆ABC ~ ∆AMP (ii) ∆ABC ~ ∆AMP(Proved) ∴ Corresponding sides are in proportion. LC and LP are corresponding angles. ∴ Adjacent sides are CA, PA. Similarly BC and MP are adjacent sides. ∴ \(\frac{CA}{PA} = \frac{BC}{MP}\)

467.

In the given figure. ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ∆ABC ~ ∆AMP

Answer»

Data: ∆ABC and ∆AMP are two right triangles, right angled at B and M respectvely.

To Prove: i) ∆ABC ~ ∆AMP

(ii) \(\frac{CA}{PA} = \frac{BC}{MP}\)

(i) In ∆ABC and ∆AMP,

∠ABC = ∠AMP = 90° (data)

∠CAB = ∠MAP (common)

∴ ∠ACB = ∠MPA

∴ These are equiangular triangles.

Similarity criterion for ∆ is A.A.A.

∴ ∆ABC ~ ∆AMP

(ii) ∆ABC ~ ∆AMP(Proved)

∴ Corresponding sides are in proportion.

LC and LP are corresponding angles.

∴ Adjacent sides are CA, PA.

Similarly BC and MP are adjacent sides.

∴ \(\frac{CA}{PA} = \frac{BC}{MP}\)

Discussion

468.	In the following figure. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF
Answer» Data: E Is a point on side CB produced of an isosceles ∆ABC with AB = AC. AD ⊥ BC and EF ⊥ AC, To Prove: ∆ABD ~ ∆ECF In ∆ABD and ∆ECF. ∠ADB = ∠EFC = 90° (data) ∠ABD = ∠FCE (∵ ∠B = ∠C) ∠BAD = ∠FEC ∴ Equiangular triangles. ∴ Similarity criterion for triangles is A.A.A. ∴ ∆ABD ~ ∆ECF.

468.

In the following figure. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF

Answer»

Data: E Is a point on side CB produced of an isosceles ∆ABC with AB = AC. AD ⊥ BC and EF ⊥ AC,

To Prove: ∆ABD ~ ∆ECF

In ∆ABD and ∆ECF.

∠ADB = ∠EFC = 90° (data)

∠ABD = ∠FCE (∵ ∠B = ∠C)

∠BAD = ∠FEC

∴ Equiangular triangles.

∴ Similarity criterion for triangles is A.A.A.

∴ ∆ABD ~ ∆ECF.

Discussion

469.	CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively.If ∆ABC ~ ∆EFG, show that (i) \(\frac{CD}{GH} = \frac{AC}{FG}\)(ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF
Answer» Data : CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H 11e on sides AB and FE of ∆ABC and ∆EFG respectively. ∆ABC ~ ∆EFG To Prove: (i) \(\frac{CD}{GH} = \frac{AC}{FG}\) (ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF Proof: ∆ABC ~ ∆EFG (data given) ∴ Their corresponding sides are in proportion. ∴ \(\frac{AB}{EF} = \frac{BC}{FG} = \frac{AC}{EG}\) ∠B = ∠F, ∠A = ∠E, ∠C = ∠G. (i) In ∆ADC and ∆EHG, ∠A = ∠E . ∠ACD = ∠EGH ∴ Their sides are in proportion. ∴ \(\frac{CD}{GH} = \frac{AC}{FG}\) (ii) In ∆DCB and ∆HGE. ∴ \(\frac{CD}{GH} = \frac{AC}{FG}\) ∴ ∆DCB ~ ∆HGE (iii) In ∆DCA and ∆HGE, ∴ \(\frac{DC}{GH} = \frac{AD}{EH} = \frac{AC}{EG}\) ∴ ∆DCA ~ ∆HGE

469.

CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively.If ∆ABC ~ ∆EFG, show that (i) \(\frac{CD}{GH} = \frac{AC}{FG}\)(ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF

Answer»

Data : CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H 11e on sides AB and FE of ∆ABC and ∆EFG respectively.

∆ABC ~ ∆EFG

To Prove:

(i) \(\frac{CD}{GH} = \frac{AC}{FG}\)

(ii) ∆DCB ~ ∆HGE

(iii) ∆DCA ~ ∆HGF

Proof: ∆ABC ~ ∆EFG (data given)

∴ Their corresponding sides are in proportion.

∴ \(\frac{AB}{EF} = \frac{BC}{FG} = \frac{AC}{EG}\)

∠B = ∠F, ∠A = ∠E, ∠C = ∠G.

(i) In ∆ADC and ∆EHG,

∠A = ∠E . ∠ACD = ∠EGH

∴ Their sides are in proportion.

