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

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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.

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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 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 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 : 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 : 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 : 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 : 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 : 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 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 : 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 : 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 : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

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Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

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 : 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 : 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 : 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.

Description : Two parallelograms are on equal bases and between the same parallels. -Maths 9th

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 : 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.

Description : Two parallelograms are on equal bases and between the same parallels. -Maths 9th

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 : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

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Description : What is the ratio of the area of a circum circle of equilateral triangle to the area of the square with same side length as equilateral triangle? -Maths 9th

Last Answer : (d) π : 3.Let each side of the equilateral Δ be a units. Then, circumradius of the circle = \(rac{ ext{side}}{\sqrt3}\) = \(rac{a}{\sqrt3}\) units∴ Area of circumcircle = \(\pi\bigg(rac{a}{\sqrt3}\bigg)^2\) = \( ... units∴ Required ratio = \(rac{rac{\pi{a}^2}{3}}{a^2}\) = \(rac{\pi}{3}\) = π : 3.

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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

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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

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Description : Find the area of an isosceles triangle having base x cm and equal side y cm. -Maths 9th

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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

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Description : The base of an isosceles triangle is 24cm and its area is 192cm^2. Find its perimeter -Maths 9th

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Description : In figure, if parallelogram ABCD and rectangle ABEM are of equal area, then -Maths 9th

Last Answer : (c) In rectangle ABEM, AB = EM [sides of rectangle] and in parallelogram ABCD, CD = AB On adding, both equations, we get AB + CD = EM + AB (i) We know that, the perpendicular distance between two ... AB+BE + EM+ AM [∴ CD = AB = EM] Perimeter of parallelogram ABCD > perimeter of rectangle ABEM

Description : In figure, if parallelogram ABCD and rectangle ABEM are of equal area, then -Maths 9th

Last Answer : (c) In rectangle ABEM, AB = EM [sides of rectangle] and in parallelogram ABCD, CD = AB On adding, both equations, we get AB + CD = EM + AB (i) We know that, the perpendicular distance between two ... AB+BE + EM+ AM [∴ CD = AB = EM] Perimeter of parallelogram ABCD > perimeter of rectangle ABEM

Description : The area of parallelogram PQRS is 88 cm sq. A perpendicular from S is drawn to intersect PQ at M. If SM = 8 cm, then find the length of PQ. -Maths 9th

Last Answer : Given area of parallelogram = 88 cm² And SM = 8cm Area of a parallelogram = height × base (Height is the measurement of a perpendicular drawn from one side to other) Here, Area of PQRS = SM × PQ 88cm² = 8cm × PQ 11cm = PQ

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 : Two adjacent angles of a ||gm are in the ratio 2:3. Find all the four angles of the parallelogram. -Maths 9th

Last Answer : Solution :-

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

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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 : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled 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).

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle.

Description : If the angles of a triangle are in the ratio 5:3:7, then the triangle is -Maths 9th

Last Answer : (a) Given, the ratio of angles of a triangle is 5 : 3 : 7. Let angles of a triangle be ∠A,∠B and ∠C. Then, ∠A = 5x, ∠B = 3x and ∠C = 7x In ΔABC, ∠A + ∠B + ∠C = 180° [since, sum of ... 36° and ∠C =7x = 7 x 12° = 84° Since, all angles are less than 90°, hence the triangle is an acute angled triangle

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 : Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Last Answer : We know that diagnol of a parallelogram bisect it in two triangles of equal area area of triangle =1÷2×b×h so. 1÷2×b×h+1÷2×b×h=b×h Hence,proved

Description : In the figure, the area of parallelogram ABCD is -Maths 9th

Last Answer : (c) We know that, area of parallelogram is the product of its any side and the corresponding altitude (or height). Here, when AB is base, then height is DL. Area of parallelogram = AB x DL and when AD is ... = DC x DL and when BC is base, then height is not given. Hence, option (c) is correct.