Description : The perimeter of a triangular field is 420 m and its sides are in the ratio 6:7:8. Find the area of the triangular field. -Maths 9th
Last Answer : Let sides of △ are=6x,7x and 8x Perimeter=6x+7x+8x=21x 21x=410 x=20 Sides are 120,140 and 160 m Area =S(S−A)(S−B)(S−C) [Heron's Formula] S=2120+140+160=210 m A=210(210−120)(210−140)(210−160)=210015 sq. m
Description : The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th
Last Answer : (c) 67.5 mGiven 2s = a + b + c ⇒ 240 = 78 + 50 + Third side ⇒ Third side = 240 m - 128 m = 112 m.∴ Area of Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)= \(\sqrt{120(120-78)(120-50)(120-112)}\)= \(\sqrt{120 imes42 imes70 ... (rac{1}{2}\)x b x h ∴ \(rac{1}{2}\) x 50 x h = 1680 ⇒ h = \(rac{1680}{25}\) = 67.5 m.
Description : Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs. 7 per m2. -Maths 9th
Last Answer : Sides of the triangle are a=50m,b=65m,c=65m Area of triangle, by Heron's formula =s(s−a)(s−b)(s−c)where, s=2a+b+cs=250+65+65s=90 Area of triangle = 90(40)(25)(25)Area of triangle = 1500m2 Cost of laying grass = Area ×7 Cost of laying grass =1500×7 Cost of laying grass = Rs 10500
Description : The sides of a triangular field are 41 m, 40 m and 9 m. -Maths 9th
Last Answer : Let a = 41m, b = 40 m, c = 9 m. s = (a + b + c)/2 = (41 + 40 +9)/2 = 90/2 ⇒ s = 45 m Area of the triangular field = root under( √s(s - a)(s - b)(s -c)) = root under( ... x 5 x 36 ) = 180 m2 = 1800000 cm2 Number of rose beds = Total area / Area needed for one rose bed = 1800000/900 = 2000
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
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 : 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).
Description : What is the volume of a right prism standing on a triangular base of sides 5 cm, 5 cm and 8 cm whose lateral surface area is 828 cm^2 ? -Maths 9th
Last Answer : Lateral surface area of a prism = Perimeter of base Height ⇒ 840 = (5 + 5 + 8) Height ⇒ Height = 8401884018 = 46 cm. = Semi perimeter of the triangular base = 182182 = 9 cm ∴ Area of triangle = 9(9- ... 4 1 = 12 cm2 ∴ Required volume of prism = Area of base Height = (12 46) cm3 = 552cm3
Description : The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m² per year. -Maths 9th
Last Answer : Let the sides of the triangular will be a = 122m, b = 12cm, c = 22m Semi-perimeter, s = a+b+c2 (122+120+224)m = 2642 m = 132m The area of the triangular side wall Rent for 1 year (i.e. 12 months) per m2 = ... = Rs. 5000 x 312 = Rent for 3 months for 1320 m2 = Rs. 5000 x 312 x 1320 = Rs. 16,50,000.
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 : 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
Description : A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th
Last Answer : Vol. of △ ular prism = Area of △ ular base × height. ∴ Area of triangular base = area of triangle PQR By heron's formula. S=S(s−a)(s−b)(s−c)where S=2a+b+c∴Areaof△PQR= S=23+4+5=6 S=6(6−3)(6−4)(6−5)=3×2×3×2×1=6cm2 ∴ vol. of Prism =6×10 =60cm3Answer.
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 : If two adjacent sides of a kite are 5cm and 7cm, find its perimeter. -Maths 9th
Last Answer : Solution :-
Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th
Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2
Description : A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs. -Maths 9th
Last Answer : Cost of painting =
Description : A rhombus shaped sheet with perimeter 40 and digonals are 12 cm is painted on bith sides at the rate of rs 5 per metre square. Find the cost of painting -Maths 9th
Last Answer : Let ABCD be a rhombus, then AB=BC=CD=DA=x Perimeter of rhombus =40cm ⇒4x=40cm⇒x=10cm ∴AB=BC=CD=DA=10cm In △ABC,S=2a+b+c=210+10+12=16cm ar△ABC=16(16−10)(16−10)(16−12)=16×6×6×4=48cm2ar.ABCD=2×48=96cm2 Cost of painting the sheet =Rs(5×96×2)=Rs960 [Both sides]
Description : Perimeter of the rhombus is 100 m and its diagonal is 40m. Find the area of rhombus. -Maths 9th
Last Answer : Perimeter of rhombus =4 side ⇒ 100=4 side ⇒ side= 4 100 ⇒ side=25 We know diagonals of a rhombus divides the rhombus in two equilateral triangle. Now, we are going to find area of 1 equilateral triangle. Semi perimeter = ... ) = 45 5 20 20 = 90000 =300m 2 ⇒ Area of rhombus =2 300m 2 =600m 2
Description : If a rectangle and a parallelogram are equal in area and have the same base and are situated on the same side, and the ratio of the perimeter -Maths 9th
Last Answer : answer:
Description : The perimeter of an equilateral triangle is 60 m. The area is -Maths 9th
Last Answer : Let each side of an equilateral be x. Then, perimeter of an equilateral triangle = 60 m x + x + x = 60 ⇒ 3x = 60 ⇒ x = 60/3 = 20 m Area of an equilateral triangle = √3/4 (Side)2 = (√3/4) x 20 x 20 = 100 √3 m2 Thus, the area of triangle is 100√3 m2.