∴ \(\frac{CD}{GH} = \frac{AC}{FG}\)

(ii) In ∆DCB and ∆HGE.

∴ \(\frac{CD}{GH} = \frac{AC}{FG}\)

∴ ∆DCB ~ ∆HGE

(iii) In ∆DCA and ∆HGE,

∴ \(\frac{DC}{GH} = \frac{AD}{EH} = \frac{AC}{EG}\)

∴ ∆DCA ~ ∆HGE

Discussion

470.	Find the height of an equilateral triangle of side 12 cm
Answer» Correct Answer - `6sqrt(3) cm`

Discussion

471.	The height of an equilateral triangle having each side 12 cm, isA. `6sqrt(2) cm`B. `6sqrt(3) cm`C. `3sqrt(6) cm`D. `6sqrt(6) cm`
Answer» Correct Answer - B

Discussion

472.	If AB = QR, BC = PR and CA = PQ, then A) ΔABC ≅ ΔPQR B) ΔCBA ≅ ΔPRQ C) ΔBAC ≅ ΔRPQ D) ΔPQR ≅ ΔBCA
Answer» B) ΔCBA ≅ ΔPRQ

Discussion

473.	Take a triangular piece of paper. Choose three different colors or signs to mark the three comers of the triangle on both sides of the paper. Fold the paper at the midpoints of two sides as observe?
Answer» The three angles of the triangle form a straight angle. ∴ m∠A + m∠B + m∠C = 180° Hence, the sum of the measures of the angles of a triangle is 180°.

473.

Take a triangular piece of paper. Choose three different colors or signs to mark the three comers of the triangle on both sides of the paper. Fold the paper at the midpoints of two sides as observe?

Answer»

The three angles of the triangle form a straight angle.

∴ m∠A + m∠B + m∠C = 180°

Hence, the sum of the measures of the angles of a triangle is 180°.

Discussion

474.	If a, b and c are the lengths of sides of a triangle, which of the following is TRUE? A) a + b>c, a - b>c B) a + b < c, a - bc, a-b < cC) a + b < c, a – b > cD) ∠B < ∠A < ∠C
Answer» D) ∠B > ∠A > ∠C

Discussion

475.	The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.A. 10 cm, 15 cmB. 15 cm, 20 cmC. 12 cm, 17 cmD. 13 cm, 18 cm
Answer» Correct Answer - B

Discussion

476.	Take a triangular piece of paper and make three different types of marks near the three angles. Take a point approximately at the center of the triangle. From this point, draw three lines that meet the three sides. Cut the paper along those lines. Place the three angles side by side as shown. See how the three angles of a triangle together form a straight angle, or, an angle that measures 180°.
Answer» The three angles of the triangle form a straight angle. Hence, the sum of the measures of the angles of a triangle is 180°.

476.

Take a triangular piece of paper and make three different types of marks near the three angles. Take a point approximately at the center of the triangle. From this point, draw three lines that meet the three sides. Cut the paper along those lines. Place the three angles side by side as shown. See how the three angles of a triangle together form a straight angle, or, an angle that measures 180°.

Answer»

The three angles of the triangle form a straight angle.

Hence, the sum of the measures of the angles of a triangle is 180°.

Discussion

477.	In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.
Answer» i. ∠NET = 100° and ∠EMR = 140° ∠EMN + ∠EMR = 180° ∴ z +140° =180° ∴ z = 180° -140° ∴ z = 40° ii. Also, ∠NET + ∠NEM = 180° [Angles in a linear pair] ∴ 100° + y = 180° ∴ y = 180° – 100° ∴ y = 80° iii. In ∆ENM, ∴ ∠ENM + ∠NEM + ∠EMN = 180° [Sum of the measures of the angles of a triangle is 180°] ∴ x +80°+ 40°= 180° ∴ x = 180° – 80° – 40° ∴ x = 60° ∴ x = 60°, = 80°, z = 40°

477.

In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.

Answer»

i. ∠NET = 100° and ∠EMR = 140°

∠EMN + ∠EMR = 180°

∴ z +140° =180°

∴ z = 180° -140°

∴ z = 40°

ii. Also, ∠NET + ∠NEM = 180° [Angles in a linear pair]

∴ 100° + y = 180°

∴ y = 180° – 100°

∴ y = 80°

iii. In ∆ENM,

∴ ∠ENM + ∠NEM + ∠EMN = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ x +80°+ 40°= 180°

