The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) -Maths 9th

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, –2). The third vertex is (x, y) -Maths 9th
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Let A(x1, y1) = (3, 4), B(x2, y2) ≡ (0, 5), C(x3, y3) ≡ (2, –1)and D(x4, y4) ≡ (3, –2) be the vertices of quadrilateral ABCD.Area of quad. ABCD = \(rac{1}{2}\) |{(x1 y2 – x2 y1) + (x2y3 – x3y2) + (x3y4 – x4y3) + (x4y1 – x1y4)}|= \(rac{1}{2}\) |{(3 × 5 – 0 × 4) + (0 × (–1) – 2 × 5) + (2 × (–2) – 3 × (–1)) + (3 × 4 – 3 × (–2))}|= \(rac{1}{2}\) |{(15 – 0) + (0 – 10) + (– 4 + 3) + (12 + 6)}|= \(rac{1}{2}\) |{15 – 11 + 0 + 18}| = \(rac{1}{2}\)x 22 = 11 sq. units.
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Description : If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of third side of the triangle cannot be -Maths 9th

Last Answer : (d) Given, the length of two sides of a triangle are 5 cm and 1.5 cm, respectively. Let sides AB = 5 cm and CA = 1.5 cm We know that, a closed figure formed by three intersecting lines ( ... options (a), (b) and (c) satisfy the above inequality but option (d) does not satisfy the above inequality.

A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Description : A point within an equilateral triangle whose perimeter is 30 m is 2 m from one side and 3 m from another side. Find its distance from third side. -Maths 9th

Last Answer : answer:

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

Description : The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

Last Answer : Let the parallelogram be ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

Description : The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. -Maths 9th

Last Answer : Let the parallelogram be ABCD, in which ∠ADC and ∠ABC are obtuse angles. Now, DE and DF are two altitudes of parallelogram and angle between them is 60°.

Let the vertex of an angle ABC be located outside a circle -Maths 9th

Description : Let the vertex of an angle ABC be located outside a circle -Maths 9th

Last Answer : Given: AD = CE To prove: ∠ABC = 1/2(∠DOE - ∠AOC) In △AOD and △COE AD = CE (Given) AO = OC and DO = OE (Radii of same circle) ∴ △AOD ≅ △COE (By SSS congruence criterion) ⇒ ∠1 = ∠3, ∠2 = ∠4 (CPCT) ... = 1/2(4z + 4x - 360°) ....(vii) From (iv) and (vii), we have ∠BAC = 1/2(∠DOE - ∠AOC)

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC