There are two concentric hexagons. Each of the side of both the hexagons are parallel. -Maths 9th

1 Answer

Answer :

In the second figure, ΔOPQ is an equilateral triangle as a regular hexagon is divided into six equilateral triangles. ∴ ∠OPC = 60º⇒ \(rac{OC}{OP}\) = sin 60° ⇒ OC = OP sin 60° ⇒ OC = 8 x \(rac{\sqrt3}{2}\) cm = 4√3 cm.Now in similar Δs OPC and OAD, \(rac{OC}{OD}\) = \(rac{OP}{OA}\) ⇒ \(rac{4\sqrt3}{6\sqrt3}\) = \(rac{8}{OA}\)⇒ OA = 12 cm ⇒ AB = OA = 12 cm(∵ OD = OC + 2√3 = 4√3  + 2√3 = 6√3 cm)∴ Required area = Area of outer hexagon – Area of inner hexagon= \(rac{3\sqrt3}{2}\)(122 - 82) cm2 = 120√3 cm2.

Related questions

Description : In Fig. 10.25, a line intersect two concentric circles with centre O at A, B, C and D, Prove that AB = CD. -Maths 9th

Last Answer : Solution :- Let OP be perpendicular from O on line l. Since the perpendicular from the centre of a circle to a chord,bisects the chord.Therefore, AP = DP ...(i) BP = CP ...(ii) Subtracting (ii) from (i), we get AP - BP = DP - CP ⇒ AB = CD

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.

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 : 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.

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 : A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Last Answer : Solution :- The two lines will not be always parallel as the sum of the two equal angles will not always be 180°. Lines will be parallel when each of the equal angles is equal to 90°.

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

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)

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)

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

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)

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)

Description : The area of a trapezium is 39cm2 . The distance between its parallel sides is 6cm. If one of the parallel sides is 5cm, then find the other parallel side. -Maths 9th

Last Answer : area of trapezium = 39 cm sq height of trpezium = 6 cm one || side = 5 cm let other be x area of trapezium = 1/2 × ( sum of || sides) × h 39 = 1/2 × ( 5+x) × 6 39×2 = (5+x) × 6 78/6 = 5+x 13-5 = x 8 = x x = 8cm

Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Last Answer : answer:

Description : ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

Last Answer : answer:

Description : In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Last Answer : answer:

Description : Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Last Answer : answer:

Description : STL file format is represented by interaction of ______ a.lines and hexagons b.lines and rectangles c.lines and triangles d.lines and circles

Last Answer : c.lines and triangles

Description : Can regular pentagons and regular hexagons tile a flat surface?

Last Answer : Need answer

Description : How many sides do 2 pentagons 3 nonagons and 2 hexagons have in all?

Last Answer : They have 49 sides.

Description : How many sides do 3 pentagons 5 heptagons and 4 hexagons have in all?

Last Answer : 74 sides

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Last Answer : hope its clear

Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm

Description : For what value of x will the lines l and m be parallel to each other ? -Maths 9th

Last Answer : Solution :-

Description : In a trapezium ABCD, AB is parallel to CD and the diagonals intersect each other at O. In this case, the ratio OA/OC is equal to: -Maths 9th

Last Answer : answer:

Description : -For two shafts in parallel or for two concentric shafts a. T = T1 + T2 b. T = T1 = T2 c. T = T1 – T2 d. T = (T1.T2)^1/2

Last Answer : a. T = T1 + T2

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

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

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

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

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.

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.

Description : Find the area of an isosceles triangle, whose equal sides are of length 15 cm each and third side is 12 cm. -Maths 9th

Last Answer : We have, Three sides13cm,13cm and 20cm. By using Heron's formula We need to get the semi-perimeter s= 2 a+b+c​ = 2 13+13+20​ = 2 46​ =23 Now, put the heron's formula, s= s(s−a)(s−b)(s−c)​ = 23(23−13)(23−13)(23−20)​ = 23×10×10×3​ =10 23×3​ =83.07cm 2

Description : If each side of a triangle is doubled,... -Maths 9th

Last Answer : Let a, b, c be the sides of the given triangle and s be its semi-perimeter. Then, s = (a + b + c)/2 ...(i) ∴ Area of the given triangle = root under ( √s(s - a)(s - b)(s - c)) = △, say According to ... - c)) = 4 root under ( √ s(s - a)(s - b)(s - c)) = 4 △ Therefore, the required ratio is 4:1

Description : If a circular sheet of perimeter 2πr touching each side of a given quadrilateral sheet of perimeter 2p -Maths 9th

Last Answer : Let ABCD be the quadrilateral from which a circular sheet is cut off touching each side of the quadrilateral. Also, given AB + BC + CD + DA = 2p ...(i) Circumference of the circle = 2πr ⇒ ... = pr.∴ Required remaining area = Area of quadrilateral - Area of circle = pr - πr2 = r(p - πr).

Description : The percentage increase in the area of a triangle if each of its side is doubled is -Maths 9th

Last Answer : (c) 300%.Let a, b, c be the sides of the original triangle and s be its semi-perimeter.Then, s = \(rac{1}{2}\)(a+b+c)Let s1 be the semi-perimeter of the new triangle. Then,s1 = \(rac{1}{2}\) (2a + 2b ... increase in area = \(\bigg(rac{A_1-A}{A} imes100\bigg)\)%= \(rac{(4A-A)}{A} imes100\)% = 300%.

Description : Find the area of regular octagon with each side ‘a’ cm. -Maths 9th

Last Answer : (a) 2a2 (1 + √2) Let ABCDEFGH be the regular octagon of side a cm. Now if we produce the sides of the octagon on both the sides, we get a square PQRS. Given, BC = DE = FG = HA = a cm. Also, BQ = QC = ... = Area of square PQRS - Total area of shaded isosceles Δs = a2 (3 + 2√2) - a2= 2a2 (1 + √2) cm2.

Description : ABCD is a square of side a cm. AB, BC, CD and AD are all chords of circles with equal radii each. -Maths 9th

Last Answer : (b) \(\bigg[a^2+4\bigg[rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg]\bigg]\)As shown in the given figures, if a' is each side of the square, then ∠DOC = 120º ⇒ ∠ODC = ∠OCD = 30ºNow in fig. (iii), \(rac{ ... of square + Total area of 4 segments = \(a^2+4\bigg(rac{\pi{a}^2}{9}-rac{a^2}{4\sqrt3}\bigg).\)

Description : Eighteen quests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and -Maths 9th

Last Answer : Since four particular guests want to sit on particular side and three others on the other side. So, we are left with 11 guests out of which we choose 5 for side A in ways and remaining 6 for ... of the table can be arranged among themselves in 9! Ways. Hence, the total number of arrangements =

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 : A point is selected at random inside an equilateral triangle. From this point a perpendicular is dropped to each side. -Maths 9th

Last Answer : answer:

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Last Answer : Let AB and CD be two parallel chords having P and Q as their mid-points, respectively. Let O be the centre of the circle. Join OP and OQ and draw OX | | AB | | CD. Since, Pis the mid-point of AB. ⇒ OP ... 90° Now, ∠POX + ∠XOQ = 90° + 90° = 180° so, POQ is a straight line . Hence proved

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Description : A transversal intersects two parallel lines. -Maths 9th

Last Answer : Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP, respectively.

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

Description : If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel. -Maths 9th

Last Answer : According to question prove that the two chords are parallel.