∴ x = 180° – 80° – 40°

∴ x = 60°

∴ x = 60°, = 80°, z = 40°

Discussion

478.	In the adjoining figure, line AB \|\| line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.
Answer» i. ∠BAD = 70°, ∠DER = 40° [Given] line AB \|\| line DE and seg AD is their transversal. ∴ ∠EDA = ∠BAD [Alternate Angles] ∴ ∠EDA = 70° ….(i) In ∆DRE, ∠EDR + ∠DER + ∠DRE = 180° [Sum of the measures of the angles of a triangle is 180°] ∴ 70°+ 40° +∠DRE = 180° [From (i) and D – R – A] ∴ ∠DRE = 180° -70° -40° ∴ ∠DRE = 70° ii. ∠DRE + ∠ARE = 180° [Angles in a linear pair] ∴ 70° + ∠ARE = 180° ∴ ∠ARE = 180°-70° ∴ ∠ARE =110° ∴ ∠DRE = 70°, ∠ARE = 110°

478.

In the adjoining figure, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.

Answer»

i. ∠BAD = 70°, ∠DER = 40° [Given]

line AB || line DE and seg AD is their transversal.

∴ ∠EDA = ∠BAD [Alternate Angles]

∴ ∠EDA = 70° ….(i)

In ∆DRE,

∠EDR + ∠DER + ∠DRE = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 70°+ 40° +∠DRE = 180° [From (i) and D – R – A]

∴ ∠DRE = 180° -70° -40°

∴ ∠DRE = 70°

ii. ∠DRE + ∠ARE = 180° [Angles in a linear pair]

∴ 70° + ∠ARE = 180°

∴ ∠ARE = 180°-70°

∴ ∠ARE =110°

∴ ∠DRE = 70°, ∠ARE = 110°

Discussion

479.	In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.
Answer» i. ∠NET = 100° and ∠EMR = 140° ∠EMN + ∠EMR = 180° ∴ z +140° = 180° ∴ z = 180° -140° ∴ z = 40° ii. Also, ∠NET + ∠NEM = 180° [Angles in a linear pair] ∴ 100° + y = 180° ∴ y = 180° – 100° ∴ y = 80° iii. In ∆ENM, ∴ ∠ENM + ∠NEM + ∠EMN = 180° [Sum of the measures of the angles of a triangle is 180°] ∴ x + 80°+ 40°= 180° ∴ x = 180° – 80° – 40° ∴ x = 60° ∴ x = 60°, y = 80°, z = 40°

479.

In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.

Answer»

i. ∠NET = 100° and ∠EMR = 140°

∠EMN + ∠EMR = 180°

∴ z +140° = 180°

∴ z = 180° -140°

∴ z = 40°

ii. Also, ∠NET + ∠NEM = 180° [Angles in a linear pair]

∴ 100° + y = 180°

∴ y = 180° – 100°

∴ y = 80°

iii. In ∆ENM,

∴ ∠ENM + ∠NEM + ∠EMN = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ x + 80°+ 40°= 180°

∴ x = 180° – 80° – 40°

∴ x = 60°

∴ x = 60°, y = 80°, z = 40°

Discussion

480.	Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.
Answer» i. ∠c + 100° = 180° [Angles in a linear pair] ∴ ∠c = 180° – 100° ∴ ∠c = 80° ii. ∠b = 70° [Vertically opposite angles] iii. ∠a + ∠b +∠c = 180° [Sum of the measures of the angles of a triangle is 180°] ∠a + 70° + 80° = 1800 ∴ ∠a = 180° – 70° – 80° ∴ ∠a = 30° ∴ ∠a = 30°, ∠b = 70°,∠ c = 80°

480.

Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.

Answer»

i. ∠c + 100° = 180° [Angles in a linear pair]

∴ ∠c = 180° – 100°

∴ ∠c = 80°

ii. ∠b = 70° [Vertically opposite angles]

iii. ∠a + ∠b +∠c = 180° [Sum of the measures of the angles of a triangle is 180°]

∠a + 70° + 80° = 1800

∴ ∠a = 180° – 70° – 80°

∴ ∠a = 30°

∴ ∠a = 30°, ∠b = 70°,∠ c = 80°

Discussion

481.	In the adjoining figure, line AB \|\| line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.
Answer» i. ∠B AD = 70°, ∠DER = 40° [Given] line AB \|\| line DE and seg AD is their transversal. ∴ ∠EDA = ∠BAD [Alternate Angles] ∴ ∠EDA = 70° ….(i) In ∆DRE, ∠EDR + ∠DER + ∠DRE = 180° [Sum of the measures of the angles of a triangle is 180°] ∴ 70°+ 40° +∠DRE = 180° [From (i) and D – R – A] ∴ ∠DRE = 180° - 70° - 40° ∴ ∠DRE = 70° ii. ∠DRE + ∠ARE = 180° [Angles in a linear pair] ∴ 70° + ∠ARE = 180° ∴ ∠ARE = 180°- 70° ∴ ∠ARE =110° ∴ ∠DRE = 70°, ∠ARE = 110°

481.

