The triangular side walls of a flyover is used for -Maths 9th

The triangular side walls of a flyover is used for -Maths 9th
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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

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.

The triangular side walls of a flyover have been used for advertisements. -Maths 9th

Description : The triangular side walls of a flyover have been used for advertisements. -Maths 9th

Last Answer : Rent paid by company is

The triangular side walls of a flyover have been used for advertisements. -Maths 9th

Description : The triangular side walls of a flyover have been used for advertisements. -Maths 9th

Last Answer : Rent paid by company is

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

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.

There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN -Maths 9th

Description : There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN -Maths 9th

Last Answer : Let the sides of the wall be a = 15m, b = 11m, c = 6m Semi-perimeter, Thus, the required area painted in colour = 20√2 m2

There is a slide in a park,one of its side walls has been painted.... -Maths 9th

Description : There is a slide in a park,one of its side walls has been painted.... -Maths 9th

Last Answer : Let the dimensions of triangular shape wall be a = 15 m, b = 6 m, c = 11 m ∴ s = (a + b + c)/2 = (15 + 6 + 11)/2 = 32/2 = 16 m Area painted = Area of triangle = root under(√s(s - a)(s - b)(s - c) = root under (√16(16 - 15)(16 - 6)(16 - 11)) = root under (√16 x 1 x 10 x 5) = 20√2 m2

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

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

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. -Maths 9th

Description : An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. -Maths 9th

Last Answer : Let the sides of each triangular piece be a = 20 cm, b = 50 cm, c = 50 cm

The edges of a triangular board are 6 cm, 8 cm and 10 cm. -Maths 9th

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.

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

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+c​s=250+65+65​s=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

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

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

The edges of a triangular board are 6 cm, 8 cm and 10 cm. -Maths 9th

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.

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

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+c​s=250+65+65​s=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

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

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

An umbrella is made by stitching 10 triangular pieces of cloth.... -Maths 9th

Description : An umbrella is made by stitching 10 triangular pieces of cloth.... -Maths 9th

Last Answer : The sides of triangular pieces are 20 cm, 50 cm and 50 cm. Let a = 20 cm, b = 50 cm, c = 50 cm ∴ Semi - perimeter, s = (a + b + c)/2 = (20 + 50 + 50)/2 s = 60 cm ∴ Area of ... 40 x 10 x 10) = 200√6 cm2 Cloth of each design required = Area of 5 triangular pieces = 5 x 200√6 = 1000√6 cm2

The sides of a triangular field are 41 m, 40 m and 9 m. -Maths 9th

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

A floral design on a floor is made up of 16 tiles which are triangular, ... -Maths 9th

Description : A floral design on a floor is made up of 16 tiles which are triangular, ... -Maths 9th

Last Answer : Measures of the sides of the triangular tile are 28 cm, 9 cm and 35 cm. Let a = 28 cm, b = 9 cm, c = 35 cm Semi-perimeter, s = (a + b + c)/2 = (28 + 9 + 35)/2 = 36 cm ∴ Area of one ... 1411.2 cm2 Hence, cost of polishing the tiles at the rate of ₹ 1/2 per cm2 = ₹ 1/2 x 1411.2 = ₹ 705.60

Triangular pieces of cardboards were cutout by some people -Maths 9th

Description : Triangular pieces of cardboards were cutout by some people -Maths 9th

Last Answer : The two cutouts may not be congruent. For example all equilateral triangles have equal angles but may have different sides. Environmental concern, cooperative, caring, social.

An umbrella is made by stitching 10 triangular -Maths 9th

Description : An umbrella is made by stitching 10 triangular -Maths 9th

Last Answer : The sides of triangular pieces are 20 cm, 50 cm and 50 cm. Let, a = 20 cm, b = 50 cm, c = 50 cm ∴ Semi-perimeter, s = (a + b + c)/2 = (20 + 50 + 50)/2 s = 60 cm ∴ Area of ... √6 = 1000 √6 cm2 Cloth of each design required for 20 umbrellas = 20 x 1000 √6 = 20,000 √6 cm2 Helpful, caring, loving.

The perimeter of a triangular field is 240 m. If two of its sides are 78 m and 50 m, -Maths 9th

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.

Inside a triangular park, there is a flower bed forming a similar triangle. Around the flower bed runs a uniform path of such a width that the -Maths 9th

Description : Inside a triangular park, there is a flower bed forming a similar triangle. Around the flower bed runs a uniform path of such a width that the -Maths 9th

Last Answer : (d) 3 : 1Let ABC be the triangular flower bed of side lengths a, b and c respectively. Then Area of ΔABC = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(rac{a+b+c}{2}\)Now according to the given condition,ΔPQR ... of ΔABC - Area of ΔABC= 3 (Area of ΔABC)∴ Reqd. Ratio = Area of Path : Area of ΔABC = 3 : 1.

