Plane Geometry
- The diagonals AC and BD of a cyclic quadrilateral ABCD intersect each other at the point P. Then, it is always true that
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On the basis of question we draw a figure of cyclic quadrilateral ABCD ,
∴ AP . PC = BP. DPCorrect Option: B
On the basis of question we draw a figure of cyclic quadrilateral ABCD ,
∴ AP . PC = BP. DP
This theorem is true for cyclic quadrilateral ABCD in which diagonals AC and BD intersect each other at the point P .
- A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC ⊥ BD, ∠CAD = θ. Then the angle ∠ABC =
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On the basis of question we draw a figure of cyclic quadrilateral ABCD,
∠B + ∠D = 180° and ∠A + ∠C = 180°
⇒ ∠BAC = ∠BCA
∠DAC = ∠DCA
∴ ∠DAB = ∠DCB = 90° , ∠DAC = θCorrect Option: C
On the basis of question we draw a figure of cyclic quadrilateral ABCD,
∠B + ∠D = 180° and ∠A + ∠C = 180°
⇒ ∠BAC = ∠BCA
∠DAC = ∠DCA
∴ ∠DAB = ∠DCB = 90° , ∠DAC = θ
∴ ∠ADE = 90° – θ = ∠CDE
∴ ∠ABC = 180° – 2(90° – θ)
∠ABC = 180° – 180° + 2θ = 2θ
- The perimeters of two similar triangles ∆ ABC and ∆ PQR are 36 cm and 24 cm respectively. If PQ = 10 cm, then AB is
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As per the given in question , we draw figures of two similar triangles PQR and ABC
In similar triangles ∆ ABC and ∆ PQR,AB = BC = AC PQ QR PR = AB + BC + AC PQ + QR + PR ⇒ AB = 36 = 3 10 24 2
Correct Option: A
As per the given in question , we draw figures of two similar triangles PQR and ABC
In similar triangles ∆ ABC and ∆ PQR,AB = BC = AC PQ QR PR = AB + BC + AC PQ + QR + PR ⇒ AB = 36 = 3 10 24 2 ⇒ AB = 3 × 10 = 15 cm. 2
- If the measure of three angles of a triangle are in the ratio 2 : 3 : 5, then the triangle is :
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Let Angles of triangle = 2k°, 3k° and 5k°
∴ 2k° + 3k° + 5k° = 180°
⇒ 10k° = 180°k° = 180 = 18° 10
Correct Option: A
Let Angles of triangle = 2k°, 3k° and 5k°
∴ 2k° + 3k° + 5k° = 180°
⇒ 10k° = 180°k° = 180 = 18° 10
Angles of triangle = 2k = 2 × 18 = 36°,
3k = 3 × 18 = 54°,
5k = 5 × 18 = 90°,
Hence, it is a right angled triangle.
- ∆ABC is a right angled triangle with AB = 6 cm, AC = 8 cm, ∠BAC = 90°. Then the radius of the incircle is
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As per the given in question , we draw a figure right-angled triangle BAC
Given , AB = 6 cm, AC = 8 cm , ∠BAC = 90°
From ∆ ABC,
BC = √AB² + AC²
BC = √6² + 8²
BC = √36 + 64
BC = √100 = 10 cm.∴ Semi–perimeter of ∆ ABC = s = 6 + 8 + 10 = 24 = 12 cm. 2 2 Area of ∆ ABC = 1 × AC × AB = 1 × 8 × 6 = 24 Sq. cm. 2 2
Correct Option: B
As per the given in question , we draw a figure right-angled triangle BAC
Given , AB = 6 cm, AC = 8 cm , ∠BAC = 90°
From ∆ ABC,
BC = √AB² + AC²
BC = √6² + 8²
BC = √36 + 64
BC = √100 = 10 cm.∴ Semi–perimeter of ∆ ABC = s = 6 + 8 + 10 = 24 = 12 cm. 2 2 Area of ∆ ABC = 1 × AC × AB = 1 × 8 × 6 = 24 Sq. cm. 2 2 ∴ In-radius = ∆ = 24 = 2cm. s 12