In the adjoining figure, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.

Answer»

i. ∠B AD = 70°, ∠DER = 40° [Given]

line AB || line DE and seg AD is their transversal.

∴ ∠EDA = ∠BAD [Alternate Angles]

∴ ∠EDA = 70° ….(i)

In ∆DRE,

∠EDR + ∠DER + ∠DRE = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 70°+ 40° +∠DRE = 180° [From (i) and D – R – A]

∴ ∠DRE = 180° - 70° - 40°

∴ ∠DRE = 70°

ii. ∠DRE + ∠ARE = 180° [Angles in a linear pair]

∴ 70° + ∠ARE = 180°

∴ ∠ARE = 180°- 70°

∴ ∠ARE =110°

∴ ∠DRE = 70°, ∠ARE = 110°

Discussion

482.	Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.
Answer» i. ∠c + 100° = 180° [Angles in a linear pair] ∴ ∠c = 180° – 100° ∴ ∠c = 80° ii. ∠b = 70° [Vertically opposite angles] iii. ∠a + ∠b +∠c = 180° [Sum of the measures of the angles of a triangle is 180°] ∠a + 70° + 80° = 180° ∴ ∠a = 180° – 70° – 80° ∴ ∠a = 30° ∴ ∠a = 30°, ∠b = 70°, ∠c = 80°

482.

Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.

Answer»

i. ∠c + 100° = 180° [Angles in a linear pair]

∴ ∠c = 180° – 100°

∴ ∠c = 80°

ii. ∠b = 70° [Vertically opposite angles]

iii. ∠a + ∠b +∠c = 180° [Sum of the measures of the angles of a triangle is 180°]

∠a + 70° + 80° = 180°

∴ ∠a = 180° – 70° – 80°

∴ ∠a = 30°

∴ ∠a = 30°, ∠b = 70°, ∠c = 80°

Discussion

483.	In the adjoining figure, line AB \|\| line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.
Answer» Given: line AB \|\| line CD and line PQ is the transversal. ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively. To prove: m ∠PTQ = 90° Proof: ∠TPB = (1/2) ∠TPQ = ∠BPQ …(i) [Ray PT bisects ∠BPQ] ∠TQD = ∠TQP = (1/2) ∠PQD ….(ii) [Ray QT bisects ∠PQD] line AB \|\| line CD and line PQ is their transversal. [Given] ∴∠BPQ + ∠PQD = 180° [Interior angles] ∴ (1/2) (∠BPQ) + (1/2) (∠PQD) = (1/2) x 180° [Multiplying both sides by (1/2)] ∠TPQ + ∠TQP = 90° In ∆PTQ, ∠TPQ + ∠TQP + ∠PTQ = 180° [Sum of the measures of the angles of a triangle is 180°] ∴ 90° + ∠PTQ = 180° [From (iii)] ∴ ∠PTQ = 180° – 90° = 90° ∴ m ∠PTQ = 90°

483.

In the adjoining figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.

Answer»

Given: line AB || line CD and line PQ is the transversal.

ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively.

To prove: m ∠PTQ = 90°

Proof: ∠TPB = (1/2) ∠TPQ = ∠BPQ …(i) [Ray PT bisects ∠BPQ]

∠TQD = ∠TQP = (1/2) ∠PQD ….(ii)

[Ray QT bisects ∠PQD]

line AB || line CD and line PQ is their transversal. [Given]

∴∠BPQ + ∠PQD = 180° [Interior angles]

∴ (1/2) (∠BPQ) + (1/2) (∠PQD) = (1/2) x 180°

[Multiplying both sides by (1/2)]

∠TPQ + ∠TQP = 90°

In ∆PTQ,

∠TPQ + ∠TQP + ∠PTQ = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 90° + ∠PTQ = 180° [From (iii)]

∴ ∠PTQ = 180° – 90° = 90°

∴ m ∠PTQ = 90°

Discussion

484.	In the adjoining figure, line AB \|\| line CD and line PQ is the transversal.Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.Given: line AB \|\| line CD and line PQ is the transversal. ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively. To prove: m∠PTQ = 90°
Answer» ∠TPB = ∠TPQ = 1/2∠BPQ …(i) [Ray PT bisects ∠BPQ] ∠TQD = ∠TQP = 1/2∠PQD ….(ii) [Ray QT bisects ∠PQD] line AB \|\| line CD and line PQ is their transversal. [Given] ∴∠BPQ + ∠PQD = 180° [Interior angles] ∴ \(\frac{1}2\)(∠BPQ) + \(\frac{1}2\)(∠PQD) =\(\frac{1}2\) x 180° [Multiplying both sides by \(\frac{1}2\)] ∠TPQ + ∠TQP = 90° In ∆PTQ, ∠TPQ + ∠TQP + ∠PTQ = 180° [Sum of the measures of the angles of a triangle is 180°] ∴ 90° + ∠PTQ = 180° [From (iii)] ∴ ∠PTQ = 180° – 90° = 90° ∴ m∠PTQ = 90°

484.

