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(English Medium) ICSE Class 9 - CISCE Question Bank Solutions

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In an isosceles triangle ABC; AB = AC and D is the point on BC produced.

Prove that: AD2 = AC2 + BD.CD.

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

In figure AB = BC and AD is perpendicular to CD.
Prove that: AC2 = 2BC. DC.

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

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ABC is a triangle, right-angled at B. M is a point on BC.

Prove that: AM2 + BC2 = AC2 + BM2

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

Diagonals of rhombus ABCD intersect each other at point O.

Prove that: OA2 + OC2 = 2AD2 - `"BD"^2/2`

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

In the following figure, OP, OQ, and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.

Prove that: AR2 + BP2 + CQ2 = AQ2 + CP2 + BR2


[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

O is any point inside a rectangle ABCD.
Prove that: OB2 + OD2 = OC2 + OA2.

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC2 - AB2 = BC2 + CD2 + DA2

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

In a rectangle ABCD,
prove that: AC2 + BD2 = AB2 + BC2 + CD2 + DA2.

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

In triangle ABC, ∠B = 90o and D is the mid-point of BC.

Prove that: AC2 = AD2 + 3CD2.

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

M andN are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at Q.
Prove that:
(i) PM2 + RN2 = 5 MN2
(ii) 4 PM2 = 4 PQ2 + QR2
(iii) 4 RN2 = PQ2 + 4 QR2(iv) 4 (PM2 + RN2) = 5 PR2

[13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Chapter: [13] Pythagoras Theorem [Proof and Simple Applications with Converse]
Concept: undefined >> undefined

Construct a quadrilateral ABCD, when:
AB = 3.2 cm, BC = 5.2 cm, CD = 6.2 cm, DA = 4.2 cm and BD = 5.2 cm.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

Construct a quadrilateral ABCD, when:
AB = 7.2 cm, BC = 5.8 cm, CD = 6.3 cm, AD = 4.3 cm and angle A = 75°.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

Construct a quadrilateral ABCD, when:
∠ A = 90o, AB = 4.6 cm, BD = 6.4 cm, AC = 6.0 cm and CD = 4.2 cm.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

Construct a quadrilateral ABCD, when:
AB = 3.8 cm, AC = 4.8 cm, AD = 2.8 cm, angle A = 105° and angle B = 60°.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

Construct a quadrilateral ABCD, when:
BC = 7.5 cm AC = 5.8 cm, AD = 3.6 cm, CD = 4.2 cm and angle A = 120°.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

Construct a quadrilateral ABCD, when:
AD = AB = 4 cm, BC = 2.8 cm, CD = 2.5 cm and angle BAD = 45°.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

Construct a quadrilateral ABCD, when:
AB = 6.3 cm, BC = CD = 4.2 cm and ∠ABC = ∠BCD = 90°.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

Using ruler and compasses only, construct the quadrilateral ABCD, having given AB = 5 cm, BC = 2.5 cm CD = 6 cm, ∠BAD = 90o and diagonal BD = 5.5 cm.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

Using ruler and compasses only, construct the quadrilateral ABCD, having given AB = 5 cm, BC = 2.5 cm, CD = 6 cm. angle BAD = 90o and the diagonal AC = 5.5 cm.

[15] Construction of Polygons (Using Ruler and Compass Only)
Chapter: [15] Construction of Polygons (Using Ruler and Compass Only)
Concept: undefined >> undefined

The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of its altitudes be 6: 5: 4. Find the lengths of its sides.
Let the sides be x cm, y cm, and (37 - x - y) cm. Also, let the lengths of altitudes be 6a cm, 5a cm, and 4a cm.

[16] Area Theorems [Proof and Use]
Chapter: [16] Area Theorems [Proof and Use]
Concept: undefined >> undefined
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