1 Answer

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 angles of the triangle ABC (and always lies inside the triangle!). Radius is the distance from the center to either side of the triangle. how to calculate, you will find in "Sample solutions of the 2nd series of the summer part of KMS 2009 / 2010.pdf example no.6" in Slovak   

Related questions

Description : What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? -Maths 9th

Last Answer : (d) 6 cmLet a = 35 cm, b = 44 cm, c = 75 cm. Thens = \(rac{a+b+c}{2}\) = \(rac{34+44+75}{2}\) cm = 77 cm∴ Area if triangle = \(\sqrt{77(77-35)(77-44)(77-75)}\) cm2= \(\sqrt{77 imes42 ... ) cm2 = 462 cm2∴ Radius of incircle = \(rac{ ext{Area}}{ ext{semi-perimeter}}\) = \(rac{462}{77}\) cm = 6 cm.

Description : Is the shortest distance from the center of the inscribed circle to the triangle sides is the circles?

Last Answer : It is its inradius.

Description : In the following figure a wire bent in the form of a regular polygon of `n` sides is inscribed in a circle of radius `a`. Net magnetic field at centre

Last Answer : In the following figure a wire bent in the form of a regular polygon of `n` sides is inscribed in a circle of ... (ni)/(a)mu_(0)tan.(pi)/(n)` D.

Description : An isosceles triangle of vertical angle `2theta` is inscribed in a circle of radius `a` . Show that the area of the triangle is maximum when `theta=pi

Last Answer : An isosceles triangle of vertical angle `2theta` is inscribed in a circle of radius `a` . Show that the area of ... pi/4` C. `pi/6` D. None of these.

Description : If the area of a circle, inscribed in an equilateral triangle is 4π cm^2, then what is the area of the triangle? -Maths 9th

Last Answer : (a) 12√3 cm2Since area of circle = 4π ⇒ πr2 = 4π ⇒ r = 2 cmIn ΔOAD,tan 30° = \(rac{OD}{AD}\) ⇒ \(rac{1}{\sqrt3}\) = \(rac{2}{AD}\)⇒ AD = 2√3 cm ∴ AB = 2AD = 4√3 cm∴ Area of equilateral ΔABC = \(rac{\sqrt3}{4}\) (AB)2= \(rac{\sqrt3}{4}\) (4√3)2 = 12√3 cm2.

Description : If the sides of a triangle are 3 cm, 4 cm and 5 cm, then what is the radius of the circum-circle? -Maths 9th

Last Answer : Semi-perimeter of triangle (s) = \(rac{3+4+5}{2}\)cm = 6 cm∴ Area of triangle A = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{6 imes3 imes2 imes1}\) cm2 = 6 cm2∴ Radius of circum-circle = \(rac{abc}{4( ext{Area of}\,\Delta)}\) = \(rac{3+4+5}{4 imes60}\) cm = 2.5 cm

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.

Description : If two chords intersect inside a circle the angles formed are called inscribed angles.?

Last Answer : yes

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

Last Answer : answer:

Description : A circle is inscribed in a polygon. As the number of sides increases, the difference in areas of circle and polygon _________.

Last Answer : A circle is inscribed in a polygon. As the number of sides increases, the difference in areas of circle and polygon _________.

Description : A circle is inscribed in an equilateral triangle of side a. What is the area of any square inscribed in this circle? -Maths 9th

Last Answer : (c) \(rac{a^2}{6}.\)If a' is length of the side of ΔABC, thenArea of ΔABC = \(rac{\sqrt3}{4}\,a^2\)semi-perimeter of ΔABC = \(rac{3a}{2}\)∴ Radius of in-circle = \(rac{ ext{Area}}{ ext{semi-perimeter}}\) = \( ... {( ext{diagonal})^2}{2}\) = \(rac{\big(rac{a}{\sqrt3}\big)^2}{2}\) = \(rac{a^2}{6}.\)

Description : Find the area of an equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid-points -Maths 9th

Last Answer : (d) \(rac{3\sqrt3a^2}{32}\)Let AB = a be the side of the outermost square.Then AG = AH = \(rac{a}{2}\)⇒ GH = \(\sqrt{rac{a^2}{4}+rac{a^2}{4}}\) = \(rac{a}{\sqrt2}\)∴ Diameter of circle = \(rac{a} ... rac{\sqrt3}{2}\) = \(rac{\sqrt3a^2}{32}\)∴ Area of ΔPQR = 3 (Area of ΔPOQ) = \(rac{\sqrt3a^2}{32}\)

Description : Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Last Answer : STEP1: Draw a circle with centre at point O and radius 5 cm. STEP2: Draw its cord AB. STEP3: With centre A as centre and radius more than half of AB, draw two arcs, one on each side ... is perpendicular bisector of AB which is chord of circle, Hence, it passes through the centre of the circle.

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

Last Answer : According to question the radius of the circle passing through the points A, B and C .

Description : In a sphere of radius 2 cm a cone of height 3 cm is inscribed. What is the ratio of volumes of the cone and sphere ? -Maths 9th

Last Answer : answer:

Description : The area of the circle that can be inscribed in a square of side 6 cm is (a) 36 π cm 2 (b) 18 π cm 2 (c) 12 π cm 2 (d) 9 π cm 2

Last Answer : (d) 9 π cm 2

Description : If the lengths of a sides of a triangle are in the ratio 4 : 5 : 6 and the in-radius of the triangle is 3 cm, -Maths 9th

Last Answer : (b) 7.5 cm.Area of a triangle = \(rac{1}{2}\)x base x height= In-radius x semi-perimeter of the Δ \(\big[ ext{Using r =}rac{\Delta}{s}\big]\)Let the sides of triangle be 4x, 5x and 6x respectively. Given: In-radius = 3 ... x \(rac{15x}{2}\) = \(rac{6x}{2}\) x h ⇒ h = \(rac{45}{6}\) = 7.5 cm.

Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm

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

Last Answer : The enclosure is the longest side of the triangle, and the two shorter sides are the enclosure.

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

Description : A square is inscribed in an isosceles right triangle, so that the square and the triangle have one angle common. -Maths 9th

Last Answer : Given In isosceles triangle ABC, a square ΔDEF is inscribed. To prove CE = BE Proof In an isosceles ΔABC, ∠A = 90° and AB=AC …(i) Since, ΔDEF is a square. AD = AF [all sides of square are equal] … (ii) On subtracting Eq. (ii) from Eq. (i), we get AB – AD = AC- AF BD = CF ….(iii)

Description : A square is inscribed in an isosceles right triangle, so that the square and the triangle have one angle common. -Maths 9th

Last Answer : Given In isosceles triangle ABC, a square ΔDEF is inscribed. To prove CE = BE Proof In an isosceles ΔABC, ∠A = 90° and AB=AC …(i) Since, ΔDEF is a square. AD = AF [all sides of square are equal] … (ii) On subtracting Eq. (ii) from Eq. (i), we get AB – AD = AC- AF BD = CF ….(iii)

Description : In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Last Answer : answer:

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:

Description : An electron is travelling at right angles to a uniform magnetic field of flux density 1.2 mT with a speed of 8 × 106 m s-1, the radius of circular path followed by electron is A. 3.8 cm B. 3.7 cm C. 3.6 cm D. 3.5 cm

Last Answer : 3.8 cm

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.

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.

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​

Description : Two concentric coils each of radius equal to `2 pi cm` are placed at right angles to each other ` 3 ampere and 4 ampere` are the currents flowing in e

Last Answer : Two concentric coils each of radius equal to `2 pi cm` are placed at right angles to each other ` 3 ampere and 4 ... ` C. `7xx10^(-5)` D. `10^(-5)`

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.

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.

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

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)

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

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

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

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

Description : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the circle solve the following :

Last Answer : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the ... D. `x^(2)+y^(2)+6x+1=0`

Description : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the circle solve the following :

Last Answer : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the ... 0` D. `x^(2)+y^(2)-8y=0`

Description : 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

Last Answer : (A) a = S.R/L

Description : Prove that angles opposite to equal sides of a triangle are equal. -Maths 9th

Last Answer : Solution :-

Description : If the angles of a triangle are in the ratio 1 : 2 : 3, then find the ratio of the corresponding opposite sides. -Maths 9th

Last Answer : answer:

Description : If two sides of one triangle are equal to two sides of another triangle and the contained angles are supplementary, show that the two sides are equal in area -Maths 9th

Last Answer : If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.

Description : In what way does knowing the measure of the three sides of a triangle tell you more than just knowing the measures of the three angles?

Last Answer : Ion kn

Description : What is the missing side of a triangle if two sides are 18.5cm and 11.1cm whose opposite angles add up to 60 degrees?

Last Answer : From the given information the missing side must be oppositeangle 120 degrees and by using the cosine rule in trigonometry themissing side works out as 25.9cm

Description : What is the length of the 3rd side of a triangle having sides of 17.1cm and 28.8cm whose opposite angles add up to 72 degrees?

Last Answer : The largest angle then is 108 degrees that is opposite the 3rdside which is the longest side and by using the cosine rule intrigonometry it is 37.77cm in length rounded to two decimalplaces

Description : In what way does knowing the measure of the three sides of a triangle tell you more than just knowing the measures of the three angles?

Last Answer : Ion kn

Description : If two angles of a triangle are congruent then two sides opposite those angles are congruent?

Last Answer : That is correct as in the case of an isosceles triangle

Description : What are the angles and area of a triangle with sides of 2.60cm by 2.85cm by 4.70cm?

Last Answer : Let angle A be opposite side a and between sides b and c; using the cosine rule you can find each angle:a² = b² + c² - 2bc cos A→ A = arc cos((b² + c² - a²)/2bc)→ Angles are:arc cos((2.85² + 4.7² - 2.6²)/(2 ... = 180° - 60.1° = 119.1°Area triangle = ½bc sin A ≈ ½ 2.85 4.7 sin(28.9°) ≈ 3.24 cm²