In the adjoining figure, line AB || line CD and line PQ is the transversal.Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.Given: line AB || line CD and line PQ is the transversal. ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively. To prove: m∠PTQ = 90°

Answer»

∠TPB = ∠TPQ = 1/2∠BPQ …(i) [Ray PT bisects ∠BPQ]

∠TQD = ∠TQP = 1/2∠PQD ….(ii) [Ray QT bisects ∠PQD]

line AB || line CD and line PQ is their transversal.

[Given]

∴∠BPQ + ∠PQD = 180° [Interior angles]

∴ \(\frac{1}2\)(∠BPQ) + \(\frac{1}2\)(∠PQD) =\(\frac{1}2\) x 180° [Multiplying both sides by \(\frac{1}2\)]

∠TPQ + ∠TQP = 90°

In ∆PTQ,

∠TPQ + ∠TQP + ∠PTQ = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ 90° + ∠PTQ = 180° [From (iii)]

∴ ∠PTQ = 180° – 90°

= 90°

∴ m∠PTQ = 90°

Discussion

485.	In the given figure, if ar (ΔPOB) = 32 cm2, then ar(ΔQOA) is(A) 40 cm2(B) 50 cm2(C) 75 cm2(D) 150 cm2
Answer» Correct option is: (B) 50 cm²

Discussion

486.	In figure, ΔABC ~ ΔDEF. AB = 5 cm, area (ΔABC) = 20 cm2, area (ΔDEF) = 45 cm2 then DE =(A) 2.0 cm (B) 4.5 cm (C) 7.5 cm (D) 9.5 cm
Answer» Correct option is: (C) 7.5 cm

Discussion

487.	In ∆ ABC,(a) AB + BC > AC(b) AB + BC < AC(c) AB + AC < BC(d) AC + BC < AB
Answer» (a) AB + BC > AC The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

487.

In ∆ ABC,(a) AB + BC > AC(b) AB + BC < AC(c) AB + AC < BC(d) AC + BC < AB

Answer»

(a) AB + BC > AC

The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

Discussion

488.	In ∆PQR, PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.
Answer» In ∆PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm [Given] Since, 12 > 10 > 8 ∴ QR > PQ > PR ∴ ∠QPR > ∠PRQ > ∠PQR [Angle opposite to greater side is greater] ∴ In ∆PQR, ∠QPR is the greatest angle and ∠PQR is the smallest angle.

488.

In ∆PQR, PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.

Answer»

In ∆PQR,

PQ = 10 cm, QR = 12 cm, PR = 8 cm [Given]

Since, 12 > 10 > 8

∴ QR > PQ > PR

∴ ∠QPR > ∠PRQ > ∠PQR

[Angle opposite to greater side is greater]

∴ In ∆PQR, ∠QPR is the greatest angle and ∠PQR is the smallest angle.

Discussion

489.	(i) In fig, if AB IICD , find the value of x.
Answer» Diagonal of trapezium divide each other proportiona AO/OC=BO/OD 4/4X-2=x+1/2x+4 4x²- 2x+4x-2=8x+16 4x²+2x-2-8x-16=0 4x²-6x-18=0 2(2x²- 3x-9)=0 2x²- 3x-9=0 2x²-6x+3x-9=0 2x(x-3)+3(x-3)=0 (x-3)(2x+3)=0 x-3=0 x=3 or,2x+3=0 2x=-3 x= -3/2 x=-3/2 is not possible So x=3

489.

(i) In fig, if AB IICD , find the value of x.

