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 : 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 : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th
Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.
Description : In parallelogram PQRS, PQ = 10cm. The altitudes corresponding to the sides PQ and SP are respectively 6cm and 8cm. Find SP. -Maths 9th
Last Answer : Solution :-
Description : In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th
Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th
Last Answer : This is the sketch of the question but its hard to answer.
Description : Find the area of the trapezium PQRS with height PQ given in the figure given below. -Maths 9th
Last Answer : The area of trapezium =
Description : If PQRS is trapezium such that PQ > RS and L, M are the mid-points of the diagonals PR and QS respectively then what is LM equal to? -Maths 9th
Last Answer : if pqrs is a trapezium so pq and RS are parallel is you draw a diagonal you will divide the trapezium into two parts such that two equal triangle so
Description : PQRS is a rectangle in which PQ = 2PS. T and U are the mid-points of PS and PQ respectively. QT and US intersect at V. -Maths 9th
Last Answer : answer:
Description : In the given truss bridge parallelogram ABCD and PQRS are congruent if AB 24 feet what is PQ?
Last Answer : In the given truss bridge parallelogram ABCD and PQRS are congruent if AB 24 feet what is PQ?
Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th
Last Answer : According to question parallelogram ABCD such that AP = CQ.
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 : 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 : In the given figure , two opposite angles of a parallelograms PQRS are (3x - 4)° and (56 - 3x)° . Find all the angles of given parallelogram . -Maths 9th
Last Answer : We know that opposite angles of a paralleloram are equal . ∴ ∠P = ∠R ⇒ 3x - 4 = 56 - 3x ⇒ 6x = 60 ⇒ x = 10 Thus, ∠P = ∠R = 3 10 - 4 = 26° Also, ∠P + ∠Q = 180° [ ... Hence, the four angles of the parallelogram PQRS are 26°, 154°,26° and 154°. We should care our earth to have good eco balance.
Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th
Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2
Description : O is any point on the diagonal PR of a parallelogram PQRS. -Maths 9th
Last Answer : Join QS. Let diagonals PR and QS intersect each other at T. We know, that diagonals of a parallelogram bisect each other . ∴ T is the mid - point of QS. Since a median of a triangle divides it into two triangles of equal ... ar(△PTS) + ar( △STO) = ar(△PQT) = ar( △ QTO ) ⇒ ar(△PSO) = ar(△PQO)
Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th
Last Answer : From the adjoining figure, we have The field PQRS is divided into three parts △PAQ, △APS and △AQR. Now, △PAQ and ||gm PQRS are on the same base and lie between the same parallels. ∴ ar(△PAQ) = 1 / 2 ar(||gm ... , she can sow wheat in △APS and △AQR, pulses in △PAQ or vice - versa .
Description : O is any point on the diagonal PR of a parallelogram PQRS (figure). -Maths 9th
Last Answer : According to question prove that ar(ΔPSO) = ar(ΔPQO).
Description : PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of △ASR = 90 cm2. Find this statement is true or false. -Maths 9th
Last Answer : Solution :- As diagonal of the parallelogram divides it into two triangles of equal area. Since, area (△SRQ ) = 1/2 area(PQRS) area (△SRQ ) = 1/2 x 180 ... = 90 cm2 (Given) This is not possible unless area (△SRQ ) = area (△ASR ) So, the given statement is false.
Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th
Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.
Description : In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. -Maths 9th
Last Answer : Given In a parallelogram PSDA, points 0 and R are on PS such that PQ = QR = RS and PA || QB || RC. To prove ar (PQE) = ar (CFD) Proof In parallelogram PABQ, and PA||QB [given] So, ... = ΔDCF [by ASA congruence rule] ∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area] Hence proved.
Description : If two parallelogram PQAD and PQBC arw on the opposite sides of PQ prove that ABCD is a paralellogram -Maths 9th
Last Answer : This answer was deleted by our moderators...
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 a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. -Maths 9th
Last Answer : Let the smallest side of the triangle be x cm long So, second side =(x+4)cm and third side =(2x−6)cm Given, perimeter =50cm Therefore, x+(x+4)+(2x−6)=50 ⇒4x=52 ⇒x=13cm So, first side of the triangle is ... =25cm Therefore, area of Δ=s(s−a)(s−b)(s−c) =25(25−13)(25−17)(25−20) =25 12 8 5 = 2030 cm2
Last Answer : According to question find the area of triangle.
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 : 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 : The lengths of two adjacent sides of a parallelogram are 17 cm and 12 cm. -Maths 9th
Last Answer : For △BCD: Let a = 17 cm, b = 12 cm, c = 25 cm So its semi-perimeter, s = (a + b + c)/2 = (17 + 12 + 25)/2 = 27 cm ∴ Area of △BCD = root under(√(s -a)(s - b)(s - c)) = ... △BCD = 2 x 90 = 180 cm2 Also, area of parallelogram ABCD = DC x AE ∴ 180 = 12 x AE ⇒ AE = 180/12 = 15 cm
Description : Find the area of a parallelogram given in the figure. Also, find the length of the altitude from vertex A on the side DC. -Maths 9th
Last Answer : Weknowthatthediagonalofaparallelogram(∥gm)dividesitintotwocongruenttriangles.SoAreaof∥gmABCD=2 Areaof△BCD.AccordingtoHeron′sformulathearea(A)oftrianglewithsidesa,b&cisgivenasA=2[s(s−a)(s−b) ... 90=180Areaof∥gm=base heightHeightofaltitudefromvertexAonsideCDoftheof∥gm=baseCDareaof∥gmABCD =12180 =15cm
Last Answer : =3 x 3 x 5 x 2 cm2 Area of parallelogram ABCD = 2 x 90 = 180 cm2 (ii) Let altitude of a parallelogram be h. Also, area of parallelogram ABCD =Base x Altitude ⇒ 180 = DC x h [from Eq. (ii)] ... h ∴ h = 180/12= 15 cm Hence, the area of parallelogram is 180 cm2 and the length of altitude is 15 cm.
Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th
Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm
Description : The quadrilateral formed by joining the mid-points of the side of quadrilateral PQRS, taken in order, is a rhombus, if -Maths 9th
Last Answer : (d) Given, the quadrilateral ABCD is a rhombus. So, sides AB, BC, CD and AD are equal.
Description : ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th