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# NCERT Solutions(Part - 1) - The Triangle and Its Properties Class 7 Notes | EduRev

## Mathematics (Maths) Class 7

Created by: Praveen Kumar

## Class 7 : NCERT Solutions(Part - 1) - The Triangle and Its Properties Class 7 Notes | EduRev

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.
All you need of Class 7 at this link: Class 7

Exercise 6.1

Question 1:

In APQR, D is the mid-point of  ____________
PD is_____________
Is QM = MR? 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 ΔABC, BE is a median.

(b) In ΔPQR, PQ and PR are altitudes of the triangle.

(c) In ΔXYZ, YL is an altitude in the exterior of the triangle.

(a) Here, BE is a median in ΔABC and AE = EC. (b) Here, PQ and PR are the altitudes of the ΔPQR and RP ⊥ QP. (c) YL is an altitude in the exterior of ΔXYZ. Question 3:

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

Isosceles triangle means any two sides are same.

Take ΔABC 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:      (i) x = 50° + 70° = 120°
(ii) x = 65° + 45° = 110°
(iii) x = 30° + 40° = 70°
(iv) x = 60°+60° = 120°
(v) x = 50° + 50° = 100°
(vi) x = 60° + 30° = 90°

Question 2:

Find the value of the unknown interior angle x in the following figures:      (i)  x + 50° = 115°   ⇒    x = 115° - 50° = 65°
(ii)  70°+ x = 100°    ⇒   x = 100°- 70° =  30°
(iii)  x + 90° = 125°  ⇒  x = 120°- 90° =  35°
(iv)  60°+ x = 120°   ⇒  x = 120°- 60° = 60°
(v)   30° + x = 80°   ⇒   x = 80°- 30° = 50°
(vi)  x + 35°= 75°   ⇒ x = 75°- 35° = 40°

Exercise 6.3

Question 1:

Find the value of unknown x in the following diagrams:      (i) In ΔABC,
∠ BAC + ∠ ACB + ∠ ABC = 180°  [ By angle sum property of a triangle]
⇒ x + 50°+ 60° = 180°
⇒  x  + 110° = 180°
⇒  x = 180°-110° = 70°

(ii) In ΔPQR,
∠ RPQ + ∠ PQR + ∠ RPQ = 180° [By angle sum property of a triangle]
⇒  90°+30°+ x = 180°
⇒  x+ 120° = 180°
⇒   x= 180°-120°= 60°

(iii) In ΔXYZ,
∠ ZXY + ∠ XYZ + ∠ YZX = 180° [By angle sum property of a triangle]
⇒   30° + 110° + x = 180°
⇒  x + 140° = 180°
⇒ x = 180°-140° = 40°

(iv)  In the given isosceles triangle,
x+x + 50° = 180°   [By angle sum property of a triangle]
⇒  2x+50°= 180°
⇒  2x = 180°- 50°
⇒  2x = 130°
⇒ x = 130°/2 = 65°

(v) In the given equilateral triangle,
x +x+x = 180° [By angle sum property of a triangle]
⇒ 3x = 180°
⇒ x = 180°/3 = 60°

(vi) In the given right angled triangle,
x + 2x+90° = 180°    [By angle sum property of a triangle)
⇒  3x+90° = 180°
⇒  3* = 180°- 90°
⇒ 3x = 90°
⇒ x = 90°/3 = 30°

Question 2:

Find the values of the unknowns x and y in the following diagrams:      (i) 50° + x = 120°    [Exterior angle property of a Δ]
⇒   x = 120°-50° = 70°
Now, 50° + x + y =180°    [Angle sum property of a Δ]
⇒  50° + 70° + y = 180°
⇒  120° + y= 180°
⇒  y = 180° -120° = 60°

(ii) y = 80° ..... (i)    [Vertically opposite angle]
Now, 50° + x + y =180°    [Angle sum property of a Δ]
⇒  50° + 80°+y = 180°    [From equation (i)]
⇒ 130° + y = 180°
⇒ y = 180° -130° = 50°

(iii) 50°+ 60° = x    (Exterior angle property of a Δ]
x = 110°
Now 50° + 60°+ y = 180°    [Angle sum property of a Δ]
⇒  110° + y = 180°
⇒  y = 180° - 110°
⇒  y = 70°

(iv) x = 60° ..... (i) [Vertically opposite angle]
Now, 30° + x + y = 180°    [Angle sum property of a Δ ]
⇒  50° + 60° + y = 180°    [From equation (i)]
⇒ 90° + y = 180°
⇒  y = 180° - 90° = 90°

(v) y = 90°   ....(i)    [Vertically opposite angle]
Now, y + x + x = 180°    [Angle sum property of a Δ]
⇒  90° + 2x = 180°    [From equation (i)]
⇒ 2x = 180°- 90°
⇒  2x = 90°
⇒ x = 90°/2 = 45°

(vi) x = y  ...(i)   [Vertically opposite angle]
Now, x + x + y = 180°    [Angle sum property of a Δ]
⇒ 2x+x = 180°    [From equation (i)]
⇒ 3x = 180°
⇒ x = 180°/3 = 60°

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