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

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Mathematics
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In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 = AD2 + BC x DE + `(1)/(4)"BC"^2`

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

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 = AD2 - BC x CE + `(1)/(4)"BC"^2`

[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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In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2AD2 + `(1)/(2)"BC"^2`

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

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 - AB2 = 2BC x ED

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

In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2(AD2 + CD2)

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

A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. 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

AD is perpendicular to the side BC of an equilateral ΔABC. Prove that 4AD2 = 3AB2.

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

In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.
Prove that: 9AQ2 = 9AC2 + 4BC2 

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

In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.
Prove that: 9BP2 = 9BC2 + 4AC2

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

In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.
Prove that : 9(AQ2 + BP2) = 13AB2 

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

In the given figure, PQ = `"RS"/(3)` = 8cm, 3ST = 4QT = 48cm.
SHow that ∠RTP = 90°.

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

In a right-angled triangle ABC,ABC = 90°, AC = 10 cm, BC = 6 cm and BC produced to D such CD = 9 cm. Find the length of AD.

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

In the given figure. PQ = PS, P =R = 90°. RS = 20 cm and QR = 21 cm. Find the length of PQ correct to two decimal places.

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

In a right-angled triangle PQR, right-angled at Q, S and T are points on PQ and QR respectively such as PT = SR = 13 cm, QT = 5 cm and PS = TR. Find the length of PQ and PS.

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

PQR is an isosceles triangle with PQ = PR = 10 cm and QR = 12 cm. Find the length of the perpendicular from P to QR.

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

In a square PQRS of side 5 cm, A, B, C and D are points on sides PQ, QR, RS and SP respectively such as PA = PD = RB = RC = 2 cm. Prove that ABCD is a rectangle. Also, find the area and perimeter of the rectangle.

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

For the set of numbers given below, find mean: 5, 7, 8, 4, 6

[19] Mean and Median (For Ungrouped Data Only)
Chapter: [19] Mean and Median (For Ungrouped Data Only)
Concept: undefined >> undefined

For the set of numbers given below, find mean: 3, 0, 5, 2, 6, 2

[19] Mean and Median (For Ungrouped Data Only)
Chapter: [19] Mean and Median (For Ungrouped Data Only)
Concept: undefined >> undefined

Calculate man of the following: 4, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 11, 3

[19] Mean and Median (For Ungrouped Data Only)
Chapter: [19] Mean and Median (For Ungrouped Data Only)
Concept: undefined >> undefined

The weight of the seven members of a family, in kilograms are given below:
20, 52, 56, 72, 64, 13, 80.
Find mean weight.

[19] Mean and Median (For Ungrouped Data Only)
Chapter: [19] Mean and Median (For Ungrouped Data Only)
Concept: undefined >> undefined
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