Plane Geometry
- In ∆ ABC, AC = BC and ∠ABC = 50°, the side BC is produced to D so that BC = CD then the value of ∠BAD is
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We draw a figure triangle ABC whose the side BC is produced to D so that BC = CD ,
Given , AC = BC and ∠ABC = ∠BAC = 50°
∴ ∠ACB = 180° – 100° = 80°
∴ ∠ACD = 180° – 80° = 100Correct Option: C
We draw a figure triangle ABC whose the side BC is produced to D so that BC = CD ,
Given , AC = BC and ∠ABC = ∠BAC = 50°
∴ ∠ACB = 180° – 100° = 80°
∴ ∠ACD = 180° – 80° = 100∠CAD = ∠CDA = 80° = 40° 2
∴ ∠BAD = ∠BAC + ∠CAD
∠BAD = 50° + 40° = 90°
- ∆ ABC is a triangle, PQ is line segment intersecting AB in P and AC in Q and PQ || BC. The ratio of AP : BP = 3 : 5 and length of PQ is 18 cm. The length of BC is
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On the basis of question we draw a figure of triangle ABC in which PQ is line segment intersecting AB in P and AC in Q ,
Here , AP : BP = 3 : 5 and PQ = 18 cm
PQ || BC
∴ ∠APQ = ∠ABC
∠AQP = ∠ACB
By AA – similarity theorem,
∆APQ ~ ∆ABC∴ AB = BC AP PQ ⇒ AB - AP = BC - PQ AP PQ ⇒ BP = BC - PQ AP PQ ⇒ 5 = BC - 18 3 18
Correct Option: B
On the basis of question we draw a figure of triangle ABC in which PQ is line segment intersecting AB in P and AC in Q ,
Here , AP : BP = 3 : 5 and PQ = 18 cm
PQ || BC
∴ ∠APQ = ∠ABC
∠AQP = ∠ACB
By AA – similarity theorem,
∆APQ ~ ∆ABC∴ AB = BC AP PQ ⇒ AB - AP = BC - PQ AP PQ ⇒ BP = BC - PQ AP PQ ⇒ 5 = BC - 18 3 18 ⇒ BC - 18 = 5 × 18 = 30 3
⇒ BC = 30 + 18 = 48 cm.
- The mid-points of sides AB and AC of a triangle ABC are respectively X and Y. If (BC + XY) = 12 units, then the value of (BC – XY) is :
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As per the given in question , we draw a figure triangle ABC in which X and Y are the mid-points of sides AB and AC ,
Point X is the mid-point of AB. Point Y is the mid-Point of AC.
∴ XY || BC
∠AXY = ∠ABC
∠AYX = ∠ACB
By AA–similarity,
∆AXY ~ ∆ABC∴ AX = XY AB BC ⇒ AX = XY ⇒ BC = 2 2AX BC XY
By componendo and dividendo,BC + XY = 2 + 1 BC - XY 2 - 1 ⇒ 12 = 3 BC -XY
Correct Option: D
As per the given in question , we draw a figure triangle ABC in which X and Y are the mid-points of sides AB and AC ,
Point X is the mid-point of AB. Point Y is the mid-Point of AC.
∴ XY || BC
∠AXY = ∠ABC
∠AYX = ∠ACB
By AA–similarity,
∆AXY ~ ∆ABC∴ AX = XY AB BC ⇒ AX = XY ⇒ BC = 2 2AX BC XY
By componendo and dividendo,BC + XY = 2 + 1 BC - XY 2 - 1 ⇒ 12 = 3 BC -XY ⇒ BC – XY = 12 = 4 units. 3
- If the difference between the measures of the two smaller angles of a right angled triangle is 8°, then the smallest angle is :
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We draw a figure right angled triangle ABC ,
From triangle , ∠A = 90°
∴ ∠B + ∠C = 90°
Let, ∠B = y° and ∠C = (y + 8)°
∴ y + y + 8° = 90°
⇒ 2y = 90° – 8° = 82°Correct Option: B
We draw a figure right angled triangle ABC ,
From triangle , ∠A = 90°
∴ ∠B + ∠C = 90°
Let, ∠B = y° and ∠C = (y + 8)°
∴ y + y + 8° = 90°
⇒ 2y = 90° – 8° = 82°⇒ y = 82 = 41° = smallest angle 2
- Let O be the orthocentre of the triangle ABC. If ∠BOC = 150°, Then ∠BAC is
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We draw a figure triangle ABC whose O be the orthocentre ,
O ⇒ Ortho-centre
∠BOC = 150°
∠BOC = 180° – ∠ACorrect Option: A
We draw a figure triangle ABC whose O be the orthocentre ,
O ⇒ Ortho-centre
∠BOC = 150°
∠BOC = 180° – ∠A
⇒ ∠BAC = 180° – 150° = 30°
