what- A triangle is formed by the intersection of the lines y = 2x + 4, y = -x – 2, and x = 1.Is the triangle equilateral, isosceles, or scalene Graph the lines on grid paper to find the vertices of the triangle?

1 Answer

Answer :

scalene

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Last Answer : isosceles

Description : The points (-4, 0), (4, 0), (0, 3) are the vertices of a: (а) Right triangle (b) Isosceles triangle (c) Equilateral triangle (d) Scalene triangle

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Description : Name the type of triangle formed. (a) Right angled (b) Equilateral (c) Isosceles (d) Scalene

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Description : Can an isosceles triangle be scalene?

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Description : Can a isosceles triangle can be scalene.?

Last Answer : An isosceles triangle can never be a scalene triangle but it cantake the shape of a right angle triangle

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

Description : what- Isosceles triangle SIX has congruent sides SI and IX and vertices S at (x, 5), I at(-2, 2), and X at (4, -1), where x > 0.What is the value of x?

Last Answer : 4

Description : O is a point in the interior of a square ABCD such that OAB is an equilateral triangle.Show that △OCD is an isosceles triangle. -Maths 9th

Last Answer : Solution :-

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).

Description : In the diagram AB and AC are the equal sides of an isosceles triangle ABC, in which is inscribed equilateral triangle DEF. -Maths 9th

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Description : A pointed arch which forms isosceles or equilateral triangle, is generally known as (A) Three centred arch (B) Two centred arch (C) Lancet arch (D) Bull's eye arch

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Description : Prove that the points (1, –1) ((-1/2),(1/2)) and (1, 2) are the vertices of an isosceles triangle. -Maths 9th

Last Answer : (x, y) is equidistant from the points (2, 1) and (1, –2) ⇒ Distance between (x, y) and (2, 1) = Distance between (x, y) and (1, –2)⇒ \(\sqrt{(x-2)^2+(y-1)^2}\) = \(\sqrt{(x-1)^2+(y+2)^2}\)⇒ x2 – 4x + 4 + y2 – 2y + 1 = x2 – 2x + 1 + y2 + 4y + 4⇒ – 4x + 2x – 2y – 4y = 0 ⇒ –2x – 6y = 0 ⇒ x + 3y = 0

Description : An equilateral triangle is cut from its three vertices to form a regular hexagon. What is the percentage of area wasted? -Maths 9th

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Description : Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Last Answer : (c) \(rac{1}{10}\)Let S be the sample space.Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon= 6C3 = \(rac{6 imes5 imes4}{3 imes2}\) = 20.Let A : Event that the ... Required probability = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

Last Answer : NEED ANSWER

Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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Description : Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Last Answer : (a) 45º 3x + y - 7 = 0 ⇒ y = -3x + 7 ⇒ Slope (m1) = -3 x + 2y + 9 = 0 ⇒ y = \(rac{-x}{2}\) - \(rac{9}{2}\) ⇒ Slope (m2) = \(-rac{1}{2}\)If θ is the angle between the given lines, then tan θ = \(\ ... \bigg|rac{-rac{5}{2}}{1+rac{3}{2}}\bigg|\)= \(\bigg|rac{-rac{5}{2}}{rac{5}{2}}\bigg|\) = 1∴ θ = 45°.

Description : A student wrote the equations of the lines a and b drawn in the following graph as y =1 and 2x + 3y =6. Is he right? -Maths 9th

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Description : Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

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

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