Answer»

Diagonal of trapezium divide each other proportiona

AO/OC=BO/OD

4/4X-2=x+1/2x+4

4x²- 2x+4x-2=8x+16

4x²+2x-2-8x-16=0

4x²-6x-18=0

2(2x²- 3x-9)=0

2x²- 3x-9=0

2x²-6x+3x-9=0

2x(x-3)+3(x-3)=0

(x-3)(2x+3)=0

x-3=0

x=3

or,2x+3=0

2x=-3

x= -3/2

x=-3/2 is not possible

So x=3

Discussion

490.	D, E and F are the points on sides BC, CA and AB respectively of Δ ABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C . If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD.
Answer» in⊿ ABC CF bisects ∠A ∴ AF/FB=AE/AC AF/5-AF=4/8 2AF=5-AF 2AF+AF=5 3AF=5 AF=5/3 cm ⊿ABC, BE bisects ∠B ∴ AE/AC=AB/BC 4-CE/CE=5/8 5CE=32-8CE 5CE+8CE=32 13CE=32 CE=32/13 cm Similarly BD/DC=AB/AC BD/8-BD=5/4 4BD=40-5BD 4BD+5BD=40 9BD=40 BD=40/9 cm

490.

D, E and F are the points on sides BC, CA and AB respectively of Δ ABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C . If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD.

Answer»

in⊿ ABC

CF bisects ∠A

∴ AF/FB=AE/AC

AF/5-AF=4/8

2AF=5-AF

2AF+AF=5

3AF=5

AF=5/3 cm

⊿ABC, BE bisects ∠B

∴ AE/AC=AB/BC

4-CE/CE=5/8

5CE=32-8CE

5CE+8CE=32

13CE=32

CE=32/13 cm

Similarly

BD/DC=AB/AC

BD/8-BD=5/4

4BD=40-5BD

4BD+5BD=40

9BD=40

BD=40/9 cm

Discussion

491.	In.fig AD bisects ∠A , AB = 12 cm, AC = 20 cm and BD = 5 cm, determine CD
Answer» AD bisects ∠A ∴ AB/AC=BD/CD 12/20=5/CD CD =100/12 CD=8.33 cm

491.

In.fig AD bisects ∠A , AB = 12 cm, AC = 20 cm and BD = 5 cm, determine CD

Answer»

AD bisects ∠A

∴ AB/AC=BD/CD

12/20=5/CD

CD =100/12

CD=8.33 cm

Discussion

492.	In Fig, check whether AD is the bisector of∠A of Δ ABC in each of the following: (i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm (ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm (iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm (iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm (v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm
Answer» (i) BD/DC=AB/AC 1.5/3.5=5/10 15/3510/10=1/2 3/7=1/2 Not bisects* (ii) 1.6/2.4=4/6 16/24=2/3 2/3=2/3 bisects (iii) BD/CD=AB/AC BD/BC-BD=AB/AC BD/24-6=8/24 6/18=1/3 1/3=1/3 bisects (iv) 1.5/2= 6/8 3/4=3/4 bisects (v) BD/CD=AB/AC BD/BC-BD=AB/AC BD/9-2.5=5/12 2.5/6.5=5/12 5/13=5/12 Not bisects

492.

In Fig, check whether AD is the bisector of∠A of Δ ABC in each of the following: (i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm (ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm (iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm (iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm (v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm

Answer»

(i) BD/DC=AB/AC

1.5/3.5=5/10

15/35*10/10=1/2

3/7=1/2

Not bisects

(ii) 1.6/2.4=4/6

16/24=2/3

2/3=2/3

bisects

(iii) BD/CD=AB/AC

BD/BC-BD=AB/AC

BD/24-6=8/24

6/18=1/3

1/3=1/3

bisects

(iv) 1.5/2= 6/8

3/4=3/4

bisects

(v) BD/CD=AB/AC

BD/BC-BD=AB/AC

BD/9-2.5=5/12

2.5/6.5=5/12

5/13=5/12

Not bisects

Discussion

493.	If ∆PQR is congruent to ∆STU (Fig.), then what is the length of TU?(a) 5 cm (b) 6 cm (c) 7 cm (d) cannot be determined
Answer» (b) 6 cm From the question it is given that, Δ PQR ≅ Δ STU So, PQR ↔ STU ∴ QR = TU TU = 6 cm

493.

If ∆PQR is congruent to ∆STU (Fig.), then what is the length of TU?(a) 5 cm (b) 6 cm (c) 7 cm (d) cannot be determined

Answer»

(b) 6 cm

From the question it is given that,

Δ PQR ≅ Δ STU

So,

PQR ↔ STU

∴ QR = TU

TU = 6 cm

Discussion

494.	D and E are points on the sides AB and AC respectively of a ∆ABC such that DE \|\| BC. Find the value of x, when AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm.
Answer» From figure, D and E are the points on the sides AB and AC respectively and DE \|\| BC then AD/DB = AE/EC AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm. AD/DB = AE/EC 4/(x-4) = 8/(3x-19) 4(3x-19) = 8(x-4) Solving, we get x = 11 cm

494.

