RD Sharma 2014 Solutions for Class 7 Math Chapter 15 Properties Of Triangles are provided here with simple step-by-step explanations. These solutions for Properties Of Triangles are extremely popular among class 7 students for Math Properties Of Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2014 Book of class 7 Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2014 Solutions. All RD Sharma 2014 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Page No 15.12:

Question 1:

Two angles of a triangle are of measures 105° and 30°. Find the measure of the third angle.

Answer:

Let the third angle be x.Sum of all the three angles of a triangle =180° 105° + 30° + x = 180°or,  x = 180° - 135° x=45°The third angle is 45°.

Page No 15.12:

Question 2:

One of the angles of a triangle is 130°, and the other two angles are equal. What is the measure of each of these equal angles?

Answer:

Let the other two angles be x.Sum of all the three angles of a triangle =180°i.e. 130° + x + x = 180° 2x = 180° - 130° 2x = 50° x = 50°2x = 25°Therefore, the other two angles are 25° each.

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Question 3:

The three angles of a triangle are equal to one another. What is the measure of each of the angles?

Answer:

Let each angle of the triangle be x.Sum of all the three angles of a triangle =180°i.e. x+x+x=180° 3x=180° x=180°3x=60°Therefore, each angle is equal to 60°.

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Question 4:

If the angles of a triangle are in the ratio 1 : 2 : 3, determine three angles.

Answer:

If the angles of a triangle are in ratio 1:2:3, then let us take the first angle to be x.Which means that the second angle will be 2x and the third angle will be 3x.Sum of all the three angles of a triangle = 180°  x + 2x + 3x = 180° 6x = 180° x = 180°6 x = 30° 2x = 2 × 30° = 60°3x = 3 × 30° = 90°Therefore, the first angle is equal to 30°, the second angle is equal to 60°, and the third angle is equal to 90°.

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Question 5:

The angles of a triangle are (x − 40)°, (x − 20)° and 12x-10°. Find the value of x.

Answer:

Sum of all the three angles of a triangle = 180°(x - 40)° + (x - 20)° + (x2 - 10)° = 180°x + x + x2 - 40° - 20° - 10° = 180°x + x + x2 - 70° = 180°5x2 = 180° + 70°5x2 = 250°x = 25 × 250°x = 100°Hence, we can conclude that x is equal to 100°.

Page No 15.12:

Question 6:

The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°, find the three angles.

Answer:

Let the first angle of the triangle be x.Therefore, we can say that the second angle of the triangle will be (x + 10°) and the third angle of the triangle will be (x + 10° + 10°).We know that the sum of all the three angles of a triangle is equal to 180°. x + (x + 10°) + (x + 10° + 10°) = 180°x + x + x + 10° + 10° + 10° = 180°3x+30°=180°3x=180°-30°3x=150°x =150°3x = 50°Now, (x + 10°) = 50° + 10°(x + 10°) = 60°And, (x + 10° + 10°) = 50° + 10° + 10°(x + 10° + 10°) = 70°Hence, we can say that the three angles of the triangle are 50°, 60°and 70°.

Page No 15.12:

Question 7:

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.

Answer:

Let the two equal angles of the triangle be x.Hence, the third angle of the triangle will be (x+30°).Sum of all the three angle of a triangle=180° x+x+(x+30°)=180° x+x+x+30°=180°3x+30°=180°3x=180°-30°3x=150°x=150°3x=50°(x+30°)=50°+30°(x+30°)=80°Hence, we can conclude that the angles of the triangle are 50°, 50° and 80°.

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Question 8:

If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.

Answer:

Let the three angles of the triangle be a, b and c.Given: a=b+c Also, the sum of all the three angle of a triangle = 180°Or, a+b+c=180°a+a=180°   (a=b+c)2a=180°a=180°2a=90°Hence, we can conclude that the given triangle is a right angle triangle.

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Question 9:

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

Answer:

Let the three angles of the triangle be a, b and c.We know: a<b+c                 .....(i)   (Given)Which means: b<a+cOr, c<a+bWe also know that the sum of all the angles of a triangle is equal to 180°.Which means: a+b+c=180°Or, b+c=180°-aPutting the value of b+c in equation (i):a<180°-a2a<180°a<90°Similarly:b<90°c<90°Hence, we can conclude that the given triangle is an acute triangle.

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Question 10:

In each of the following, the measures of three angles are given. State in which cases, the angles can possibly be those of a triangle:
(i) 63°, 37°, 80°
(ii) 45°, 61°, 73°
(iii) 59°, 72°, 61°
(iv) 45°, 45°, 90°
(v) 30°, 20°, 125°

Answer:

(i) We know that the sum of all the three angles of a triangle is equal to 180°.Now, let us find the sum of 63°, 37° and 80°.63°+37°+80°=180°The sum of 63°, 37° and 80° is equal to 180°.Hence, we can say that the given angles can be those of a triangle.
(ii) We know that the sum of all the three angles of a triangle is equal to 180°.Now, let us find the sum of 45°, 61° and 73°.45°+61°+73°=179° The sum of 45°, 61° and 73° is not equal to 180°.Hence, we can say that the given angles cannot be those of a triangle.
(iii) We know that the sum of all the three angles of a triangle is equal to 180°.Now, let us find the sum of 59°, 72° and 61°.59°+72°+61°=192° The sum of 59°, 72° and 61° is not equal to 180°.Hence, we can say that the given angles cannot be those of a triangle.
(iv) We know that the sum of all the three angles of a triangle is equal to 180°.Now, let us find the sum of 45°, 45° and 90°.45°+45°+90°=180°The sum of 45°, 45° and 90° is equal to 180°.Hence, we can say that the given angles can be those of a triangle.
(v) We know the sum of all the three angles of a traingle is equal to 180°.Now, let us find the sum of 30°, 20° and 125°.30°+20°+125°=175° The sum of 30°, 20° and 125° is not equal to 180°.Hence, we can say that the given angles cannot be those of a triangle.
Therefore, we can conclude that in (i) and (iv), the angles can be those of a triangle.

