Q1. In APQR, D is the midpoint of
____________
PD is_____________
Is QM = MR?
Ans: Given: QD = DR
is altitude.
PD is median.
No, QM (eq) MR as D is the midpoint of QR.
Q2. Draw rough sketches for the following:
(a) In ΔABC, BE is a median.
Ans: Here, BE is a median in ΔABC and AE = EC.
(b) In ΔPQR, PQ and PR are altitudes of the triangle.
Ans: Here, PQ and PR are the altitudes of the ΔPQR and RP ⊥ QP.
(c) In ΔXYZ, YL is an altitude in the exterior of the triangle.
Ans: YL is an altitude in the exterior of ΔXYZ.
Q3. Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same.
Ans:
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.
Q1. Find the value of the unknown exterior angle x in the following diagrams:
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 50° + 70°
= x = 120°
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 65° + 45°
= x = 110°
Ans: We Know That, An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 30° + 40°
= x = 70°
Ans: We Know That, An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 60° + 60°
= x = 120°
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 50° + 50°
= x = 100°
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x = 30° + 60°
= x = 90°
Q2. Find the value of the unknown interior angle x in the following figures:
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x + 50° = 115°
By transposing 50° from LHS to RHS it becomes – 50°
= x = 115° – 50°
= x = 65°
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= 70o + x = 100°
By transposing 70° from LHS to RHS it becomes – 70°
= x = 100° – 70°
= x = 30°
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right angled triangle. So the angle opposite to the x is 90°.
= x + 90° = 125°
By transposing 90° from LHS to RHS it becomes – 90°
= x = 125° – 90°
= x = 35°
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
= x + 60° = 120°
By transposing 60° from LHS to RHS it becomes – 60°
= x = 120° – 60°
= x = 60°
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right angled triangle. So the angle opposite to the x is 90°.
= x + 30° = 80°
By transposing 30° from LHS to RHS it becomes – 30°
= x = 80° – 30°
= x = 50°
Ans: We Know That,
An exterior angle of a triangle is equal to the sum of its interior opposite angles.
The given triangle is a right angled triangle. So the angle opposite to the x is 90°.
= x + 35° = 75°
By transposing 35° from LHS to RHS it becomes – 35°
= x = 75° – 35°
= x = 40°
Q1. Find the value of unknown x in the following diagrams:
Ans: In ΔABC,
∠ BAC + ∠ ACB + ∠ ABC = 180° [ By angle sum property of a triangle]
⇒ x + 50°+ 60° = 180°
⇒ x + 110° = 180°
⇒ x = 180°110° = 70°
Ans: In ΔPQR,
∠ RPQ + ∠ PQR + ∠ RPQ = 180° [By angle sum property of a triangle]
⇒ 90° + 30°+ x = 180°
⇒ x + 120° = 180°
⇒ x= 180°  120°= 60°
Ans: In ΔXYZ,
∠ ZXY + ∠ XYZ + ∠ YZX = 180° [By angle sum property of a triangle]
⇒ 30° + 110° + x = 180°
⇒ x + 140° = 180°
⇒ x = 180°  140° = 40°
Ans: 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°
Ans: In the given equilateral triangle,
x + x + x = 180° [By angle sum property of a triangle]
⇒ 3x = 180°
⇒ x = 180° / 3 = 60°
Ans: 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°
Q2. Find the values of the unknowns x and y in the following diagrams:
Ans: 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°
Ans: 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°
Ans: 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°
Ans: 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°
Ans: 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°
Ans: 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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