Which of the following are zero dimensional or one dimensional figuresA. RayB. SegmentC. AngleD. PlaneE. Point F. Line?

Which of the following are zero dimensional or one dimensional figuresA. RayB. SegmentC. AngleD. PlaneE. Point F. Line?
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A. RayB. SegmentE. PointF. Line
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ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Last Answer : Solution :-

A piece of paper is in the shape of a right-angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. -Maths 9th

Description : A piece of paper is in the shape of a right-angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. -Maths 9th

Last Answer : (d) 14.365Given, ST || RQ∴ \(rac{ ext{Area of ΔSPT}}{ ext{Area of ΔRPQ}}\) = \(rac{ST^2}{RQ^2}\)Also, given ST = \(\bigg(1-rac{35}{100}\bigg)RQ\) = (0.65) RQ∴ \(rac{ST}{RQ}\) = 0.65 ⇒ \(\bigg(rac ... ΔRPQ}}\) = 0.4225 ⇒ \(rac{ ext{Area of ΔSPT}}{{34}}\) = 0.4225⇒ Area of ΔSPT = 0.4225 x 34 = 14.365

If length ‘L’, force ‘F’ and time ‘T’ are taken as fundamental quantities, what would be the dimensional equation of mass and density?

Description : If length ‘L’, force ‘F’ and time ‘T’ are taken as fundamental quantities, what would be the dimensional equation of mass and density?

Last Answer : If length ‘L’, force ‘F’ and time ‘T’ are taken as fundamental quantities, what would be the dimensional equation of mass and density?

One dimensional fluid flow implies the (A) Flow in straight lines only (B) Uniform flow (C) Steady uniform flow (D) Flow in which transverse components are zero

Description : One dimensional fluid flow implies the (A) Flow in straight lines only (B) Uniform flow (C) Steady uniform flow (D) Flow in which transverse components are zero

Last Answer : (D) Flow in which transverse components are zero

A one dimensional flow is one which (A) Is uniform flow (B) Is steady uniform flow (C) Takes place in straight lines (D) Involves zero transverse component of flow

Description : A one dimensional flow is one which (A) Is uniform flow (B) Is steady uniform flow (C) Takes place in straight lines (D) Involves zero transverse component of flow

Last Answer : Answer: Option D

Is it possible for a right angled triangle with sides 3 and 4 units long to have a hypotenuse 6 units in length?

Description : Is it possible for a right angled triangle with sides 3 and 4 units long to have a hypotenuse 6 units in length?

Last Answer : answer:I'm not quite getting you. It isn't actually a triangle when the hypotenuse has these indentations, right? The hypotenuse isn't a straight line as you describe it. If the other sides are 3 ... and 5.00001, you don't have a straight line. Unless I'm misunderstanding what you're suggesting.

The centre of gravity of a plane lamina will not be at its geometrical centre if it is a a.Square b.Equilateral triangle c.Circle d.Rectangle e.Right angled triangle

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Last Answer : e. Right angled triangle

A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Description : A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Last Answer : When rt. ∠ed △ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm . Slant height of the cone is 5 cm. Volume of cone = 1 / 3 πr2h 1 / 3 22 / 7 3 3 4 = ... Total surface area of the solid = πr2 + πrl = πr (r + l ) = 22 / 7 3 8 = 75.43 cm2

On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Last Answer : According to question ∠BAC = ∠BDC.

A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Description : A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Last Answer : When rt. ∠ed △ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm . Slant height of the cone is 5 cm. Volume of cone = 1 / 3 πr2h 1 / 3 22 / 7 3 3 4 = ... Total surface area of the solid = πr2 + πrl = πr (r + l ) = 22 / 7 3 8 = 75.43 cm2

On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Last Answer : According to question ∠BAC = ∠BDC.

triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

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triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Last Answer : This answer was deleted by our moderators...