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Question 11:

The angles of a triangle are in the ratio 3 : 4 : 5. Find the smallest angle.

Answer:

If the angles of the given triangle are in the ratio 3:4:5, then let us take the smallest angle as 3x.This means that the second angle will be 4x and the third angle will be 5x.We know that the sum of all the three angles of a triangle is equal to 180°. 3x+4x+5x=180°12x=180° x=180°12x=15°Now, 3x=3×15°=45°Therefore, we can conclude that the smallest angle is 45°.

Page No 15.12:

Question 12:

Two acute angles of a right triangle are equal. Find the two angles.

Answer:

Let each of the two acute angles of the given triangle be x.We know that the third angle is 90°.  (Given)We also know that the sum of all the three angles of a triangle is equal to 180°.Which means: x+x+90°=180°2x=180°-90°x=90°2x=45°Hence, we can conclude that each of the two acute angles is equal to 45°.

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Question 13:

One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?

Answer:

Let the three angles of the given triangle be a, b and c.We know: a>b+c                  .....(i)   (Given)We also know that the sum of all the angles of a triangle is equal to 180°.a+b+c=180° b+c=180°-aPutting the value of b+c from equation (i):a>180°-a2a>180°a>90°Thus, the angle is more than 90°.Hence, we can conclude by saying that the given triangle is an obtuse triangle.



Page No 15.13:

Question 14:

In the six cornered figure, AC, AD and AE are joined. Find ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA.

Answer:

We have to find FAB+ABC+BCD+CDE+DEF+EFA  ......(i) From the figure, we have:  FAB=FAE+EAD+DAC+CABBCD=ACB+ACDCDE=ADC+ADEDEF=AED+AEFPutting the  values of FAB,BCD,CDE,DEF in equation (i):(FAE+EAD+DAC+CAB) + ABC+(ACB+ACD) + (ADC+ADE) + (AED+AEF) + EFA(ABC+ACB+CAB) + (FAE+AEF+EFA) + (FAE+AEF+EFA) + (ADC+ACD+DAC)   .....(ii)We know that the sum of the three angles of a triangle is equal to 180°. Hence we can say the following:ABC+ACB+CAB =180°(angles of ABC)FAE+AEF+EFA =180°(angles of AFE)AED+ADE+EAD =180°(angles of AED)ADC+ACD+DAC =180°(angles of ADC)Putting these values in equation (ii):180°+180°+180°+180°Hence, the sum of the given angles is 720° 

Page No 15.13:

Question 15:

Find x, y, z (whichever is required) in the figures given below:

Answer:


(i) In ABC and ADE, we have:ADE=ABC  (corresponding angles)x°=40°AED=ACB  (corresponding angles)y°=30°We know that the sum of all the three angles of a triangle is equal to 180°. x°+y°+z°=180°  (Angles of ADE)Which means: 40°+30°+z°=180° z°=180°-70°z°=110°Therefore, we can conclude that the three angles of the given triangle are 40°, 30° and 110°.

(ii) We can see that in ADC, ADC is equal to 90°. (ADC is a right triangle)We also know that the sum of all the angles of a triangle is equal to 180°.Which means: 45°+90°+y°=180° (Sum of the angles of ADC)135°+y°=180°y°=180°-135°y°=45°We can also say that in ABC, ABC+ACB+BAC is equal to 180°. (Sum of the angles of ABC)40°+y°+(x°+45°)=180°40°+45°+x°+45°=180°      (y°=45°)x°=180°-130°x°=50°Therefore, we can say that the required angles are 45° and 50°.

(iii) We know that the sum of all the angles of a triangle is equal to 180°.Therefore, for ABD:ABD+ADB+BAD=180° (Sum of the angles of ABD) 50°+x°+50°=180° 100°+x°=180°x°=180°-100°x°=80°For ABC:ABC+ACB+BAC=180° (Sum of the angles of ABC)50°+z°+(50°+30°)=180°50°+z°+50°+30°=180°      z°=180°-130°z°=50°Using the same argument for ADC:ADC+ACD+DAC=180° (Sum of angles of ADC)y°+z°+30°=180° y°+50°+30°=180°    (z° =50°)y°=180°-80°y°=100°Therefore, we can conclude that the required angles are 80°, 50° and 100°.

(iv) In ABC and ADE:ADE=ABC  (Corresponding angles)y°=50°Also, AED=ACB  (Corresponding angles)z°=40°We know that the sum of all the three angles of a triangle is equal to 180°.Which means: x°+50°+40°=180°  (Angles of ADE)x°=180°-90°x°=90°Therefore, we can conclude that the required angles are 50°, 40° and 90°.

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Question 16:

If one angle of a triangle is 60° and the other two angles are in the ratio 1 : 2, find the angles.

Answer:

We know that one of the angles of the given triangle is 60°. (Given)We also know that the other two angles of the triangle are in the ratio 1:2.Let one of the other two angles be x.Therefore, the second one will be 2x.We know that the sum of all the three angles of a triangle is equal to 180°.60°+x°+2x°=180°3x°=180°-60°3x°=120°x°=120°3x°=40°2x°=2×402x°=80°Hence, we can conclude that the required angles are 40° and 80°.

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Question 17:

If one angle of a triangle is 100° and the other two angles are in the ratio 2 : 3, find the angles.

Answer:

We know that one of the angles of the given triangle is 100°.We also know that the other two angles are in the ratio 2:3.Let one of the other two angles be 2x.Therefore, the second angle will be 3x.We know that the sum of all three angles of a triangle is 180°.100°+2x°+3x°=180°5x°=180°-100°5x°=80°x°=80°5x°=16°2x°=2×162x°=32°3x°=3×163x°=48°Thus, the required angles are 32° and 48°.

Page No 15.13:

Question 18:

In a ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate the angles.

Answer:

We know that for the given triangle, 3A is equal to 6C. A=2C                               ...(i)We also know that for the same triangle, 4B is equal to 6C. B=64C                        ...(ii)We know that the sum of all three angles of a triangle is 180°.Therefore, we can say that:A+B+C=180°  (Angles of ABC)        ...(iii)On putting the values ofA and B in equation (iii), we get: 2C+64C+C=180°184C=180°C=40°From equation (i), we have:A=2C=2×40A=80°From equation (ii), we have:B=64C=64×40°B=60°A=80°, B=60°, C=40°Therefore, the three angles of the given triangle are 80°, 60° and 40°.

