The document RD Sharma Solutions - Ex-15.3, (Part - 2), Properties Of Triangles, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**11. In Fig, ABC is a right triangle right angled at A. D lies on BA produced and DE perpendicular to BC intersecting AC at F. If âˆ AFE = 130Â°, find**

**(i) âˆ BDE**

**(ii) âˆ BCA**

**(iii) âˆ ABC**

**Solution.**

(i)

Here,

âˆ BAF + âˆ FAD = 180Â° (Linear pair)

âˆ FAD =180Â°-âˆ BAF =180Â°â€“ 90Â° = 90Â°

Also,

âˆ AFE = âˆ ADF + âˆ FAD (Exterior angle property)

âˆ ADF + 90Â° = 130Â°

âˆ ADF = 130Â°âˆ’90Â°=40Â°

(ii) We know that the sum of all the angles of a triangle is 180Â°.

Therefore, for Î”BDE, we can say that :

âˆ BDE + âˆ BED + âˆ DBE = 180Â°.

âˆ DBE= 180Â°â€“ âˆ BDE âˆ BED = 180Â°âˆ’90Â°âˆ’40Â°= 50Â°â€”â€”(i)

Also,

âˆ FAD = âˆ ABC + âˆ ACB (Exterior angle property)

90Â° = 50Â° + âˆ ACB

Or,

âˆ ACB = 90Â°â€“ 50Â° = 40Â°

(iii) âˆ ABC = âˆ DBE =50Â° [From (i)]

**12. ABC is a triangle in which âˆ B=âˆ C and ray AX bisects the exterior angle DAC. If âˆ DAX=70Â°. Find âˆ ACB.**

**Solution.**

Here,

âˆ CAX = âˆ DAX (AX bisects âˆ CAD)

âˆ CAX =70Â°

âˆ CAX +âˆ DAX + âˆ CAB =180Â°

70Â°+ 70Â° + âˆ CAB =180Â°

âˆ CAB =180Â° â€“140Â°

âˆ CAB =40Â°

âˆ ACB +âˆ CBA + âˆ CAB =180Â° (Sum of the angles of Î”ABC)

âˆ ACB +âˆ ACB+ 40Â° =180Â° (âˆ C= âˆ B)

2âˆ ACB= 180Â°â€“ 40Â°

**13. The side BC of Î”ABC is produced to a point D. The bisector of âˆ A meets side BC in L. If âˆ ABC=30Â°and âˆ ACD=115Â°, find âˆ ALC**

**Solution.**

âˆ ACD and âˆ ACL make a linear pair.

âˆ ACD+ âˆ ACB = 180Â°

115Â° + âˆ ACB =180Â°

âˆ ACB = 180Â°â€“ 115Â°

âˆ ACB = 65Â°

We know that the sum of all angles of a triangle is 180Â°.

Therefore, for Î”ABC, we can say that :

âˆ ABC + âˆ BAC + âˆ ACB = 180Â°

30Â° + âˆ BAC + 65Â° = 180Â°

Or,

Using the above rule for Î”ALC, we can say that :

âˆ ALC + âˆ LAC + âˆ ACL = 180Â°

Or,

**14. D is a point on the side BC of Î”ABC. A line PDQ through D, meets side AC in P and AB produced at Q. If âˆ A = 80Â°, âˆ ABC = 60Â° and âˆ PDC = 15Â°, find**

**(i) âˆ AQD**

**(ii) âˆ APD**

**Solution.**

âˆ ABD and âˆ QBD form a linear pair.

âˆ ABC + âˆ QBC =180Â°

60Â° + âˆ QBC = 180Â°

âˆ QBC = 120Â°

âˆ PDC = âˆ BDQ (Vertically opposite angles)

âˆ BDQ = 75Â°

In Î”QBD :

âˆ QBD + âˆ QDB + âˆ BDQ =180Â° (Sum of angles of Î”QBD)

120Â°+ 15Â° + âˆ BQD = 180Â°

âˆ BQD = 180Â°â€“ 135Â°

âˆ BQD=45Â°

âˆ AQD = âˆ BQD = 45Â°

In Î”AQP:

âˆ QAP + âˆ AQP + âˆ APQ = 180Â° (Sum of angles of Î”AQP)

80Â° + 45Â° + âˆ APQ = 180Â°

âˆ APQ= 55Â°

âˆ APD = âˆ APQ

**15. Explain the concept of interior and exterior angles and in each of the figures given below. Find x and y**

**Solution.**

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:

âˆ ACB + x = 180Â° (Linear pair)

75Â°+ x = 180Â°

Or,

x = 105Â°

We know that the sum of all angles of a triangle is 180Â°.

Therefore, for Î”ABC, we can say that :

âˆ BAC+ âˆ ABC +âˆ ACB = 180Â°

40Â°+y+ 75Â° = 180Â°

Or,

y = 65Â°

(ii)

x +80Â° = 180Â° (Linear pair)

= x = 100Â°

In Î”ABC:

x+ y+ 30Â° = 180Â° (Angle sum property)

100Â° + 30Â° + y = 180Â°

= y = 50Â°

(iii)

We know that the sum of all angles of a triangle is 180Â°.

Therefore, for Î”ACD, we can say that :

30Â° + 100Â° + y = 180Â°

Or,

y = 50Â°

âˆ ACB + 100Â° = 180Â°

âˆ ACB = 80Â° â€¦ (i)

Using the above rule for Î”ACD, we can say that :

x+ 45Â° + 80Â° = 180Â°

= x = 55Â°

(iv)

We know that the sum of all angles of a triangle is 180Â°.

Therefore, for Î”DBC, we can say that :

30Â° + 50Â° + âˆ DBC = 180Â°

âˆ DBC = 100Â°

x + âˆ DBC = 180Â° (Linear pair)

x = 80Â°

And,

y = 30Â° + 80Â° = 110Â° (Exterior angle property)

**16. Compute the value of x in each of the following figures**

**Solution.**

(i) From the given figure, we can say that :

âˆ ACD + âˆ ACB = 180Â° (Linear pair)

Or,

âˆ ACB = 180Â°â€“ 112Â° = 68Â°

We can also say that :

âˆ BAE + âˆ BAC = 180Â° (Linear pair)

Or,

âˆ BAC = 180Â°â€“ 120Â° = 60Â°

We know that the sum of all angles of a triangle is 180Â°.

Therefore, for Î”ABC:

x+ âˆ BAC + âˆ ACB = 180Â°

x = 180Â°â€“ 60Â°â€“ 68Â° = 52Â°

= x = 52Â°

(ii) From the given figure, we can say that :

âˆ ABC + 120Â° = 180Â° (Linear pair)

âˆ ABC = 60Â°

We can also say that :

âˆ ACB+ 110Â° = 180Â° (Linear pair)

âˆ ACB = 70Â°

We know that the sum of all angles of a triangle is 180Â°.

Therefore, for Î”ABC :

x+ âˆ ABC + âˆ ACB = 180Â°

= x = 50Â°

(iii)

From the given figure, we can see that :

âˆ BAD = âˆ ADC = 52Â° (Alternate angles)

We know that the sum of all the angles of a triangle is 180Â°.

Therefore, for Î”DEC:

x + 40Â°+ 52Â° = 180Â°

= x = 88Â°

(iv) In the given figure, we have a quadrilateral whose sum of all angles is 360Â°.

Thus,

35Â° + 45Â° + 50Â° + reflexâˆ ADC = 360Â°

Or,

reflexâˆ ADC = 230Â°

230Â° + x = 360Â° (A complete angle)

= x = 130Â°