D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC. Find the value of x, when AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm.

Answer»

From figure, D and E are the points on the sides AB and AC respectively and DE || BC

then AD/DB = AE/EC

AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm.

AD/DB = AE/EC

4/(x-4) = 8/(3x-19)

4(3x-19) = 8(x-4)

Solving, we get x = 11 cm

Discussion

495.	D and E are points on the sides AB and AC respectively of a ∆ABC such that DE \|\| BC. Find the value of x, when AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm.
Answer» From figure, D and E are the points on the sides AB and AC respectively and DE \|\| BC then AD/DB = AE/EC AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm. x/(x-2) = (x+2)/(x-1) x(x-1) = (x+2)(x-2) Solving above equation, we get x = 4 cm

495.

D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC. Find the value of x, when AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm.

Answer»

From figure, D and E are the points on the sides AB and AC respectively and DE || BC

then AD/DB = AE/EC

AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm.

x/(x-2) = (x+2)/(x-1)

x(x-1) = (x+2)(x-2)

Solving above equation, we get

x = 4 cm

Discussion

496.	If ∆ABC and ∆DBC are on the same base BC, AB = DC and AC = DB (Fig.), then which of the following gives a congruence relationship?(a) ∆ ABC ≅ ∆ DBC (b) ∆ ABC ≅ ∆CBD(c) ∆ ABC ≅ ∆ DCB (d) ∆ ABC ≅ ∆BCD
Answer» (c) ∆ ABC ≅ ∆ DCB Consider the ΔABC and ΔDCB, From the question it is given that, AB = DC and AC = DB BC = BC … [because common side] Therefore, ∆ ABC ≅ ∆ DCB

496.

If ∆ABC and ∆DBC are on the same base BC, AB = DC and AC = DB (Fig.), then which of the following gives a congruence relationship?(a) ∆ ABC ≅ ∆ DBC (b) ∆ ABC ≅ ∆CBD(c) ∆ ABC ≅ ∆ DCB (d) ∆ ABC ≅ ∆BCD

Answer»

(c) ∆ ABC ≅ ∆ DCB

Consider the ΔABC and ΔDCB,

From the question it is given that, AB = DC and AC = DB

BC = BC … [because common side]

Therefore, ∆ ABC ≅ ∆ DCB

Discussion

497.	Two triangles DEF and GHK are sucht that `angle D=48^(@) and angle H=57^(@)`. If `Delta DEF~Delta GHK` then find the measure of `angleF`.
Answer» Correct Answer - `75^(@)` `angle D=48^(@), angle E= angle H=57^(@) [ :. Delta DEF~Delta GHK]` `:. angleF=180^(@)-(angleD+angleE)=180^(@)-(48^(@)+57^(@))=75^(@)`

Discussion

498.	`Delta ABC~ DeltaDEF` such that `ar (Delta ABC)= 36 cm^(2) and ar (Delta DEF) 49 cm^(2)` . Then, the ratio of their corresponding sides isA. `36:49`B. `6:7`C. `7:6`D. `sqrt(6):sqrt(7)`
Answer» Correct Answer - B `((AB)/(DE))^(2)=(ar(Delta ABC))/(ar (Delta DEF))=(36)/(49)=((6)/(7))^(2) rArr (AB)/(DE)=(6)/(7)` `:.` the ratio of the corresponding sides is `6:7`

Discussion

499.	State the AA-similarity criterion.
Answer» If two angles are correspondingly equal to the two angles of another triangle, then the two triangles are similar.

Discussion

500.	State the AA-similarity criterion.
Answer» In two triangles, if two angles of the one triangle are equal to the corresponding angles of the other triangle, then the triangles are similar.

Discussion

Explore topic-wise InterviewSolutions in .

State whether the statement are True or False.It is possible to have a triangle in which each angle is equal to 60°

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Fill in the blanks to make the statements true.In Fig., ∆PQR ≅ ∆ ________

Fill in the blanks to make the statements true.In the given figure, ∆PQR ≅ ∆ ________

Fill in the blanks to make the statements true.In the given figure, ∆ ________ ≅ ∆PQR

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Triangle DEF is right-angled at E, EG is perpendicular to DF and GH is perpendicular to EF. Prove that:1) GH2=EH.HF2) EG2=DG.GF

From the given figure, what is the additional information needed to prove ΔOBE ≅ ΔOCD ?A) BE = CD B) ΔBOE = ΔCOD C) ΔOBE = ΔOCD D) BD = CE

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ABCD is a trapezium in which AB||DC and its diagonals intersect each other at the point O. Show that `(A O)/(B O)=(C O)/(D O)`.