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Question 19:

Is it possible to have a triangle, in which
(i) two of the angles are right?
(ii) two of the angles are obtuse?
(iii) two of the angles are acute?
(iv) each angle is less than 60°?
(v) each angle is greater than 60°?
(vi) each angle is equal to 60°?
Give reasons in support of your answer in each case.

Answer:

(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.
(ii) No, because as we know that the sum of all three angles of a triangle is always 180°. If there are two obtuse angles, then their sum will be more than 180°, which is not possible in case of a triangle.
(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.
(iv) No, because if each angle is less than 60°, then the sum of all three angles will be less than 180°, which is not possible in case of a triangle.
Proof :
Let the three angles of the triangle be A, B and C.As per the given information, A<60°   ...(i)B<60°   ...(ii)C<60°   ...(iii)On adding (i), (ii) and (iii), we get:A+B+C<60°+60°+60°A+B+C<180°We can see that the sum of all three angles is less than 180°, which is not possible for a triangle.Hence, we can say that it is not possible for each angle of a triangle to be less than 60°.

(v) No, because if each angle is greater than 60°, then the sum of all three angles will be greater than 180°, which is not possible.
Proof :
Let the three angles of the given triangle be A, B and C.As per the given information,A>60°   ...(i)B>60°   ...(ii)C>60°   ...(iii)On adding (i), (ii) and (iii), we get:A+B+C>60°+60°+60°A+B+C>180°We can see that the sum of all three angles of the given triangle are greater than 180°, which is not possible for a triangle. Hence, we can say that it is not possible for each angle of a triangle to be greater than 60°.

(vi) Yes, if each angle of the triangle is equal to 60°, then the sum of all three angles will be 180°, which is possible in case of a triangle.
Proof :
Let the three angles of the given triangle be A, B and C.As per the given information,A= 60°   ...(i)B= 60°   ...(ii)C= 60°  ...(iii)On adding (i), (ii) and (iii), we get:A+B+C=60°+60°+60°A+B+C=180°We can see that the sum of all three angles of the given triangle is equal to 180°, which is possible in case of a triangle.Hence, we can say that it is possible for each angle of a triangle to be equal to 60°.

Page No 15.13:

Question 20:

In ∆ABC, ∠A = 100°, AD bisects ∠A and ADBC. Find ∠B.

Answer:


ConsiderABD.BAD=100°2    (AD bisects A)BAD=50°ADB=90°  (ADBC)We know that the sum of all three angles of a triangle is 180°.Thus,ABD+BAD+ADB=180°  (Sum of angles of ABD)Or,ABD+50°+90°=180°ABD=180°-140°ABD=40°

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Question 21:

In ∆ABC, ∠A = 50°, ∠B = 70° and bisector of ∠C meets AB in D. Find the angles of the triangles ADC and BDC.

Answer:


We know that the sum of all three angles of a triangle is equal to 180°.Therefore, for the given ABC, we can say that:A+B+C=180° (Sum of angles of ABC)50°+70°+C=180°C=180°-120°C=60°ACD=BCD=C2(CD bisects C and meets ABinD.)ACD=BCD=60°2=30°Using the same logic for the given ACD, we can say that:DAC+ACD+ADC=180°50°+30°+ADC=180°ADC=180°-80°ADC=100°If we use the same logic for the given BCD, we can say that:DBC+BCD+BDC=180°70°+30°+BDC=180°BDC=180°-100°BDC=80°Thus,For ADC: A = 50°, D =100°, C = 30°For BDC: B = 70°, D= 80° , C= 30°

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Question 22:

In ∆ABC, ∠A = 60°, ∠B = 80° and the bisectors of ∠B and ∠C meet at O. Find
(i) ∠C
(ii) ∠BOC.

Answer:


We know that the sum of all three angles of a triangle is 180°.Hence, for ABC, we can say that:A+B+C=180°  (Sum of angles of ABC)60°+80°+C=180°C=180°-140°C=40°For OBC:OBC=B2=80°2   (OBbisects B.)OBC=40°OCB=C2=40°2  (OC bisects C.)OCB=20°If we apply the above logic to this triangle, we can say that:OCB+OBC+BOC=180° (Sum of angles of OBC)20°+40°+BOC=180°BOC=180°-60°BOC=120°



Page No 15.14:

Question 23:

The  bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.

Answer:


We know that the sum of all three angles of a triangle is 180°.Hence, for ABC, we can say that:A+B+C=180° A+90°+C=180° A+C=180°-90°A+C=90°          ForOAC:OAC=A2    (OA bisects A.)OCA=C2    (OC bisects C.)On applying the above logic to OAC, we get:AOC+OAC+OCA=180° (Sum of angles of AOC)AOC+A2 +C2 =180°AOC+A+C2=180°AOC+90°2=180°    ( A+C=90° )AOC=180°-45°AOC=135°

Page No 15.14:

Question 24:

In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.

Answer:


In the given triangle, ACD=A+B.   (Exterior angle is equal to the sum of two opposite interior angles.)We know that the sum of all three angles of a triangle is 180°.Therefore, for the given triangle, we can say that:ABC+BCA+CAB=180°. (Sum of all angles of ABC)A+B+BCA=180°BCA=180°-(A+B)ECA=ACD2  (EC bisects ACD)ECA=A+B 2   (ACD=A+B)EBC=ABC2 =B2 (EB bisects ABC)ECB=ECA+BCAECB=A+B 2+180°-(A+B)If we use the same logic forEBC, we can say that:EBC+ECB+BEC=180° (Sum of all angles of EBC)B2+A+B 2+180°-(A+B)+BEC=180°BEC=A+B-(A+B 2)-B2BEC=A2BEC=50°2=25°

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Question 25:

In ∆ABC, ∠B = 60°, ∠C = 40°, ALBC and AD bisects ∠A such that L and D lie on side BC. Find ∠LAD.

