Description : The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is -Maths 9th
Last Answer : s= 2 4+4+2 =5 Area of the triangle Δ= s(s−a)(s−b)(s−c) = 5(5−4)(5−4)(5−2) = 15 cm 2
Description : Find the area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm. -Maths 9th
Description : In an isosceles triangle, the measure of each of equal sides is 10 cm and the angle between them is 45º. The area of the triangle is: -Maths 9th
Last Answer : (c) 25√2 cm2.ΔABC is an isosceles triangle with AB = AC = 10 cm. ∠A = 45° ∴ Area of ΔABC= \(rac{1}{2}\) x 10 x 10 x sin 45°[Using Δ = \(rac{1}{2}\) bc sin A]= \(rac{50}{\sqrt2}\) = \(rac{50}{\sqrt2}\) x \(rac{\sqrt2}{\sqrt2}\) = 25√2 cm2.
Description : Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th
Last Answer : If h is the height of the triangle, then h 2 =y 2 − 4 x 2 ⇒h= 4 4y 2 −x 2 cm ∴Area= 2 1 ×base×h = 2 x 4 4y 2 −x 2 cm 2
Description : Each of the equal sides of an isosceles triangle is 5 cm greater than the base. The perimeter is 46 Centimeters. What is the length of each side of the triangle?
Last Answer : a is one of the equal sides of the iscosceles triangle b is the base perimeter is a + a + b = 46cm a = b + 5cm subsitute a for b + 5cm in the perimeter equation b + 5cm + b + 5cm + b = 46cm ... = 12cm + 5cm a = 17cm So you have 2 sided with the length of 17cm and the base with the length of 12cm
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
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
Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th
Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]
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.
Description : The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. -Maths 9th
Last Answer : Area of the triangle =
Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th
Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.
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 : One side of an equilateral triangle is 24 cm. The mid-points of its sides are joined to form another triangle whose mid-points -Maths 9th
Last Answer : Perimeter of the largest (outermost) equilateral triangle = 3 24 = 72 cm. Now, the perimeter of the triangle formed by joining the midpoints of a given triangle will be half the perimeter of the original triangle. ∴ Required sum = 72 + ... -rac{1}{2}}\) = \(rac{72}{rac{1}{2}}\) = 72 x 2 = 144 cm.
Description : The perimeter of an isosceles triangle is 15 cm -Maths 9th
Last Answer : Yes, 2b + a = 15 ⇒ 25 + 7 = 15 ⇒ b = 14 ∴ Area of isosceles triangle = 7/4 root under( √4b2 - a2) = 7/4 root under( √4 x 42 - 72) = 7/4 root under( √64 - 49) = 7/4. √15 cm2 Curiosity, knowledge, truthfulness.
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
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 ]=
Description : Construct a triangle whose sides are 3.6 cm , 3.0 cm and 4. 8 cm. Bisect the smallest angle and .measure each part. -Maths 9th
Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. Draw a line segment BC of length 4.8 cm. From B, point A is at a distance of 3.6 cm. ... at P. Joining BP, we obtain angle bisector of ∠B. Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = ½ x 139° = 19.5°
Last Answer : To construct a triangle ABC in which AB = 3.6 cm, AC = 3.0 cm and BC = 4. 8 cm, use the following steps. 1.Draw a line segment BC of length 4.8 cm. 2.From B, point A is at a distance of 3.6 ... 3.Joining BP, we obtain angle bisector of ∠B. 4.Flere, ∠ABC=39° Thus, ∠ABD = ∠DBC = 1/2 x 139° = 19.5°
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 : what- Two sides of a triangle measure 8 cm and 15 cm.What is the least possible length of the third side if the length is a whole number?
Last Answer : 8cm
Description : An isosceles right triangle has area 8 cm2. The length of its hypotenuse is -Maths 9th
Last Answer : (a) Given, area of an isosceles right triangle = 8 cm2 Area of an isosceles triangle = 1/2 (Base x Height) ⇒ 8 = 1/2 (Base x Base) [∴ base = height, as triangle is an ... √32 cm [taking positive square root because length is always positive] Hence, the length of its hypotenuse is √32 cm.
