Description : In which of the following figures, you find two polygons on the same base and between the same parallels ? -Maths 9th
Last Answer : (d) In figures (a), (b) and (c) there are two polygons on the same base but they are not between the same parallels. In figure (d), there are two polygons (PQRA and BQRS) on the same base and between the same parallels .
Description : If a triangle and a parallelogram are on same base and between same parallels, then find the ratio of the area of the triangle to the area of parallelogram. -Maths 9th
Last Answer : If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to half of the parallelogram.
Description : If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is -Maths 9th
Last Answer : (b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
Description : If a triangle and a parallelogram are on same base and between the same parallels,then find the ratio of the area of the triangle to the area of parallelogram -Maths 9th
Last Answer : Solution :-
Description : Two parallelograms are on equal bases and between the same parallels. -Maths 9th
Last Answer : Two Parallelograms on the equal based and between the same parallels are equal in area.
Last Answer : (b) We know that, parallelogram on the equal bases and between the same parallels are equal in area. So, ratio of their areas is 1 :1.
Description : Prove that parallelogram on equal bases and between the same parallels are equal in area. -Maths 9th
Last Answer : Suppose AL and PM are the altitudes corresponding to equal bases AB and PQ of ||gm ABCD and PQRS respectively . Since the ||gm are between the same parallels PB and SC. ∴ AL = PM Now, ar(||gm ABCD) = AB AL ar(|| ... PM But, AB = PQ [given] AL = PM [proved] ∴ ar(||gm ABCD) = ar(||gm PQRS)
Description : Two cans have the same height equal to 21 cm. One can is cylindrical, the diameter of whose base is 10 cm. -Maths 9th
Last Answer : (c) 450 cm3. Required difference in capacities = 227227 x (5)2 x 21~ (10)2 x 21 = (1650 ~ 2100) cm3 = 450 cm3
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
Description : It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same? -Maths 9th
Last Answer : Let h be the height and r be the radius of a cylindrical tank. Height of cylindrical tank, h = 1m Radius = half of diameter = (140/2) cm = 70cm = 0.7m Area of sheet required = Total surface are of tank = 2πr( ... [2 (22/7) 0.7(0.7+1)] = 7.48 Therefore, 7.48 square meters of the sheet are required.
Description : A triangle and a parallelogram have the same base and the same area. -Maths 9th
Last Answer : Let a = 26 cm, b = 28 cm, c = 30 cm ∴ s = (a + b + c)/2 = (26 + 28 + 30)/2 = 84/2 = 42 ∴ Area of triangle = root under (√s(s - a )(s - b)(s - c) = root ... cm2 Now, Area of parallelogram = Area of triangle ⇒ Base x height = 336 ⇒ 28 x height = 336 ⇒ height = 336/28 = 12 cm
Description : If a triangle and a convex quadrilateral are drawn on the same base and no part of the quadrilateral is outside the triangle, -Maths 9th
Last Answer : answer:
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
Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th
Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area
Description : There are two prisms, one has equilateral triangle as a base and the other has a regular hexagon as a base. If both the prisms have equal height -Maths 9th
Description : The median of a trapezoid cuts the trapezoid into two regions whose areas are in the ratio 1 : 2. Compute the ratio of the smaller base of the -Maths 9th
Description : Cbqs (case base study ) of chapter 4 Linear Equations in Two Variables of maths class 9th -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
Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th
Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.
Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th
Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.
Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th
Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.
Description : What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m -Maths 9th
Last Answer : Solution: Height of conical tent, h = 8m Radius of base of tent, r = 6m Slant height of tent, l2 = (r2+h2) l2 = (62+82) = (36+64) = (100) or l = 10 Again, CSA of conical tent = πrl = (3.14 6 ... .2) 3] = 188.4 L-0.2 = 62.8 L = 63 Therefore, the length of the required tarpaulin sheet will be 63 m.
Description : A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. -Maths 9th
Last Answer : : Ncert solutions class 9 chapter 13-5 Let ABC be a conical tent Height of conical tent, h = 10 m Radius of conical tent, r = 24m Let the slant height of the tent be l. (i) In right triangle ... (13728/7) 70 = Rs 137280 Therefore, the cost of the canvas required to make such a tent is Rs 137280.
Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th
Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.
Description : Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m -Maths 9th
Last Answer : Radius of cone, r = 24/2 m = 12m Slant height, l = 21 m Formula: Total Surface area of the cone = πr(l+r) Total Surface area of the cone = (22/7)×12×(21+12) m2 = 1244.57m2
Description : he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th
Last Answer : Say h = height of the frame of lampshade, looks like cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use curved surface area formula to find the ... 2πrh = (2 (22/7) 10 35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required for covering the lampshade.
Description : Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th
Last Answer : Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm Slant height of cone, say l = 10 cm CSA of cone is = πrl = (22/7)×5.25×10 = 165
Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th
Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.
Description : A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30cm long, 25 cm wide and 25 cm high. -Maths 9th
Last Answer : Length of greenhouse, say l = 30cm Breadth of greenhouse, say b = 25 cm Height of greenhouse, say h = 25 cm (i) Total surface area of greenhouse = Area of the glass = 2[lb+lh+bh] = [2(30 ... (30+25+25)] (after substituting the values) = 320 Therefore, 320 cm tape is required for all the 12 edges.
Description : Find the area of a triangle with base =20cm and height are 10 cm. -Maths 9th
Last Answer : Area of a triangle = 1/2 × Base × Altitude ( height ) therefore., Area of a triangle = 1/2 × 20 cm× 10cm = 10cm ×10cm = 100 cm^2
Description : The circumference of the base of 9 m high wooden solid cone is 44 m. Find the slant height of the cone. -Maths 9th
Last Answer : Circumference of the base of a cone = 2πr
Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th
Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required
Description : How much ice-cream can be put into a cone with base radius 3.5 cm and height 12 cm ? -Maths 9th
Last Answer : Here, radius (r) = 3.5 cm and height (h) = 12 cm ∴ Amount of ice cream = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 3.5 × 3.5 × 12 = 154 cm3
Description : A cone is 8.4 cm high and the radius of its base is 2.1 cm. -Maths 9th
Last Answer : Volume of cone = Volume of sphere 1 / 3π(2.1)2 × 8.4 = 4 / 3 πr3 ⇒ r3 = (2.1)2 × 8.4 / 4 = (2.1)3 ⇒ r = 2.1 cm ∴ Radius of the sphere = 2.1 cm
Description : A cylindrical vessel can hold 154 g of water. If the radius of its base is 3.5 cm, and 1 cm3 of water weighs lg,find the depth of water. -Maths 9th
Last Answer : Since 1 cm3 of water weighs 1 g. ∴ Volume of cyclinder vessel = 154 cm3 πr2h = 154 h = 154 × 7 / 22 × 3.5 × 3.5 h ;= 4 cm Hence, the depth of water is 4 cm.
Description : How many metres of 5 m wide cloth will be required to make a conical tent, the radius of whose base is 3.5 m and height is 12 m ? -Maths 9th
Last Answer : l = √h2 + r2 = √(3.5)2 + (12)2 = √12.25 + 144 = √156.25 = 12.5 m Curved surface area = πrl = 22 / 7 × 3.5 × 12.5 = 137.5 m2 Area of cloth = 137.5 m2 Length of cloth required = C.S.A. / Width l = 137.5 / 5 = 27.5 m
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 : 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