Answer:


We know that the sum of all angles of a triangle is 180°.Therefore, for ABC, we can say that:A+B+C=180° Or, A+60°+40°=180° A=80°DAC=A2  (ADbisects A)DAC=80°2=40°If we use the above logic on ADC, we can say that:ADC+DCA+DAC=180°. (Sum of all the angles of ADC)ADC+40°+40°=180° ADC=180°-80°=100°ADC=ALD+LAD  (Exterior angle is equal to the sum of two Interior opposite angles.)100°=90°+LAD     (ALBC)LAD =10°  

Page No 15.14:

Question 26:

Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 35° and ∠CDB = 55°, find ∠BOD.

Answer:


We know that ACBD and AB cuts AC and BD at A and B, respectively.CAB=DBA (Alternate interior angles)DBA=35°We also know that the sum of all three angles of a triangle is 180°.Hence, for OBD, we can say that:DBO+ODB+BOD=180° 35°+55°+BOD=180° (DBO=DBA andODB=CDB)BOD=180°-90°=BOD=90°

Page No 15.14:

Question 27:

In Fig., ∆ABC is right angled at A. Q and R are points on line BC and P is a point such that QP || AC and RP || AB. Find ∠P.

Answer:

In the given triangle, ACQP and BR cuts AC and QP at  C and Q, respectively.QCA=CQP (Alternate interior angles)Because RPAB and BR cuts AB and RP at B and R, respectively, ABC=PRQ (alternate interior angles).We know that the sum of all three angles of a triangle is 180°.Hence, for ABC, we can say that:ABC+ACB+BAC=180°ABC+ACB+90°=180° (Right angled at A)ABC+ACB=90°Using the same logic for PQR, we can say that:PQR+PRQ+QPR=180° ABC+ACB+QPR=180° (ABC=PRQ and QCA=CQP )Or,90°+QPR=180°   (ABC+ACB=90°)QPR=90°



Page No 15.19:

Question 1:

In Fig., ∠CBX is an exterior angle of ∆ABC at B. Name

(i) the interior adjacent angle
(ii) the interior opposite angles to exterior ∠CBX.
Also, name the interior opposite angles to an exterior angle at A.

Answer:

(i) The interior angle adjacent to exterior CBX is ABC.(ii) The interior angles opposite to exterior CBX are BAC and ACB.Also, the interior angles opposite to exterior BAY are ABC and ACB.



Page No 15.20:

Question 2:

In Fig, two of the angles are indicated. What are the measures of ∠ACX and ∠ACB?

Answer:

InABC, A = 50° and B = 55°.Because of the angle sum property of the triangle, we can say that:A +B +C = 180° 50°+55°+C = 180°Or, C = 75°i.e. ACB =75°ACX = 180°-ACB =180°-75° = 105° (Linear pair)

Page No 15.20:

Question 3:

In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angles is 55°. Find all the angles of the triangle.

Answer:


We know that the sum of interior opposite angles is equal to the exterior angle.Hence, for the given triangle, we can say that:ABC+BAC=BCO   55°+BAC=95°Or, BAC=95°-55°=BAC=40°We also know that the sum of all angles of a triangle is 180°.Hence, for the given ABC, we can say that:ABC+BAC+BCA=180°  55°+40°+BCA=180°Or,BCA=180°-95°= BCA=85°

Page No 15.20:

Question 4:

One of the exterior angles of a triangle is 80°, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?

Answer:

Let us assume that A and B are the two interior opposite angles. We know that A is equal to B.We also know that the sum of interior opposite angles is equal to the exterior angle.Hence, we can say that:A+B=80°  Or,A+A=80°   (A=B)2A=80°A=80°2=40°A=B=40°Thus, each of the required angles is of 40°.

Page No 15.20:

Question 5:

The exterior angles, obtained on producing the base of a triangle both ways are 104° and 136°. Find all the angles of the triangle.

Answer:


In the given figure, ABE and ABC form a linear pair.ABE + ABC=180°ABC=180°-136°ABC=44°We can also see that ACD and ACB form a linear pair.ACD + ACB=180°ACB=180°-104°ACB=76°We know that the sum of interior opposite angles is equal to the exterior angle. Therefore, we can say that:BAC+ABC=104°BAC=104°-44=60°Thus,ACB = 76° andBAC= 60°

Page No 15.20:

Question 6:

In Fig., the sides BC, CA and BA of a ∆ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°; find all the angles of the ∆ABC.

Answer:

In ABC, BAC and EAF are vertically opposite angles.Hence, we can say that:BAC = EAF = 45° Considering the exterior angle property, we can say that:BAC + ABC = ACD = 105°ABC = 105°-45° = 60°Because of the angle sum property of the triangle, we can say that:ABC +ACB +BAC = 180°ACB = 75°Therefore, the angles are 45°, 60° and 75°.

Page No 15.20:

Question 7:

In Fig., ACCE and ∠A :∠B : ∠C = 3 : 2 : 1, find the value of ∠ECD.

Answer:

In the given triangle, the angles are in the ratio 3:2:1.Let the angles of the triangle be 3x, 2x and x.Because of the angle sum property of the triangle, we can say that:3x+2x+x = 180°6x = 180°Or, x = 30°     ...(i)Also, ACB +ACE +ECD = 180°x+ 90°+ECD = 180°  (ACE = 90°)ECD = 60°  [From  (i)]

Page No 15.20:

Question 8:

A student when asked to measure two exterior angles of ∆ABC observed that the exterior angles at A and B are of 103° and 74° respectively. Is this possible? Why or why not?

Answer:

Here,Internal angle at A+ External angle at A=180° Internal angle at A+ 103°=180° Internal angle at A=77° Internal angle at B+ External angle at B=180° Internal angle at B+ 74°=180° Internal angle at B=106° Sum of internal angles at A and B=77°+106°=183°It means that the sum of internal angles at A and B is greater than 180°, which cannot be possible.

Page No 15.20:

Question 9:

In Fig., AD and CF are respectively perpendiculars to sides BC and AB of ∆ABC. If ∠FCD = 50°, find ∠BAD.