Last Answer : This answer was deleted by our moderators...
Description : An isosceles right triangle has area 8 cm2 . Find the length of its hypotenuse. -Maths 9th
Last Answer : Area = 1/2a2 ⇒ 1/2a2 = 8 ⇒ a2 = 16 cm ⇒ a = 4 cm Hypotenuse = √2a = √2.4 = 4√2 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 : The sides of a triangle are 8 cm, 15 cm and 17 cm. Find its area. -Maths 9th
Last Answer : Let a = 8cm, b = 15cm, c = 17cm s = (a + b + c)/2 = (8 + 15/ + 17)/2 = 40/2 = 20cm ∴ Area = root under √s(s - a)(s - b)(s - c) = root under √20(20 - 8)(20 - 15)(20 - 17) = root under √20 x 12 x 5 x 3 = 60 cm2
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
Description : Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. -Maths 9th
Last Answer : Steps of Construction (i) Draw BC = 12 cm. (ii) Construct ÐCBY = 90°. (iii) From ray BY, cut-off line segment BD = 18 cm. (iv) Join CD. (v) Draw the perpendicular bisector of CD intersecting BD at A. (vi ... = AC Now, BD = BA + AD ⇒ BD = AB + AC Hence, △ABC is the required triangle.
Description : The perimeter of an isosceles triangle is 32 cm. -Maths 9th
Last Answer : Let each of the equal side of isosceles triangle = 3x cm and base of isosceles triangle = 2x cm ∴ Perimeter = 3x + 3x + 2x 32 = 8x ⇒ x = 4 ∴ Sides are 3 x 4,3 x 4, 2 x 4 i.e., 12 cm, 12 cm, 8 cm Now, ... c)) = under root(√16(16 - 12)(16 - 12)(16 - 8)) = under root (√16 x 4 x 4 x 8) = 32√2 cm2
Description : Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side -Maths 9th
Last Answer : Solution :-
Description : If circles are drawn taking two sides of a triangle as diameter, prove that the point of intersection of these circles lie on the third side. -Maths 9th
Last Answer : Solution :- Given: Two circles are drawn on sides AB and AC of a △ABC as diameters. The circles intersects at D. To prove: D lies on BC Construction: Join A and D Proof: ∠ADB = 90° (Angle in the semi-circle ... + 90° => ∠ADB + ∠ADC = 180° => BDC is a straight line. Hence, D lies On third side BC.
Description : what- The length of the shortest side of the isosceles triangle is 6 inches.Find the length of the two congruent sides?
Last Answer : 10 in
Description : If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles. -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.
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 : The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude -Maths 9th
Last Answer : s= 2 a+b+c = 2 35+54+61 =75 Area, A= s(s−a)(s−b)(s−c) = 75(75−35)(75−54)(75−61) =420 5 cm 2 Now, Area of the triangle is also given as A= 2 1 ×a×h Where, h is the longest altitude. Therefore, 2 1 ×a×h=420 5 Hence, h=24 5 cm
Last Answer : The length of its longest altitude
Description : The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th
Last Answer : Let, a = 7 cm, b = 13 cm, c = 12 cm ∴ s = (a + b + c)/2 = (7 +13 +12)/2 = 32/2 = 16 cm Area of △ ABC = under root( √s(s -a) (s - b)(s -c)) = under root( √16(16 - 7)(16 - 13)(16 - 12) = ... 24 √3 cm2 Also, Area of △ ABC = 1/2AC.BD 24 √3 = 1/2 x 12 x BD ⇒ BD = (24 √3 x 2)/12 = 4 √3 cm
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
Description : The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th
Last Answer : Let the smallest side of the triangle be x cm long So, second side =(x+4)cm and third side =(2x−6)cm Given, perimeter =50cm Therefore, x+(x+4)+(2x−6)=50 ⇒4x=52 ⇒x=13cm So, first side of the triangle is ... =25cm Therefore, area of Δ=s(s−a)(s−b)(s−c) =25(25−13)(25−17)(25−20) =25 12 8 5 = 2030 cm2
Last Answer : According to question find the area of triangle.
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).