In the given figure. ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ∆ABC ~ ∆AMP

In the following figure. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF

CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively.If ∆ABC ~ ∆EFG, show that (i) \(\frac{CD}{GH} = \frac{AC}{FG}\)(ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF

Find the height of an equilateral triangle of side 12 cm

The height of an equilateral triangle having each side 12 cm, isA. `6sqrt(2) cm`B. `6sqrt(3) cm`C. `3sqrt(6) cm`D. `6sqrt(6) cm`

If AB = QR, BC = PR and CA = PQ, then A) ΔABC ≅ ΔPQR B) ΔCBA ≅ ΔPRQ C) ΔBAC ≅ ΔRPQ D) ΔPQR ≅ ΔBCA

Take a triangular piece of paper. Choose three different colors or signs to mark the three comers of the triangle on both sides of the paper. Fold the paper at the midpoints of two sides as observe?

If a, b and c are the lengths of sides of a triangle, which of the following is TRUE? A) a + b&gt;c, a - b&gt;c B) a + b &lt; c, a - bc, a-b &lt; cC) a + b &lt; c, a – b &gt; cD) ∠B &lt; ∠A &lt; ∠C

The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.A. 10 cm, 15 cmB. 15 cm, 20 cmC. 12 cm, 17 cmD. 13 cm, 18 cm

In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.

In the adjoining figure, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.

In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.

Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.

In the adjoining figure, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.

Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.

In the adjoining figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.

In the given figure, if ar (ΔPOB) = 32 cm2, then ar(ΔQOA) is(A) 40 cm2(B) 50 cm2(C) 75 cm2(D) 150 cm2

In figure, ΔABC ~ ΔDEF. AB = 5 cm, area (ΔABC) = 20 cm2, area (ΔDEF) = 45 cm2 then DE =(A) 2.0 cm (B) 4.5 cm (C) 7.5 cm (D) 9.5 cm

In ∆ ABC,(a) AB + BC &gt; AC(b) AB + BC &lt; AC(c) AB + AC &lt; BC(d) AC + BC &lt; AB

In ∆PQR, PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.

(i) In fig, if AB IICD , find the value of x.

D, E and F are the points on sides BC, CA and AB respectively of Δ ABC such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C . If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD.

In.fig AD bisects ∠A , AB = 12 cm, AC = 20 cm and BD = 5 cm, determine CD

If ∆PQR is congruent to ∆STU (Fig.), then what is the length of TU?(a) 5 cm (b) 6 cm (c) 7 cm (d) cannot be determined

D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC. Find the value of x, when AD = 4 cm, DB = (x – 4) cm, AE = 8 cm and EC = (3x – 19) cm.

D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC. Find the value of x, when AD = x cm, DB = (x – 2) cm, AE = (x + 2) cm and EC = (x – 1) cm.

If ∆ABC and ∆DBC are on the same base BC, AB = DC and AC = DB (Fig.), then which of the following gives a congruence relationship?(a) ∆ ABC ≅ ∆ DBC (b) ∆ ABC ≅ ∆CBD(c) ∆ ABC ≅ ∆ DCB (d) ∆ ABC ≅ ∆BCD

Two triangles DEF and GHK are sucht that `angle D=48^(@) and angle H=57^(@)`. If `Delta DEF~Delta GHK` then find the measure of `angleF`.

`Delta ABC~ DeltaDEF` such that `ar (Delta ABC)= 36 cm^(2) and ar (Delta DEF) 49 cm^(2)` . Then, the ratio of their corresponding sides isA. `36:49`B. `6:7`C. `7:6`D. `sqrt(6):sqrt(7)`

State the AA-similarity criterion.

State the AA-similarity criterion.

In triangle ABC, AB = 5 cm, BC = 4 cm, CA = 3 cm, which of the following is TRUE? A) ∠A = ∠B = ∠C B) ∠B > ∠A > ∠C C) ∠A < ∠B < ∠C D) ∠B < ∠A < ∠C

Fill in the blanks to make the statements true.If ∆POR and ∆XYZ are congruent under the correspondence QPR ⟷ XYZ, then(i) ∠R = (ii) QR = (iii) ∠P = (iv) OP = (v) ∠Q = (vi) RP =

If a, b and c are the lengths of sides of a triangle, which of the following is TRUE? A) a + b>c, a - b>c B) a + b < c, a - bc, a-b < cC) a + b < c, a – b > cD) ∠B < ∠A < ∠C

In ∆ ABC,(a) AB + BC > AC(b) AB + BC < AC(c) AB + AC < BC(d) AC + BC < AB