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

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

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

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.

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Last Answer : Length of a room (l) = 5m Breadth (b) = 4 m and height (h) = 3 m ∴ Area of 4 walls = 2(l + b) x h = 2(5 + 4) x 3 = 6 x 9 = 54 m² and area of ceiling = l x b = 5 x ... ∴ Total area = 54 + 20 = 74 m2 Rate of white washing = 7.50 per m² ∴ Total cost = ₹74 x 7.50 = ₹555

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

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.

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th

Last Answer : Length (l) of room = 5m Breadth (b) of room = 4m Height (h) of room = 3m It can be observed that four walls and the ceiling of the room are to be white washed. Total area to be white washed = Area of walls + ... m2 area = Rs.7.50 (Given) Cost of white washing 74 m2 area = Rs. (74 7.50) = Rs. 555

What is the name of the largest flyover in the country ?

Description : What is the name of the largest flyover in the country ?

Last Answer : The name of the largest flyover in the country is Mayor Mohammad Hanif Flyover (Poobnam Jatrabari-Gulistan Flyover).

When was the construction work of Mohakhali flyover inaugurated ?

Description : When was the construction work of Mohakhali flyover inaugurated ?

Last Answer : Construction work of Mohakhali Flyover was inaugurated on December 19 , 2001.

Which is the longest flyover in Bangladesh ?

Description : Which is the longest flyover in Bangladesh ?

Last Answer : The longest flyover in Bangladesh is the Mayor Mohammad Hanif Flyover.

Which is the longest flyover in the world ?

Last Answer : : Bang Na Expressway (Thailand , 54 km).

Which is the 2nd launched flyover in the country ?

Last Answer : 2nd Launched Flyover 'Khilgaon Flyover'

Which is the first flyover in the country ?

Last Answer : Mohakhali Flyover' is the first flyover in the country.

general The largest Flyover Which one ?

Last Answer : Bangladesh Most The largest Flyover Name Mayor Mohammed Hanif Flyover . Of Length 10 . 3 Km . Original Of the flyover Length 4 . 1 Km , Connector On the road Length . _ 5 km . This Dhaka Chankha ' rapul From The beginning By In Jatrabari The end Done .

Which of the following is an example of an intermediate goods. (a) A Tata Indica sold by a dealer of second hand car ; (b) Steel and cement used to construct a flyover ; (c) Farming crop purchased by FCI (d) All the three

Description : Which of the following is an example of an intermediate goods. (a) A Tata Indica sold by a dealer of second hand car ; (b) Steel and cement used to construct a flyover ; (c) Farming crop purchased by FCI (d) All the three

Last Answer : (b) Steel and cement used to construct a flyover ;

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Description : A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Last Answer : Here, each side of the rhombus = 30 m. Let ABCD be the given rhombus and the diagonal, BD = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m Since, a diagonal divides the rhombus into ... Area of grass for 18 cows to graze = 864 m2 ⇒ Area of grass for 1 cow to graze = 86418 m2 = 48 m2

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula. -Maths 9th

Description : A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula. -Maths 9th

Last Answer : Let each side of the equilateral triangle be a. Semi-perimeter of the triangle,

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

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

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

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

In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

Description : In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

Last Answer : The sum of angles of a triangle is180° If one aangke is of 90° then the sum of two angles is 90° It means that the angle forming 90° is biggest angle We know , Angle opposite to the longest side is largest. It means hypotenuse is the biggest side of right angled triangle

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

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).

Two cubes of side 2 cm each are joined end to end. Find the volume of the cuboid so formed. -Maths 9th

Description : Two cubes of side 2 cm each are joined end to end. Find the volume of the cuboid so formed. -Maths 9th

Last Answer : When two cubes of side 2 cm each are joined end to end then, Length (l) = (2 + 2) = 4cm Breadth (b) = 2 cm; Height (h) = 2 cm ∴ Volume of cuboid = lbh = 4 x 2 x 2 = 16 cm3

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

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

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

PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Description : PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram . -Maths 9th

Last Answer : Here, PQ = SR = 12cm Let PS = x and PS = QR ∴ x + 12 + x +12 = Perimeter 2x + 24 = 40 2x = 16 x = 8 Hence, length of each side of the parallelogram is 12cm , 8cm , 12cm and 8cm.

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

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

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

In the given figure, WXYZ is a quadrilateral with a point P on side WX. If ZY // WX, show that : -Maths 9th

Description : In the given figure, WXYZ is a quadrilateral with a point P on side WX. If ZY // WX, show that : -Maths 9th

Last Answer : ar (ZPY)=ar( ZXY) they lie between the same base and between the same parallels Similarly, ar(WZY)=ar(ZPY) ar(ZWX)=ar(XWY)

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP