Description : In right triangle ABC, right angled at C, if tan A = 1, then the value of 2 sin A cos A is (a) 0 (b) 1 (c) – 1 (d) 2
Last Answer : (b) 1
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.
Description : Name the type of triangle formed. (a) Right angled (b) Equilateral (c) Isosceles (d) Scalene
Last Answer : (d) Scalene
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
Last Answer : (b) Isosceles triangle
Description : Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th
Last Answer : Let the co-ordinates of any point on the x-axis be (x, 0). Then distance between (x, 0) and (– 4, 8) is 10 units.⇒ \(\sqrt{(x+4)^2+(0-8)^2}\) = 10 ⇒ x2 + 8x + 16 + 64 = 100 ⇒ x2 + 8x – 20 = 0 ⇒ (x + 10) (x – 2) = 0 ⇒ x = –10 or 2 ∴ The required points are (– 10, 0) and (2, 0).
Description : In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is: a 12/7 b 24/7 c 20/7 d 7/24
Last Answer : b 24/7
Description : The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is: (a) 12 units (b) 11 units (c) 5units (d) (7 + √5)units
Last Answer : (a) 12 units
Description : A (5,-1) , B (-3,-2) and C (-1,8) are the vertices of ∆ ABC . The median from A meets BC at D ,then the coordinates of the point D are: (a) (-4,6) (b) (2,7/2) (c) (1,-3/2) (d) (-2,3)
Last Answer : (d) (-2,3)
Description : The points (-2,-1),(a,0),(4,b),(1,2) are the vertices of a parallelogram taken in order , then the values of a and b are: (a) a = 1,b = 3 (b) a = 3 , b = 1 (c) a = 1 , b = 1 (d) a = 0 , b = 4
Last Answer : (a) a = 1,b = 3
Description : The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a: A(-3,-5) and(a)Square (b) Rhombus (c) Rectangle (d) Trapezium
Last Answer : (c) Rectangle
Description : The coordinates of the vertices of region bounded by lines 3x - y = 5; x+ 2y = 4 and Y-axis are: (a) A(2,0); B(5,0) and C(1,2) (b) A(2,0); B(-5,0) and C(1,2) (c) A(0,2); B(0,-5) and C(2,1) (d) A(0,2); B(0,5) and C(2,1)
Last Answer : (c) A(0,2); B(0,-5) and C(2,1)
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 : What is the value of h in a right angled triangle with 31° and 15cm?
Last Answer : Following the symbols in the image:Assuming 15cm corresponds to a (the line adjacent to the angle), then you need to use the cosine formulacos(ø) = a/hcos(31º) = 15cm/hh*cos(31º) = (15cm/h) * hh*cos(31º) = 15cm * 1h*cos(31º)/cos(31º) = 15cm/cos(31º)h*1 = 15cm/cos(31º)h = 16.398
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:
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
Description : Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th
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 : 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 : D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is (a) 2.5 (b) 3 (c) 5 (d) 6
Last Answer : (b) 3
Description : Areas of two similar triangles are 36 cm 2 and 100 cm 2 . If the length of a side of thelarger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is: (a) 12cm (b) 13cm (c) 14cm (d) 15cm
Last Answer : (a) 12cm
Description : The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th
Last Answer : In right-angled triangle ABC AB2 + BC2 = AC2 (By Pythagoras Theorem) ⇒ AB2 + 42 = 52 ⇒ AB2 = 25 – 16 = 9 5 cm ⇒ AB = 3 cm ∴ Area of △ABC = 1/2 BC x AB = 1/2 x 4 x 3 = 6cm2
Description : The centre of gravity of a plane lamina will not be at its geometrical centre if it is a a.Square b.Equilateral triangle c.Circle d.Rectangle e.Right angled triangle
Last Answer : e. Right angled triangle
Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th
Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB
Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th
Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.
Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th
Last Answer : NEED ANSWER
Last Answer : This answer was deleted by our moderators...
Description : ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th
Last Answer : Solution :-
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.
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 : A piece of paper is in the shape of a right-angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. -Maths 9th
Last Answer : (d) 14.365Given, ST || RQ∴ \(rac{ ext{Area of ΔSPT}}{ ext{Area of ΔRPQ}}\) = \(rac{ST^2}{RQ^2}\)Also, given ST = \(\bigg(1-rac{35}{100}\bigg)RQ\) = (0.65) RQ∴ \(rac{ST}{RQ}\) = 0.65 ⇒ \(\bigg(rac ... ΔRPQ}}\) = 0.4225 ⇒ \(rac{ ext{Area of ΔSPT}}{{34}}\) = 0.4225⇒ Area of ΔSPT = 0.4225 x 34 = 14.365
Description : A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th
Description : What is a right-angled triangle ?
Last Answer : : A triangle which has three right angles is called a right angled triangle. The sum of any two angles of a right-angled triangle is always greater than 900. The arms of a right-angled ... the center , and the perpendicular center of a right -angled triangle are all located inside the triangle.
Last Answer : : A triangle whose three angles are smaller than one right angle ( 90 0) is called a right-angled triangle.
Description : Is right angled triangle a right angle?
Last Answer : Yes and a right angle is 90 degrees
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
Description : The C.G. of a plane lamina will not be at its geometrical centre in the case of a (A) Right angled triangle (B) Equilateral triangle (C) Square (D) Circle
Last Answer : (A) Right angled triangle
Description : The sides of a triangle are in the ratio 3 : 5 : 7 and its perimeter is 30 cm. The length of the greatest side of the triangle in cm is (1) 6 (2) 10 (3) 14 (4) 16
Last Answer : (3) 14
Description : The values of the trigonometric ratios of an angle ———— with the lengths of the sides of the triangle, if the angle remains the same. a. vary b. Do not vary c. None of these D. Both
Last Answer : b. Do not vary
Description : An equilateral triangle has a side of length 4.68m. What it is perimeter ?
Last Answer : Since each side is equal, just multiply by 3.
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 : Without using Pythagoras’ theorem, show that the points A (0, 4), B(1, 2) and C(3, 3) are the vertices of a right angle triangle. -Maths 9th
Last Answer : Slope (m) = \(rac{(y_2-y_1)}{(x_2-x_1)}\) = \(rac{6-2}{5-1}\) = \(rac{4}{4}\) = 1Also slope (m) = tan θ, where θ is the inclination of the line to the positive direction of the x-axis in the anticlockwise direction. tan θ = 1 ⇒ θ = tan –11 = 45º.
Description : If ̳α ‘ and ̳β ‘ are the zeroes of a quadratic polynomial x 2 − 5x + b and α − β = 1, then the value of ̳b‘ is (a) – 5 (b) 6 (c) 5 (d) – 6
Last Answer : (b) 6
Description : If one zero of the quadratic polynomial x 2 + 3x + k is 2, then the value of k is (a) 10 (b) –10 (c) 5 (d) –5
Last Answer : (b) –10
Description : The point (-1,2) divides the line segment joining the points A(2,5) and B(x,y) in the ratio 3:4, then the value of x2 + y2 is : (a) 27 (b) 28 (c) 29 (d) 30
Last Answer : (c) 29
Description : If (x, y) is a solution of the following pair of linear equations in two variables, then the value of expression (√ is: x + 2y = 4 and 3x - y = 5 (a) 2 (b) 3 (c) 4 (d) √2
Last Answer : (a) 2
Description : If Log2 x - 5 Log x + 6 = 0, then what would the value(s) of x be?
Last Answer : Answer: x = e2 or e3.
Description : If (0, 0) and (2, 0) are the two vertices of a triangle whose centroid is (1, 1), then the area of the triangle is: -Maths 9th
Last Answer : (b) \(\bigg(rac{2\sqrt{13}+20\sqrt2}{\sqrt{13}+\sqrt{17}+5\sqrt2},rac{8\sqrt{13}-6\sqrt{17}}{\sqrt{13}+\sqrt{17}+5\sqrt2}\bigg)\)Let A(x1, y1), B(x2, y2), C(x3, y3) be the vertices of ΔABC the ... +6)^2}\) = \(\sqrt{4+196}\) = \(\sqrt{200}=10\sqrt{2}\)∴ Co-ordinates of incentre of Δ ABC are
Description : If A (-2, 4), B (0, 0) and C (4, 2) are the vertices of triangle ABC, then find the length of the median through the vertex A. -Maths 9th
Last Answer : D=slid ht of BC D≅(20+4,20+2) =(2,1) ∴ Length of median = Light of AD =root(−2−2)2+(4−1)2=root42+32=5 hope it helps thank u
Description : Which of the diagrams below represents the statement If it is an triangle then it has three vertices?
Last Answer : Figure Athree vertices-> triangle