Answer:


We know that the sum of all angles of a triangle is 180°.Therefore, for the given FCB, we can say that:FCB+CBF+BFC=180° 50°+CBF+90°=180°Or,CBF=180°-50°-90°=40°  ...(i)Using the above rule for ABD, we can say that:ABD+BDA+BAD=180°BAD=180°-90°-40°=50°  [From (i)]

Page No 15.20:

Question 10:

In Fig., measures of some angles are indicated. Find the value of x.

Answer:


Here,AED+120°=180°    (Linear pair)AED=180°-120°=60°We know that the sum of all angles of a triangle is 180°.Therefore, for ADE, we can say that:ADE+AED+DAE=180° 60°+ADE+30°=180°Or, ADE=180°-60°-30°=90°From the given figure, we can also say that:FDC+90°=180°    (Linear pair) FDC=180°-90°=90°Using the above rule for CDF, we can say that:CDF+DCF+DFC=180°90°+DCF+60°=180°DCF=180°-60°-90°=30°Also,DCF+x=180°    (Linear pair)30°+x=180°Or, x=180°-30°=150°



Page No 15.21:

Question 11:

In Fig., ABC is a right triangle right angled at A. D lies on BA produced and DEBC, intersecting AC at F. If ∠AFE = 130°, find

(i) ∠BDE
(ii) ∠BCA
(iii) ∠ABC

Answer:


(i) Here,BAF+FAD=180° (Linear pair)FAD=180°-BAF=180°-90°=90°Also, AFE=ADF+FAD  (Exterior angle property)ADF+90°=130°   ADF=130°-90°=40°(ii)We know that the sum of all the angles of a triangle is 180°.Therefore, for BDE, we can say that:BDE+BED+DBE=180°.DBE=180°-BDE-BED=180°-90°-40°=50°   ...(i)Also, FAD=ABC+ACB (Exterior angle property) 90°=50°+ACBOr,ACB=90°-50°=40°(iii) ABC =DBE = 50°   [From (i)]

Page No 15.21:

Question 12:

ABC is a triangle in which ∠B = ∠C and ray AX bisects the exterior angle DAC. If ∠DAX = 70°, find ∠ACB.

Answer:


Here,CAX =DAX (AX bisects CAD)CAX=70°CAX +DAX + CAB =180° 70° +70°+ CAB =180° CAB =180°-140°CAB =40°  ACB +CBA + CAB =180°   (Sum of the angles of ABC)ACB +ACB+ 40° =180°    (C=B)2ACB=180°-40°ACB=140°2ACB=70°

Page No 15.21:

Question 13:

The side BC of ∆ABC is produced to a point D. The bisector of ∠A meets side BC in L. If ∠ABC = 30° and ∠ACD = 115°, find ∠ALC.

Answer:


ACD and ACL make a linear pair.ACD + ACB=180°115°+ ACB=180°ACB=180°-115°ACB=65°We know that the sum of all angles of a triangle is 180°. Therefore, for ABC, we can say that: ABC+BAC+ACB=180° 30°+BAC+65°=180°Or,BAC=85°=LAC=BAC2=85°2Using the above rule for ALC, we can say that:ALC+LAC+ACL=180°ALC+85°2+65°=180°     (ACL=ACB)Or, ALC=180°-85°2-65°= ALC=145°2=7212°Thus,ALC = 7212°    

Page No 15.21:

Question 14:

D is a point on the side BC of ∆ABC. A line PDQ, through D, meets side AC in P and AB produced at Q. If ∠A = 80°, ∠ABC = 60° and ∠PDC = 15°, find (i) ∠AQD (ii) APD.

Answer:


ABD and QBD form a linear pair.ABC + QBC=180°60°+QBC=180°QBC=120°PDC=BDQ (Vertically opposite angles)BDQ=15°In QBD:QBD+QDB+BQD=180° (Sum of angles of QBD)120°+15°+BQD=180°BQD=180°-135°BQD=45°AQD=BQD=45°In AQP:QAP+AQP+APQ=180°  (Sum of angles of AQP)80°+45°+APQ=180°APQ=55°APD=APQ

Page No 15.21:

Question 15:

Explain the concept of interior and exterior angles and in each of the figures given below, find x and y.

Answer:

The interior angles of a triangle are the three angle elements inside the triangle.
The exterior angles are formed by extending the sides of a triangle, and if the side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Using these definitions, we will obtain the values of x and y.

(i)
From the given figure, we can see that:ACB + x = 180° (Linear pair) 75° + x = 180°Or, x = 105°We know that the sum of all angles of a triangle is 180°.Therefore, for ABC, we can say that:BAC + ABC +ACB = 180° 40°+ y+ 75° =  180°Or,y = 65°

(ii)
x + 80° =180° (Linear pair)= x = 100°In ABC:x+ y+ 30° = 180° (Angle sum property)100°+30°+y = 180°= y = 50°

(iii)We know that the sum of all angles of a triangle is 180°.Therefore, for ACD, we can say that:30°+100°+y = 180°Or, y = 50°ACB + 100° = 180°ACB = 80°   ...(i)Using the above rule for ACB, we can say that:x+45°+80° = 180°.= x = 55°(iv)We know that the sum of all angles of a triangle is 180°.Therefore, for DBC, we can say that:30°+ 50° + DBC = 180°DBC  = 100°x + DBC= 180° (Linear pair)x = 80°And,y = 30° + 80° = 110° (Exterior angle property)



Page No 15.22:

Question 16:

Compute the value of x in each of the following figures:

Answer:

(i) From the given figure, we can say that:ACD + ACB = 180° (Linear pair)Or,ACB = 180° - 112° = 68°      ...(i)We can also say that:BAE + BAC = 180°  (Linear pair)Or, BAC = 180°-120° = 60°       ...(ii)We know that the sum of all angles of a triangle is 180°.Therefore, for ABC: x+ BAC + ACB = 180°x = 180°-60°-68° = 52°= x = 52°(ii) From the given figure, we can say that:ABC + 120° = 180°  (Linear pair)ABC = 60°We can also say that:ACB +110° = 180° (Linear pair)ACB = 70°We know that the sum of all angles of a triangle is 180°.Therefore, for ABC:x+ ABC + ACB = 180°= x = 50°(iii) From the given figure, we can see that:BAD = ADC  = 52°  (Alternate angles)We know that the sum of all the angles of a triangle is 180°.Therefore, for DEC: x + 40° + 52° = 180°= x = 88°

