What are the perpendicular sides of a right triangle?

I got stuck in math in my country I don't know exactly how to call it a clamp or a clamp ??
by

1 Answer

Answer :

The enclosure is the longest side of the triangle, and the two shorter sides are the enclosure.
by

Related questions

In a right triangle we know a perpendicular 22.5 cm and a radius of a circle inscribed 4.5 cm. Determine the remaining sides and angles of the triangle.

Last Answer : I determined and what next? A circle inscribed in a triangle This is a circle that touches all sides of the triangle. The center of the circle inscribed in the triangle ABC is the intersection of the axes of the ... the 2nd series of the summer part of KMS 2009 / 2010.pdf example no.6" in Slovak

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

In an equilateral triangle if a, b and c denote the lengths of the perpendicular from A, B and C respectively on the opposite sides, then -Maths 9th

Description : In an equilateral triangle if a, b and c denote the lengths of the perpendicular from A, B and C respectively on the opposite sides, then -Maths 9th

Last Answer : b=2c=3​⇒⇒b=3​c=23​​cosA=2bcb2+c2−a2​⇒23​​=33+43​−a2​⇒233​​=415​−a2 ⇒a2=415​−43​​⇒a=1.673278 We know sinaa​=2R1​⇒R=2asina​=221​​=41​

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 : NEED ANSWER

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...

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.

How many perpendicular lines on a right angle triangle?

Description : How many perpendicular lines on a right angle triangle?

Last Answer : The 2 perpendicular sides of a right angle triangle intersect each other at right angles

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 perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Description : A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Last Answer : Let given right triangle be ABC. Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps 1.Draw the base BC = 3.5 cm 2.Make ... AB = BD - AD = BD - AC [from Eq. (i)] => BD = AB + AC Thus, our construction is justified.

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Description : A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Last Answer : Since, the given right angled triangle is revolved about the side 8 cm, it will form a Cone of radius 6cm and height 8cm. Volume of a cone = 1/3∏r2h = 1/3 3.14 6 6 8 = 301.44 cm3 Curved Surface area of a cone ... value of l in (i), we get Curved Surface area of a cone = 3.14 6 10 = 188.4 cm2

A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Description : A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm. -Maths 9th

Last Answer : Let given right triangle be ABC. Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps 1.Draw the base BC = 3.5 cm 2.Make ... AB = BD - AD = BD - AC [from Eq. (i)] => BD = AB + AC Thus, our construction is justified.

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Description : A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. -Maths 9th

Last Answer : Since, the given right angled triangle is revolved about the side 8 cm, it will form a Cone of radius 6cm and height 8cm. Volume of a cone = 1/3∏r2h = 1/3 3.14 6 6 8 = 301.44 cm3 Curved Surface area of a cone ... value of l in (i), we get Curved Surface area of a cone = 3.14 6 10 = 188.4 cm2

A right triangle ABC with sides 5 cm, -Maths 9th

Description : A right triangle ABC with sides 5 cm, -Maths 9th

Last Answer : Let ABC be a right triangle with AB = 12 cm, BC = 5 cm and AC = 13 cm. When △ABC is revolved about AB, it forms a right circular cone of radius BC = 5 cm and height AB = 12 cm. Thus, volume of cone formed = 1/3 πr2h = 1/3 x π x 52 x 12 = 100π cm3

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.

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

Last Answer : answer:

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

Last Answer : answer:

In a right triangle the hypotenuse is always shortest of the three sides?

Description : In a right triangle the hypotenuse is always shortest of the three sides?

Last Answer : False because the hypotenuse is always the longest side which isopposite the biggest angle of 90 degrees

In a right triangle the hypotenuse is always the shortest of the three sides.?

Description : In a right triangle the hypotenuse is always the shortest of the three sides.?

Last Answer : No, it is not the shortest side. The hypotenuse is the longest side. Also, it's always opposite the right angle.

Sides of triangles are (i) 3 cm, 4 cm, 6 cm. (ii) 4 cm, 5 cm, 6 cm. (iii) 7 cm, 24 cm, 25 cm (iv) 5 cm, 12 cm, 14 cm. Which of these is right triangle?(a) (i) (b) (ii) (c) (iii) (d) (iv)

Description : Sides of triangles are (i) 3 cm, 4 cm, 6 cm. (ii) 4 cm, 5 cm, 6 cm. (iii) 7 cm, 24 cm, 25 cm (iv) 5 cm, 12 cm, 14 cm. Which of these is right triangle?(a) (i) (b) (ii) (c) (iii) (d) (iv)

Last Answer : (c) (iii)

Ratios of sides of a right triangle with respect to its acute angles are knownas ————– a. Trigonometric Identities b. Trigonometric Ratios c. Trigonometry d. trigonometry formula

Description : Ratios of sides of a right triangle with respect to its acute angles are knownas ————– a. Trigonometric Identities b. Trigonometric Ratios c. Trigonometry d. trigonometry formula

Last Answer : b. Trigonometric Ratios

Find the area of a right angled triangle with sides of 90 degree unit and the functions described by L = cos y and M = sin x. a) 0 b) 45 c) 90 d) 180

Description : Find the area of a right angled triangle with sides of 90 degree unit and the functions described by L = cos y and M = sin x. a) 0 b) 45 c) 90 d) 180

Last Answer : d) 180

ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Description : ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Last Answer : Solution :-

Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Description : Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Last Answer : answer:

A quadrilateral has one pair of equal length sides and diagonals which are not perpendicular. Name the quadrilateral or quadrilaterals.?

