Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th
Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]
Description : D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is (a) 2.5 (b) 3 (c) 5 (d) 6
Last Answer : (b) 3
Description : The pair of equations ax+by+c=0 and dx+ey+c=0 represent the equations with infinitely many solutions if (a)ad=be (b)ae=bd (c)ab=de (d)ac=de
Last Answer : (b)ae=bd
Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th
Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.
Description : In the adjoining figure, if ∠BAC = 90° and AD ⊥ BC, then (а) BD.CD = BC2 (b) AB.AC = BC2 (c) BD.CD = AD2 (d) AB.AC = AD2
Last Answer : (c) BD.CD = AD2
Description : If ΔABC ~ ΔEDF and ΔABC is not similar to ΔDEF then which of the following is not true? (a) BC.EF = AC.FD (b) AB.EF = AC.DE (c) BC.DE = AB.EF (d) BC.DE = AB.FD
Last Answer : (c) BC.DE = AB.EF
Description : Match the column: 1. In and ÆB = ÆC PQ PR (A) AA similarity criterion = 2. In and A = = 3. In and (B) SSS similarity criterion (C) BPT ÆB = ÆC = BC PQ PR QR 4. In , DE ÆD = ÆE BD CE (D) SAS similarity ... 7. Areas of these triangles are in the ratio (a) 9 : 35 (b) 9 : 49 (c) 49 : 9 (d) 9 : 42
Last Answer : (b) 9 : 49
Description : In Fig. 7.21, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. -Maths 9th
Last Answer : It is given that ∠BAD=∠EAC ∠BAD+∠DAC=∠EAC+∠DAC [add ∠DAC on both sides] ∴∠BAC=∠DAE In △BAC and △DAE AB=AD (Given) ∠BAC=∠DAE (Proved above) AC=AE (Given) ∴△BAC≅△DAE (By SAS congruence rule) ∴BC=DE (By CPCT)
Description : In a parallelogram ABCD, AE is perpendicular to DC and CF is perpendicular to AD. If AB = 10 cm, AE = 6 cm and CF = 8 cm, then find AD. -Maths 9th
Last Answer : Given, Parallelogram ABCD pAE = 8cm AB = 16cm CF = 10cm In a parallelogram, we know that opposite sides are equal. Therefore, CD = AB = 16cm To find the value of AD, the base is multiplied with height. Area of parallelogram = b x h 16 x 8 = AD x 10 128 = 10AD AD = 12.8cm
Description : In the given figure, if AB = 14 cm, BD = 10 cm and DC = 8 cm, then the value of tan B is a) 4/3 b) 14/3 c) 11/3 d) 9/12
Last Answer : a) 4/3
Description : In a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, -Maths 9th
Last Answer : answer:
Description : ABCD is a parallelogram AE pependicular to DC CF perpendixular to AD AB =16 m ,AE =8m ,CF =10m ,fimd AD -Maths 9th
Last Answer : This answer was deleted by our moderators...
Description : ABCD is a parallelogram with diagonal AC If a line XZ is drawn such that XZ ∥ AB, then BX/XC = ? (a) (AY/AC) (b) DZ/AZ (c) AZ/ZD (d) AC/AY Answer: (c) AZ/ZD 13. In the given figure, value of x (in cm) is (a) 5cm (b) 3.6 cm (c) 3.2 cm (d) 10 cm
Last Answer : (a) 5cm
Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th
Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.
Description : In Fig. 8.31, D is the mid-point of AB and PC = 1/2AP = 3 cm. If AD = DB = 4 cm and DE||BP. Find AE. -Maths 9th
Last Answer : Solution :-
Description : a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th
Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.
Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th
Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF=ABFB=1 from (1) ⇒LF=LD so, DF=2DL
Description : Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th
Description : ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th
Last Answer : Here is your First of all we will draw a quadrilateral ABCD with AD = BC and join AC, BD, P,Q,R,S are the mid points of AB, AC, CD and BD respectively. In the triangle ABC, P and Q are mid points of AB and AC respectively. All sides are equal so PQRS is a Rhombus.
Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th
Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th
Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.
Description : ABC is an isosceles triangle with AB = AC and BD, CE are its two medians. Show that BD = CE. -Maths 9th
Last Answer : Given ΔABC is an isosceles triangle in which AB = AC and BD, CE are its two medians. To show BD = CE.
Description : ABCE is an isosceles trapezoid and ACDE is a rectangle. AB = 10 and EC = 20. What is the length of AE?
Last Answer : Answer: AE = 10.
Description : In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is: a 12/7 b 24/7 c 20/7 d 7/24
Last Answer : b 24/7
Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th
Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)
Description : In the figure CD, AE and BF are one-third of their respective sides. It is given that -Maths 9th
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
Description : In the given figure, AB is the diameter where AP = 12 cm and PB = 16 cm. If the value of π is taken 3, what is the perimeter of the shaded region? (a) 58 cm (b) 116 cm (c) 29 cm (d) 156 cm
Last Answer : (a) 58 cm
Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th
Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.
Description : ABCD is a rhombus and AB is produved to E and F such that AE=AB=BF prove that ED and FC are perpendicular to each other -Maths 9th
Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th
Last Answer : According to question find the area of the parallelogram ABCD.
Description : In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th
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 order to determine whether or not 'fuse 2' shown in the illustration is defective, you should connect the voltage tester leads across points _____________. EL-0062 A. AC B. AD C. BC D. BD
Last Answer : Answer: B
Description : In order to determine whether or not 'Fuse 1', shown in the illustration is defective, you should connect the voltage tester leads across points _____________. EL-0062 A. AC B. AD C. BC D. BD
Last Answer : Answer: C
Description : P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th
Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS is a rhombus.
Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th
Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,
Description : Show that in a quadrilateral ABCD,AB + BC + CD + DA > AC + BD. -Maths 9th
Description : Figure abc is rectangle. segment AB is 6cm long , segment BC is 8 cm long , and segment AC is 10 cm long. what is the area of triangle abc?
Last Answer : 28
Description : The Karnaugh map for a Boolean function is given as The simplified Boolean equation for the above Karnaugh Map is (A) AB + CD + AB’ + AD (B) AB + AC + AD + BCD (C) AB + AD + BC + ACD (D) AB + AC + BC + BCD
Last Answer : (B) AB + AC + AD + BCD
Description : How many two input AND gates and two input OR gates are required to realize Y = BD+CE+AB (A) 1, 1 (B) 4, 2 (C) 3, 2 (D) 2, 3
Last Answer : Ans: A There are three product terms, so three AND gates of two inputs are required. As only two input OR gates are available, so two OR gates are required to get the logical sum of three product terms.
Description : in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th