The document NCERT Solutions(Part - 1) - The Triangle and Its Properties Class 7 Notes | EduRev is a part of the Class 7 Course Mathematics (Maths) Class 7.

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**Exercise 6.1 **

**Question 1: **

**In APQR, D is the mid-point of ****____________ PD is_____________ Is QM = MR?**

**Answer 1: **

**Given: QD = DR**

is altitude.

PD is median.

No, QM (eq) MR as D is the mid-point of QR.

**Question 2: **

**Draw rough sketches for the following: **

**(a) In Î”ABC, BE is a median. **

**(b) In Î”PQR, PQ and PR are altitudes of the triangle. **

**(c) In Î”XYZ, YL is an altitude in the exterior of the triangle. **

**Answer 2: **

**(a) Here, BE is a median in Î”ABC and AE = EC.**

(b) Here, PQ and PR are the altitudes of the Î”PQR and RP âŠ¥ QP.

(c) YL is an altitude in the exterior of Î”XYZ.

**Question 3: **

**Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same. **

**Answer 3: **

Isosceles triangle means any two sides are same.

Take Î”ABC and draw the median when AB = AC.

AL is the median and altitude of the given triangle.

**Exercise 6.2 **

**Question 1: **

**Find the value of the unknown exterior angle x in the following diagrams: **

**Answer 1: **

(i) x = 50Â° + 70Â° = 120Â°

(ii) x = 65Â° + 45Â° = 110Â°

(iii) x = 30Â° + 40Â° = 70Â°

(iv) x = 60Â°+60Â° = 120Â°

(v) x = 50Â° + 50Â° = 100Â°

(vi) x = 60Â° + 30Â° = 90Â°

**Question 2: **

**Find the value of the unknown interior angle x in the following figures: **

**Answer 2: **

(i) x + 50Â° = 115Â° â‡’ x = 115Â° - 50Â° = 65Â°

(ii) 70Â°+ x = 100Â° â‡’ x = 100Â°- 70Â° = 30Â°

(iii) x + 90Â° = 125Â° â‡’ x = 120Â°- 90Â° = 35Â°

(iv) 60Â°+ x = 120Â° â‡’ x = 120Â°- 60Â° = 60Â°

(v) 30Â° + x = 80Â° â‡’ x = 80Â°- 30Â° = 50Â°

(vi) x + 35Â°= 75Â° â‡’ x = 75Â°- 35Â° = 40Â°

**Exercise 6.3 **

**Question 1: **

**Find the value of unknown x in the following diagrams: **

**Answer 1: **

(i) In Î”ABC,

âˆ BAC + âˆ ACB + âˆ ABC = 180Â° [ By angle sum property of a triangle]

â‡’ x + 50Â°+ 60Â° = 180Â°

â‡’ x + 110Â° = 180Â°

â‡’ x = 180Â°-110Â° = 70Â°

(ii) In Î”PQR,

âˆ RPQ + âˆ PQR + âˆ RPQ = 180Â° [By angle sum property of a triangle]

â‡’ 90Â°+30Â°+ x = 180Â°

â‡’ x+ 120Â° = 180Â°

â‡’ x= 180Â°-120Â°= 60Â°

(iii) In Î”XYZ,

âˆ ZXY + âˆ XYZ + âˆ YZX = 180Â° [By angle sum property of a triangle]

â‡’ 30Â° + 110Â° + x = 180Â°

â‡’ x + 140Â° = 180Â°

â‡’ x = 180Â°-140Â° = 40Â°

(iv) In the given isosceles triangle,

x+x + 50Â° = 180Â° [By angle sum property of a triangle]

â‡’ 2x+50Â°= 180Â°

â‡’ 2x = 180Â°- 50Â°

â‡’ 2x = 130Â°

â‡’ x = 130Â°/2 = 65Â°

(v) In the given equilateral triangle,

x +x+x = 180Â° [By angle sum property of a triangle]

â‡’ 3x = 180Â°

â‡’ x = 180Â°/3 = 60Â°

(vi) In the given right angled triangle,

x + 2x+90Â° = 180Â° [By angle sum property of a triangle)

â‡’ 3x+90Â° = 180Â°

â‡’ 3* = 180Â°- 90Â°

â‡’ 3x = 90Â°

â‡’ x = 90Â°/3 = 30Â°

**Question 2: **

**Find the values of the unknowns x and y in the following diagrams:**

**Answer 2: **

(i) 50Â° + x = 120Â° [Exterior angle property of a Î”]

â‡’ x = 120Â°-50Â° = 70Â°

Now, 50Â° + x + y =180Â° [Angle sum property of a Î”]

â‡’ 50Â° + 70Â° + y = 180Â°

â‡’ 120Â° + y= 180Â°

â‡’ y = 180Â° -120Â° = 60Â°

(ii) y = 80Â° ..... (i) [Vertically opposite angle]

Now, 50Â° + x + y =180Â° [Angle sum property of a Î”]

â‡’ 50Â° + 80Â°+y = 180Â° [From equation (i)]

â‡’ 130Â° + y = 180Â°

â‡’ y = 180Â° -130Â° = 50Â°

(iii) 50Â°+ 60Â° = x (Exterior angle property of a Î”]

x = 110Â°

Now 50Â° + 60Â°+ y = 180Â° [Angle sum property of a Î”]

â‡’ 110Â° + y = 180Â°

â‡’ y = 180Â° - 110Â°

â‡’ y = 70Â°

(iv) x = 60Â° ..... (i) [Vertically opposite angle]

Now, 30Â° + x + y = 180Â° [Angle sum property of a Î” ]

â‡’ 50Â° + 60Â° + y = 180Â° [From equation (i)]

â‡’ 90Â° + y = 180Â°

â‡’ y = 180Â° - 90Â° = 90Â°

(v) y = 90Â° ....(i) [Vertically opposite angle]

Now, y + x + x = 180Â° [Angle sum property of a Î”]

â‡’ 90Â° + 2x = 180Â° [From equation (i)]

â‡’ 2x = 180Â°- 90Â°

â‡’ 2x = 90Â°

â‡’ x = 90Â°/2 = 45Â°

(vi) x = y ...(i) [Vertically opposite angle]

Now, x + x + y = 180Â° [Angle sum property of a Î”]

â‡’ 2x+x = 180Â° [From equation (i)]

â‡’ 3x = 180Â°

â‡’ x = 180Â°/3 = 60Â°

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