Description : A quadrilateral has one pair of equal length sides and diagonals which are not perpendicular. Name the quadrilateral or quadrilaterals.?

Last Answer : trapezium

The centroid of a triangle is at the inter-section of a.perpendicular bisectors b.107 dynes c.base and perpendicular from opposite vertex d.perpendiculars down from the corners e.medians

Description : The centroid of a triangle is at the inter-section of a.perpendicular bisectors b.107 dynes c.base and perpendicular from opposite vertex d.perpendiculars down from the corners e.medians

Last Answer : e. medians

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

Description : Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

Last Answer : According to question prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side,

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

Description : Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

Last Answer : According to question prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side,

If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th

Description : If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th

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

A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Description : A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Last Answer : answer:

The medians AD and BE of the triangle with vertices A(0, b), B(0, 0) and C(a, 0) are mutually perpendicular if -Maths 9th

Description : The medians AD and BE of the triangle with vertices A(0, b), B(0, 0) and C(a, 0) are mutually perpendicular if -Maths 9th

Last Answer : (c) \(rac{b+k}{f+h}\)Let the slope of the lin passing through the points (-k, h) and (b, - f) be m1. Then m1 = \(rac{-f-h}{b+k}\) = \(-\bigg(rac{f+h}{b+k}\bigg)\)\(\bigg[Slope = rac{y_2-y_1}{x_2-x_1}\bigg]\) ... \(-rac{1}{m_1}\)= \(rac{-1}{-\big(rac{f+h}{b+k}\big)}\) = \(\bigg(rac{b+k}{f+h}\bigg)\)

The point A(0, 0), B(1, 7) and C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from -Maths 9th

Description : The point A(0, 0), B(1, 7) and C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from -Maths 9th

Last Answer : (b) 3x - y = 0 Given lines are 3x - y - 3 = 0 and 3x - y + 5 = 0. Line parallel to the given lines can be written as 3x - y + c = 0 ...(i) Let us taken a point, say ... 5c + 15 = - 3c + 15 ⇒ 8c = 0 ⇒ c = 0. Substituting c = 0 in (i), the required equation is 3x - y = 0.

Which term describes a line segment that connects a vertex of a triangle to a point on the line containing the opposite sideso that the line segment is perpendicular to that line?

Description : Which term describes a line segment that connects a vertex of a triangle to a point on the line containing the opposite sideso that the line segment is perpendicular to that line?

Last Answer : Altitude

The area of a plane triangle ABC, having its base AC and perpendicular height , is (A) ½ bh (B) ½ ba sin C (C) ½ bc sin A (D) All the above

Description : The area of a plane triangle ABC, having its base AC and perpendicular height , is  (A) ½ bh (B) ½ ba sin C (C) ½ bc sin A (D) All the above

Last Answer : (D) All the above

The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P, Q, R, S and T in such a way that the holes were parallel to each other, perpendicular to the edge and all the holes were still within the lens edge. What is the total number of edges in the lens after the holes were drilled?

Description : The figure shows the front view of a convex lens, which originally had only one edge. Five holes of different shapes, namely triangle, square, pentagon, hexagon and circle, were drilled through it at points P ... . What is the total number of edges in the lens after the holes were drilled? 

Last Answer : 57

The statement --- if forces acting on a point can be represented in magnitude and direction by the sides of a polygon taken in order then their resultant will be represented in magnitude and direction by the closing side of the polygon taken in opposite order ---- is known as a.Law of triangle of forces b.Newton's law of forces c.Law of polygon of forces d.D'Alembert's rule e.Lami's theorem

Description : The statement --- if forces acting on a point can be represented in magnitude and direction by the sides of a polygon taken in order then their resultant will be represented in magnitude and direction by the closing ... 's law of forces c.Law of polygon of forces d.D'Alembert's rule e.Lami's theorem

Last Answer : c. Law of polygon of forces

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). -Maths 9th

Last Answer : NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths ... = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m S

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Description : A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. -Maths 9th

Last Answer : Each shade of paper is divided into 3 triangles i.e., I, II, III 8 cm For triangle I: ABCD is a square [Given] ∵ Diagonals of a square are equal and bisect each other. ∴ AC = BD = 32 cm Height of AABD ... are: Area of shade I = 256 cm2 Area of shade II = 256 cm2 and area of shade III = 17.92 cm2

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Last Answer : Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm

The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

Description : The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

Last Answer : Let the sides of triangle be 3x,4x,5x Perimeter =3x + 4x + 5x=144 cm 12x=144 ∴x=12 Then sides of triangle are 3x=3 12=36 cm, 4x=4 12=48 cm, 5x=5 12=60 cm. Now, Semi perimeter, s=2 Sum of sides of ... , Area of triangle =s (s−a)(s−b)(s−c) = 72(72−36)(72−48)(72−60) = 72 36 24 12 = 864 cm2

Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Last Answer : Step-by-step explanation: ◾As we have given the two sides of triangle, let the three sides of triangle are (a) , (b), (c) . ◾And perimeter of given triangle is 10.5 cm ◾were, let us assume the sides are, ... . ◾So, the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 [Area ]=

Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 12 cm, 6 cm and 15 cm. -Maths 9th

Last Answer : Using the formulas A=s(s﹣a)(s﹣b)(s﹣c) s=a+b+c 2Solving forA A=1 4﹣a4+2(ab)2+2(ac)2﹣b4+2(bc)2﹣c4=1 4·﹣124+2·(12·6)2+2·(12·15)2﹣64+2·(6·15)2﹣154≈34.19704cm²

If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

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