Plane Geometry
- P and Q are centre of two circles with radii 9 cm and 2 cm respectively, where PQ = 17 cm. R is the centre of another circle of radius x cm, which touches each of the above two circles externally. If ∠PRQ = 90°, then the value of x is
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According to question , we draw a figure of two circles with radii 9 cm and 2 cm and R is the centre of another circle of radius x cm
Given , ∠PRQ = 90°
From figure , PR = 2 + x
PQ = 17
RQ = 9 + x
From ∆PRQ ,
∴ PQ² = PR² + RQ²
⇒ 17² = (2 + x)² + (9 + x)²
⇒ 289 = 4 + 4x + x² + 81 + 18x + x²
⇒ 289 = 2x² + 22x + 85
⇒ 2x² + 22x + 85 – 289 = 0
⇒ 2x² + 22x – 204 = 0Correct Option: B
According to question , we draw a figure of two circles with radii 9 cm and 2 cm and R is the centre of another circle of radius x cm
Given , ∠PRQ = 90°
From figure , PR = 2 + x
PQ = 17
RQ = 9 + x
From ∆PRQ ,
∴ PQ² = PR² + RQ²
⇒ 17² = (2 + x)² + (9 + x)²
⇒ 289 = 4 + 4x + x² + 81 + 18x + x²
⇒ 289 = 2x² + 22x + 85
⇒ 2x² + 22x + 85 – 289 = 0
⇒ 2x² + 22x – 204 = 0
⇒ x² + 11x – 102 = 0
⇒ x² + 17x – 6x – 102 = 0
⇒ x(x + 17) – 6 (x + 17) = 0
⇒ (x – 6) (x + 17) = 0
⇒ x = 6 as x ≠ –17
∴ x = 6 cm
- The radius of two concentric circles are 17cm and 10cm. A straight line ABCD intersects the larger circle at the point A and D and intersects the smaller circle at the points B and C. If BC = 12 cm, then the length of AD (in cm) is :
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As per the given in question , we draw a figure of two concentric circles whose radii are 17cm and 10cm
Here , BE = EC = 6 cm, OB = 10 cm, OA = 17 cm
From ∆ OBE,
OE = √OB² - BE²
OE = √10² - 6² = √16 × 4 = 8 cm
From ∆ OAE,
AE = √OA² - OE²Correct Option: C
As per the given in question , we draw a figure of two concentric circles whose radii are 17cm and 10cm
Here , BE = EC = 6 cm, OB = 10 cm, OA = 17 cm
From ∆ OBE,
OE = √OB² - BE²
OE = √10² - 6² = √16 × 4 = 8 cm
From ∆ OAE,
AE = √OA² - OE²
AE = √17² - 8²
AE = √25 × 9 = 15 cm
∴ AD = 2 × AE = 2 × 15 = 30 cm
- Triangle PQR circumscribes a circle with centre O and radius r cm such that ∠PQR = 90°. If PQ = 3 cm, QR = 4 cm, then the value of r is :
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According to question , we draw a figure of triangle PQR circumscribes a circle with centre O
Given that , ∠PQR = 90° , PQ = 3 cm, QR = 4 cm
PR² = PQ² + PR²
PR² = 3² + 4² = 2²
∴ PR = √25 = 5 cmWe know that , r = Area of triangle Semi perimeter of triangle r = 1 × 3 × 4 2 3 + 4 + 5 2
Correct Option: D
According to question , we draw a figure of triangle PQR circumscribes a circle with centre O
Given that , ∠PQR = 90° , PQ = 3 cm, QR = 4 cm
PR² = PQ² + PR²
PR² = 3² + 4² = 2²
∴ PR = √25 = 5 cmWe know that , r = Area of triangle Semi perimeter of triangle r = 1 × 3 × 4 2 3 + 4 + 5 2 r = 6 = 1 cm 6
Second method :
In the above figure, QR = r + 2 + r = 4
⇒ 2r = 4 - 2 = 2
⇒ r = 1
- If the circumradius of an equilateral triangle ABC be 8 cm, then the height of the triangle is
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As per the given in question , we draw a figure of an equilateral triangle ABC whose circumradius is 8 cm
As we know that ,Circum-radius of equilateral triangle = 2 × height 3 ∴ 8 = 2 × height 3
Correct Option: D
As per the given in question , we draw a figure of an equilateral triangle ABC whose circumradius is 8 cm
As we know that ,Circum-radius of equilateral triangle = 2 × height 3 ∴ 8 = 2 × height 3 ∴ Height = 8 × 3 = 12 cm. 2
- If the ∆ABC is right angled at B, find its circumradius if the sides AB and BC are 15 cm and 20 cm respectively.
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On the basis of question we draw a figure of a right angled triangle ABC and find its circumradius
Here , AB = 15 cm and BC = 20 cm
From ∆ABC ,
AC = √AB² + BC²
AC = √15² + 20²Correct Option: D
On the basis of question we draw a figure of a right angled triangle ABC and find its circumradius
Here , AB = 15 cm and BC = 20 cm
From ∆ABC ,
AC = √AB² + BC²
AC = √15² + 20²
AC = √225 + 400 = √625 = 25 cm.
∠ABC = 90°
As, AC = diameter
∴ Circum radius (OA) = 25 ÷ 2 = 12.5 cm.