The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Description : The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Last Answer : In right-angled triangle ABC AB2 + BC2 = AC2 (By Pythagoras Theorem) ⇒ AB2 + 42 = 52 ⇒ AB2 = 25 – 16 = 9 5 cm ⇒ AB = 3 cm ∴ Area of △ABC = 1/2 BC x AB = 1/2 x 4 x 3 = 6cm2

If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Description : If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Last Answer : Let one of the remaining sides be x cm.Then, other side = \(\sqrt{5^2-x^2}\) cm∴ Area = \(rac{1}{2} imes{x} imes\sqrt{25-x^2}\) = 6⇒ \(x\sqrt{25-x^2}\) = 12 ⇒ x2(25 - x2) = 144⇒ 25x2 - x4 = 144 ⇒ x4 - 25x2 ... (x2 - 16) (x2 - 9) = 0 ⇒ x2 = 16 or x2 = 9 ⇒ x = 4 or 3∴ The two sides are 4 cm and 3 cm.

In a right-angled triangle ABC, D is the foot of the perpendicular from B on the hypotenuse AC -Maths 9th

Description : In a right-angled triangle ABC, D is the foot of the perpendicular from B on the hypotenuse AC -Maths 9th

Last Answer : Area of ΔABC = \(rac{1}{2}\) x 3 x 4 cm2 = 6 cm2. Also, AC = \(\sqrt{3^2+4^2}\) = 5 cm.∴ Area of ΔABC = \(rac{1}{2}\) x BD x AC ⇒ 6 = \(rac{1}{2}\) BD x 5 ⇒ BD = \(rac{12}{5}\) cm.Now in ΔABD, AD = \(\ ... \(rac{1}{2}\)x AD x BD = \(rac{1}{2}\) x \(rac{9}{5}\) x \(rac{12}{5}\) = \(rac{54}{25}\) cm2.

The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Description : The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Last Answer : (b) \(rac{q^2}{4}\) (2√3 + 1).AC = q, ∠ABC = 90º ⇒ q = \(\sqrt{AB^2+BC^2}\)⇒ q = \(\sqrt{2x^2}\)⇒ q2 = 2x2 ⇒ \(x\) = \(rac{q}{\sqrt2}\)∴ Area of the re-entrant hexagon = Sum of areas of (ΔABC + ΔADC ... (rac{\sqrt3}{4}\)q2 + \(rac{\sqrt3}{8}\)q2 + \(rac{\sqrt3q^2}{8}\) = \(rac{q^2}{4}\) (2√3 + 1).

Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is -Maths 9th

Description : Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is -Maths 9th

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ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

Description : ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

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If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Description : If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

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A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th

Description : A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th

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Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th

Description : Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th

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Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

Description : Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

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How many sides does a 2520 angled polygon have?

Description : How many sides does a 2520 angled polygon have?

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What is the value of h in a right angled triangle with 31° and 15cm?

Description : What is the value of h in a right angled triangle with 31° and 15cm?

Last Answer : Following the symbols in the image:Assuming 15cm corresponds to a (the line adjacent to the angle), then you need to use the cosine formulacos(ø) = a/hcos(31º) = 15cm/hh*cos(31º) = (15cm/h) * hh*cos(31º) = 15cm * 1h*cos(31º)/cos(31º) = 15cm/cos(31º)h*1 = 15cm/cos(31º)h = 16.398

If Chris Bromham took off on his Yamaha and jumped a horizontal distance of 74.0 m across a row of cars and assuming that he started and landed at the same level and was airborne for 1.3 s, what was his initial (angled) velocity as he left the ground?

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Description : What is a right-angled triangle ?

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What is the small index of refractive of the material of a right angled prism with equal sides for which a ray of light entering one of the sides norm

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If A(5,2), B(2,-2) and C(-2,t) are the vertices of a right angled triangle with ∠B = 900 , then the value of t is: (a) -1 (b) 1 (c) -2 (d) 2

Description : If A(5,2), B(2,-2) and C(-2,t) are the vertices of a right angled triangle with ∠B = 900 , then the value of t is: (a) -1 (b) 1 (c) -2 (d) 2

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