(iv) In the given figure, we have a quadrilateral whose sum of all angles is 360°.Thus,35° + 45° + 50° + reflex ADC = 360°Or,  reflex ADC = 230°230°+ x= 360° (A complete angle)= x = 130°



Page No 15.24:

Question 1:

In each of the following, there are three positive numbers. State if these numbers could possibly be the lengths of the sides of a triangle:
(i) 5, 7, 9
(ii) 2, 10, 15
(iii) 3, 4, 5
(iv) 2, 5, 7
(v) 5, 8, 20

Answer:

(i) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side.
Here,
5+7>9,  5+9>7,  9+7>5

(ii) No, these numbers cannot be the lengths of the sides of a triangle because​ the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.

(iii) Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of triangle is always greater than the third side.
Here,
3+4 >5,  3+5>4,  4+5>3

(iv) No, these numbers cannot be the lengths of the sides of a triangle because​ the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.
Here,
2+5 = 7

(v) No, these numbers cannot be the lengths of the sides of a triangle because​ the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.​
Here,
5+8 <20

Page No 15.24:

Question 2:

In Fig., P is the point on the side BC. Complete each of the following statements using symbol ' = ', '>' or '<' so as to make it true:

(i) AP ... AB + BP
(ii) AP .... AC + PC
(iii) AP....12(AB+AC+BC)

Answer:

(i) In triangle APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side. 

(ii)  In triangle APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side. 

(iii) AP < 12(AB+AC+BC)
In triangles ABP and ACP, we can see that:
AP < AB + BP         ...(i)  (Because the sum of any two sides of a triangle is greater than the third side)
AP < AC + PC         ...(ii)  (Because the sum of any two sides of a triangle is greater than the third side)
  
On adding (i) and (ii), we have:

AP + AP < AB + BP + AC + PC
2AP < AB + AC + BC (BC = BP + PC)
AP < 12(AB+AC+BC)

Page No 15.24:

Question 3:

P is a point in the interior of ∆ABC as shown in Fig. State which of the following statements are true (T) or false (F):

(i) AP + PB < AB
(ii) AP + PC > AC
(iii) BP + PC = BC

Answer:

(i) False
We know that the sum of any two sides of a triangle is greater than the third side; it is not true for the given triangle.

(ii) True
We know that the sum of any two sides of a triangle is greater than the third side; it is true for the given triangle. 

(iii) False
We know that the sum of any two sides of a triangle is greater than the third side; it is not true for the given triangle.



Page No 15.25:

Question 4:

O is a point in the exterior of ∆ABC. What symbol '>', '<' or '=' will you use to complete the statement OA + OB .. AB? Write two other similar statements and shown that

OA+OB+OC>12(AB+BC+CA)

Answer:

Because the sum of any two sides of a triangle is always greater than the third side, in triangle OAB, we have:

OA+OB>AB   ...(i)OB+OC>BC   ...(ii)OA+OC>CA   ...(iii)On adding equations (i), (ii) and (iii), we get:OA+OB+OB+OC+OA+OC>AB+BC+CA2(OA+OB+OC)>AB+BC+CAOA+OB+OC>AB+BC+CA2

Page No 15.25:

Question 5:

In ∆ABC, ∠A = 100°, ∠B = 30°, ∠C = 50°. Name the smallest and the largest sides of the triangle.

Answer:

Because the smallest side is always opposite to the smallest angle, which in this case is 30o, it is AC.
Also, because the largest side is always opposite to the largest angle, which in this case is 100o, it is BC.



Page No 15.30:

Question 1:

State Pythagoras theorem and its converse.

Answer:

The Pythagoras Theorem: In a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides.

Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle.

Page No 15.30:

Question 2:

In right ∆ ABC, the lengths of the legs are given. Find the length of the hypotenuse.
(i) a = 6 cm, b = 8 cm
(ii) a = 8 cm, b = 15 cm
(iii) a = 3 cm, b = 4 cm
(iv) a = 2 cm, b = 1.5 cm

Answer:

According to the Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Height)2

(i) c2=a2+b2c2=62+82c2=36+64=100c=10 cm(ii) c2=a2+b2c2=82+152c2=64+225=289c=17 cm

(iii) c2=a2+b2c2=32+42c2=9+16=25c=5 cm(iv) c2=a2+b2c2=22+1.52c2=4+2.25=6.25c=2.5 cm

Page No 15.30:

Question 3:

The hypotenuse of a triangle is 2.5 cm. If one of the sides is 1.5 cm, find the length of the other side.

Answer:

Let the hypotenuse be "c" and the other two sides be "b" and "c".Using the Pythagoras theorem, we can say that:c2=a2+b22.52=1.52+b2b2=6.25-2.25=4b=2Hence, the length of the other side is 2 cm.

Page No 15.30:

Question 4:

A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches.

Answer:


Let the hypotenuse be h.Using the Pythagoras theorem, we get:3.72=1.22+h2h2=13.69-1.44=12.25 h=3.5Hence, the height of the wall is 3.5 m.

Page No 15.30:

Question 5:

If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is right-angled triangle.

Answer:

In the given triangle, the largest side is 6 cm.We know that in a right-angled triangle, the sum of the squares of the smaller sides should be equal to the square of the largest side.Therefore,32+42=9+16=25But,62=3632+4262Hence, the given triangle is not a right-angled triangle.

Page No 15.30:

Question 6:

The sides of certain triangles are given below. Determine which of them are right triangles.
(i) a = 7 cm, b = 24 cm and c = 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm

Answer:

(i) We know that in a right-angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides. Here, the larger side is c, which is 25 cm.c2=625We have:a2+b2=72+242=49+576=625=c2Thus, the given triangle is a right triangle.