Last Answer : (d) Let each side of an equilateral be x. Then, perimeter of an equilateral triangle = 60 m x + x + x = 60 ⇒ 3x = 60 ⇒ x = 60/3 = 20 m Area of an equilateral triangle = √3/4 (Side)2 = (√3/4) x 20 x 20 = 100 √3 m2 Thus, the area of triangle is 100√3 m2.
Description : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla this brress is doins centinuously infinite46. The perimeter of 7 th triangle is \( ( \) in \( cm ) \)a) \( \ ... of the 5th triangle is (in \( cm \) )a) 6b) \( 1.5 \)c) \( 0.75 \)d) 3
Last Answer : Each side of an equilateral triand 12 is \( 24 cm \) the mid - point of its sides are joined to form another tranpla ... ( 1.5 \) c) \( 0.75 \) d) 3
Description : The edges of a triangular board are 6 cm, 8 cm and 10 cm. -Maths 9th
Last Answer : s=2a+b+c=26+8+10=12 By Heron's formula, Area of the triangle =s(s−a)(s−b)(s−c)=12(12−6)(12−8)(12−10)=12(6)(4)(2)=24cm2 Cost of painting =9×24 paise =216 paise = Rs. 2.16.
Description : A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. -Maths 9th
Last Answer : S(△ABC)=60+80+402=90S(△ABC)=60+80+402=90 ar△ABDar△ABD =90(90−80)(90−60)(90−40)−−−−−−−−−−−−−−−−−−−−−−−√=90(90−80)(90−60)(90−40) =90×10×30×50−−−−−−−−−−−−−−√=90×10×30×50 =30015−−√m2=30015m2 ar□ABCE=2×ar△ABDar◻ABCE=2×ar△ABD =60015−−√m2
Last Answer : Area of the parallelogram
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 : A field is in the shape of a trapezium having parallel sides 90 m and 30 m. -Maths 9th
Last Answer : NEED ANSWER
Last Answer : In trapezium ABCD, we draw a perpendicular line CE to the line AB.
Description : The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th
Last Answer : 1 : 2 Let each side of the base of the original prism be a units and the height of the prism be h units. Then Required ratio = Vol. of original prismVol. of new prismVol. of original ... )2×h3√4×(2a)2×h234×(a)2×h34×(2a)2×h2 = 2a2h4a2h2a2h4a2h = 1 : 2.
Description : The cost of preparing a cricket ground of rectangular shape at 70 paisa per square meter is Rs.508.20. Find the perimeter of the field if its sides are in the ratio 3:2 a) 121m b) 96m c) 55m d) 110m
Last Answer : d) 110m
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 : 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 floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2 is Rs.15000, find the height of the hall. -Maths 9th
Last Answer : Let length, breadth, and height of the rectangular hall be l, b, and h respectively. Area of four walls = 2lh+2bh = 2(l+b)h Perimeter of the floor of hall = 2(l+b) = 250 m Area of four walls = 2( ... of paining the walls is Rs. 15000. 15000 = 2500h Or h = 6 Therefore, the height of the hall is 6 m.
Description : Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th
Last Answer : Solutions :- We have, Perimeter of triangle = 32 cm One of its side = 11 cm Let the second side be x And third side be x + 5 Perimeter of triangle = sum of three sides A/q => 11 + x + x + 5 ... 13 cm Now, By using heron's formula, Find the area of a triangle :- Answer : Area of triangle = 43.81 cm²
Description : The total surface area of a cube is 150sq.cm. Find the perimeter of any one of its face ? -Maths 9th
Last Answer : Take the side be s, 6s^(2)=150cm^(2) s^(2)= 150/6 =25 s^(2)=square root of 25 s=5 Area of square s^(2) 5×5=25sq.cm. Perimeter = 4×side =4×5=20cm
Description : The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th
Last Answer : Given, base of an isosceles triangle =24 cm Area of isosceles triangle =192 sq.cm Area = 21×b×h ∴192=224.h ∴h=16 cm Side=h2+122=256+144=20 cm Perimeter of triangle =2a+b =2(20)+24=64 cm
Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th
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 : If the perimeter of an equilateral triangle is 24cm, then find the area of the triangle -Maths 9th
Last Answer : Perimeter of equilateral triangle = 24cm ( there are three equal sides in an e quilateral triangle ) Each side = 24 / 3 = 8cm Area of an equilateral triangle = root 3 / 4 s square = root 3 / 4 × 8 square = root 3 / 4 x 64 = root 3 × 16 = 16 root 3
Description : The perimeter of one face of a cube is 20 cm. Find the lateral surface area of the cube. -Maths 9th
Last Answer : Let a be the edge of cube, ∴ Perimeter of one face of cube = 20 cm ⇒ 4a = 20 ⇒ a = 5 ∴ Lateral surface area of cube = 4a2 = 4 (5)2 = 100 cm2
Description : The area of a square and circle is same and the perimeter of square and equilateral triangle is same, -Maths 9th
Last Answer : (b) 9 : 4√3.Let each side of the square = a cm. Then, Area of square = a2 cm2 Also, let r be the radius of the circle. Then, πr2 = a2 Let each side of the equilateral triangle = b cm. Then 3b = 4a ⇒ ... ratio between area of circle and area of equilateral Δ is a2 : \(rac{4\sqrt3a^2}{9}\) = 9 : 4√3.