What is the length of a perpendicular line drawn from one vertex to the opposite side?

What is the length of a perpendicular line drawn from one vertex to the opposite side?
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Last Answer : Let the x-intercept = a. Then y-intercept = -1 - a The equation of the required line is \(rac{x}{a}\) + \(rac{y}{-1-a}\) = 1Given, it passes through (4, 3), so,\(rac{4}{a}\) + \(rac{3}{-1-a}\) = 1⇒ - 4 - 4a ... 2}\) - \(rac{y}{3}\) = 1When a = -2, required line is \(rac{x}{-2}\) + \(rac{y}{3}\) = 1

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Last Answer : Weknowthatthediagonalofaparallelogram(∥gm)dividesitintotwocongruenttriangles.SoAreaof∥gmABCD=2 Areaof△BCD.AccordingtoHeron′sformulathearea(A)oftrianglewithsidesa,b&cisgivenasA=2[s(s−a)(s−b) ... 90=180Areaof∥gm=base heightHeightofaltitudefromvertexAonsideCDoftheof∥gm=baseCDareaof∥gmABCD =12180 =15cm

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Last Answer : =3 x 3 x 5 x 2 cm2 Area of parallelogram ABCD = 2 x 90 = 180 cm2 (ii) Let altitude of a parallelogram be h. Also, area of parallelogram ABCD =Base x Altitude ⇒ 180 = DC x h [from Eq. (ii)] ... h ∴ h = 180/12= 15 cm Hence, the area of parallelogram is 180 cm2 and the length of altitude is 15 cm.

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Last Answer : The lines are perpendicular to a common line AB. Hence, the angle between these lines will be 90+90=180∘ Since, the angle between them is 180∘, the lines are parallel to each other.

Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. -Maths 9th

Description : Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. -Maths 9th

Last Answer : To draw a line perpendicular to AB through A and B, respectively. Use the following steps of construction. 1.Draw a line segment AB = 4 cm. 2.Taking 4 as centre and radius more than ½ AB (i.e ... [since, it sum of interior angle on same side of transversal is 180°, then the two lines are parallel]

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Last Answer : (d) \(rac{17}{\sqrt{13}};\) 17 sq. unitsEquation of line BC : y - 7 = \(rac{1-7}{5-1}(x-1)\)⇒ y - 7 = \(rac{-6}{4}(x-1)\)⇒ 2 (y - 7) = -3 (x - 1) ⇒ 2y - 14 = -3x + 3 ⇒ 3x + 2y - 17 = 0. ∴ ... \) = \(2\sqrt{13}\)∴ Required area = \(rac{1}{2}\) x \(rac{17}{\sqrt{13}}\) x \(2\sqrt{13}\) = 17 sq. units.

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Last Answer : (b) \(rac{\sqrt{17}}{2}\)Equation of the line through the points (5, 0) and (0, 3) y - 0 = \(rac{3-0}{0-5}\) (x - 5)⇒ y = \(rac{-3}{5}\)(x - 5)⇒ 5y + 3x - 15 = 0 ∴ Distance of perpendicular from ... (rac{|20+12-15|}{\sqrt{25+9}{}}\) = \(rac{17}{\sqrt{34}}\) units. = \(rac{\sqrt{17}}{2}\) units.

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If S, L and R are the arc length, long chord and radius of the sliding circle then the perpendicular distance of the line of the resultant cohesive force, is given by (A) a = S.R/L (B) a = L.S/R (C) a = L.R/S (D) None of these

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Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

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Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

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