(ii) We know that in a right-angled triangle, the square of the largest side is equal to the sum of the squares of the smaller sides. Here, the larger side is c, which is 18 cm.c2=324We have: a2+b2=92+162=81+256=337c2Thus, the given triangle is not a right triangle.

Page No 15.30:

Question 7:

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
[Hint: Find the hypotenuse of a right triangle having the sides (11 − 6) m = 5 m and 12 m]

Answer:


The distance between the tops of the poles is the distance between points A and B.We can see from the given figure that points A, B and C form a right triangle, with AB as the hypotenuse.On using the Pythagoras Theorem in ABC, we get: (11-6)2+122=AB225+144=AB2AB2=169AB=13Hence, the distance between the tops of the poles is 13 m.

Page No 15.30:

Question 8:

A man goes 15 m due west and then 8 m due north. How far is he from the starting point?

Answer:


Let O be the starting point and P be the final point.By using the Pythagoras theorem, we can find the distance OP. 152+82=OP2OP2=225+64OP2=289OP=17Hence, the required distance is 17 m.



Page No 15.31:

Question 9:

The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?

Answer:


Let the length of the ladder be L m.By using the Pythagoras theorem, we can find the length of the ladder.62+82=L2L2=36+64=100L=10Thus, the length of the ladder is 10 m. 


When the ladder is shifted:Let the height of the ladder after it is shifted be H m.By using the Pythagoras theorem, we can find the height of the ladder after it is shifted.82+H2=102H2=100-64=36H=6Thus, the height of the ladder is 6 m.

Page No 15.31:

Question 10:

A ladder 50 dm long when set against the wall of a house just reaches a window at a height of 48 dm. How far is the lower end of the ladder from the base of the wall?

Answer:


Let the distance of the lower end of the ladder from the wall be X m.On using the Pythagoras theorem, we get: X2+482=502X2=502-482=2500-2304=196X=14 dmHence, the distance of the lower end of the ladder from the wall is 14 dm.

Page No 15.31:

Question 11:

The two legs of a right triangle are equal and the square of the hypotenuse is 50. Find the length of each leg.

Answer:

Let the length of each leg of the given triangle be x units.Using the Pythagoras theorem, we get: x2+x2=(Hypotenuse)2x2+x2=502x2=50x2=25x=5Hence, we can say that the length of each leg is 5 units.

Page No 15.31:

Question 12:

Verify that the following numbers represent Pythagorean triplet:
(i) 12, 35, 37
(ii) 7, 24, 25
(iii) 27, 36, 45
(iv) 15, 36, 39

Answer:

We will check for a Pythagorean triplet by checking if the square of the largest side is equal to the sum of the squares of the other two sides.

(i) 372=1369122+352=144+1225=1369122+352=372Yes, they represent a Pythagorean triplet.(ii) 252=62572+242=49+576=62572+242=252Yes, they represent a Pythagorean triplet.(iii) 452=2025272+362=729+1296=2025272+362=452Yes, they represent a Pythagorean triplet.(iv) 392=1521152+362=225+1296=1521152+362=392 Yes, they represent a Pythagorean triplet.

Page No 15.31:

Question 13:

In a ∆ABC, ∠ABC = 100°, ∠BAC = 35° and BDAC meets side AC in D. If BD = 2 cm, find ∠C and length DC.

Answer:



We know that the sum of all angles of a triangle is 180°.Therefore, for the given ABC, we can say that: ABC+BAC+ACB=180°100°+35°+ACB=180°ACB=180°-135°ACB=45°C=45°If we apply the above rule on BCD, we can say that:BCD+BDC+CBD=180° 45°+90°+CBD=180°      (ACB=BCD and BDAC)CBD=180°-135°CBD=45°We know that the sides opposite to equal angles have equal length.Thus,BD=DCDC=2 cm

Page No 15.31:

Question 14:

In a ∆ABC, AD is the altitude from A such that AD = 12 cm, BD = 9 cm and DC = 16 cm. Examine if ∆ABC is right angled at A.

Answer:



In ADC,ADC=90° (AD is an altitude on BC.)Using the Pythagoras theorem, we get: 122+162=AC2AC2=144+256=400AC=20 cmIn ADB,ADB=90° (AD is an altitude on BC.)Using the Pythagoras theorem, we get:122+92=AB2AB2=144+81=225AB=15 cmIn ABC,BC2=252=625AB2+AC2=152+202=625AB2+AC2=BC2Because it satisfies the Pythagoras theorem, we can say that ABC is right angled at A.

Page No 15.31:

Question 15:

Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 105°. Measure AB. Is (AB)2 = (AC)2 + (BC)2? If not, which one of the following is true:
(AB)2 > (AC)2 + (BC)2 or (AB)2 < (AC)2 + (BC)2?

Answer:


Draw ABC.
Draw a line BC = 3 cm.
At point C, draw a line at 105° angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Now, join AB, which will be approximately 5.5 cm.
AC2+BC2=42+32=9+16=25AB2=5.52=30.25

(AB)2  (AC)2 + (BC)2

Here,

(AB)2 > (AC)2 + (BC)2

Page No 15.31:

Question 16:

Draw a triangle ABC, with AC = 4 cm, BC = 3 cm and ∠C = 80°. Measure AB. Is (AB)2 = (AC)2 + (BC)2? If not, which one of the following is true:
(AB)2 > (AC)2 + (BC)2 or (AB)2 < (AC)2 + (BC)2?

Answer:


First draw ABC.

Draw a line BC = 3 cm.
At point C, draw a line at 80° angle with BC.
Take an arc of 4 cm from point C, which will cut the line at point A.
Now, join AB; it will be approximately 4.5 cm.
AC2+BC2=42+32=9+16=25AB2=4.52=20.25

(AB)2  (AC)2 + (BC)2

Here,

(AB)2 < (AC)2 + (BC)2



Page No 15.4:

Question 1:

Take three non-collinear points A, B and C on a page of your notebook. Join AB, BC and CA. What figure do you get? Name the triangle. Also, name
(i) the side opposite to ∠B
(ii) the angle opposite to side AB
(iii) the vertex opposite to side BC
(iv) the side opposite to vertex B.

Answer:


The figure that we get is that of a triangle.
The name of the triangle is ABC.
(i) The side opposite to B is AC.
(ii) The angle opposite to AB is ACB.
(iii) The vertex opposite to BC is A.
(iv) The side opposite to the vertex B is AC.

Page No 15.4:

Question 2:

Take three collinear points A, B and C on a page of your note book. Join AB, BC and CA. Is the figure a triangle? If not, why?

Answer:



No, the figure is not a triangle. By definition, a triangle is a plane figure formed by three non-parallel line segments.

Page No 15.4:

Question 3:

Distinguish between a triangle and its triangular region.

Answer:

A triangle is a plane figure formed by three non-parallel line segments, whereas, its triangular region includes the interior of the triangle along with the triangle itself.

Page No 15.4:

Question 4:

In Fig., D is a point on side BC of a ∆ABC. AD is joined. Name all the triangles that you can observe in the figure. How many are they?

Answer:

We can observe the following three triangles in the given figure:
1. △ ABC
2.  ACD
​3. ​ ADB

Page No 15.4:

Question 5:

In Fig., A, B, C and D are four points, and no three points are collinear. AC and BD intersect at O. There are eight triangles that you can observe. Name all the triangles.

Answer:

The eight triangles that can be observed in the given figure are as follows:
1. AOD2. AOB3. BOC4. COD5. ACD6. ACB7. ADB8. CDB

Page No 15.4:

Question 6:

What is the difference between a triangle and triangular region?

Answer:

A triangle is a plane figure formed by three non-parallel line segments, whereas, a triangular region is the interior of a triangle along with the triangle itself.

Page No 15.4:

Question 7:

Explain the following terms:
(i) Triangle
(ii) Parts or elements of a triangle
(iii) Scalene triangle
(iv) Isosceles triangle
(v) Equilateral triangle
(vi) Acute triangle
(vii) Right triangle
(viii) Obtuse triangle
(ix) Interior of a triangle
(x) Exterior of a triangle.

Answer:

(i) A triangle is a plane figure formed by three non-parallel line segments.
(ii) The three sides and the three angles of a triangle are together known as the parts or elements of that triangle.
(iii) A scalene triangle is a triangle in which no two sides are equal.

(iv) An isosceles triangle is a triangle in which two sides are equal.

(v) An equilateral triangle is a triangle in which all three sides are equal.

(vi) An acute triangle is a triangle in which all the angles are acute (less than 90°).

(vii) A right angled triangle is a triangle in which one angle is right angled, i.e 90°.

(viii) An obtuse triangle is a triangle in which one angle is obtuse (more than 90°).

(ix) The interior of a triangle is made up of all such points that are enclosed within the triangle.
(x) The exterior of a triangle is made up of all such points that are not enclosed within the triangle.

Page No 15.4:

Question 8:

In Fig., the length (in cm) of each side has been indicated along the side. State for each triangle whether it is scalene, isosceles or equilateral:

Answer:

(i) This triangle is a scalene triangle because no two sides are equal.
(ii) This triangle is an isosceles triangle because two of its sides, viz. PQ and PR, are equal.
(iii) This triangle is an equilateral triangle because all its three sides are equal.
(iv) This triangle is a scalene triangle because no two sides are equal.
(v) This triangle is an isosceles triangle because two of its sides are equal.



Page No 15.5:

Question 9:

In Fig., there are five triangles. The measures of some of their angles have been indicated. State for each triangle whether it is acute, right or obtuse.

Answer:

(i) This is a right triangle because one of its angles is 90°.
(ii) This is an obtuse triangle because one of its angles is 120°, which is greater than 90°.
(iii) This is an acute triangle because all its angles are acute angles (less than 90°).
(iv) This is a right triangle because one of its angles is 90°.
(v) This is an obtuse triangle because one of its angles is 110°, which is greater than 90°.



Page No 15.6:

Question 10:

Fill in the blanks with the correct word/symbol to make it a true statement:
(i) A triangle has ....... sides.
(ii) A triangle has ....... vertices.
(iii) A triangle has ....... angles.
(iv) A triangle has ....... parts.
(v) A triangle whose no two sides are equal is known as .......
(vi) A triangle whose two sides are equal is known as .....
(vii) A triangle whose all the sides are equal is known as .......
(viii) A triangle whose one angle is a right angle is known as .......
(ix) A triangle whose all the angles are of measure less than 90° is known as .......
(x) A triangle whose one angle is more than 90° is known as ......

Answer:

(i) three
(ii) three
(iii) three
(iv) six (three sides + three angles)
(v) a scalene triangle
(vi) an isosceles triangle
(vii) an equilateral triangle
(viii) a right triangle
(ix) an acute triangle
(x) an obtuse triangle

Page No 15.6:

Question 11:

In each of the following, state if the statement is true (T) or false (F):
(i) A triangle has three sides.
(ii) A triangle may have four vertices.
(iii) Any three line-segments make up a triangle.
(iv) The interior of a triangle includes its vertices.
(v) The triangular region includes the vertices of the corresponding triangle.
(vi) The vertices of a triangle are three collinear points.
(vii) An equilateral triangle is isosceles also.
(viii) Every right triangle is scalene.
(ix) Each acute triangle is equilateral.
(x) No isosceles triangle is obtuse.

Answer:

(i) True.
(ii) False. A triangle has three vertices.
(iii) False. Any three non-parallel line segments can make up a triangle.
(iv) False. The interior of a triangle is the region enclosed by the triangle and the vertices are not enclosed by the triangle.
(v) True. The triangular region includes the interior region and the triangle itself.
(vi) False. The vertices of a triangle are three non-collinear points.
(vii) True. In an equilateral triangle, any two sides are equal.
(viii) False. A right triangle can also be an isosceles triangle.
(ix) False. Each acute triangle is not an equilateral triangle, but each equilateral triangle is an acute triangle.
(x) False. An isosceles triangle can be an obtuse triangle, a right triangle or an acute triangle.



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