Description : The bearings of the lines AB and BC are 146° 30' and 68° 30'. The included angle ABC is (A) 102° (B) 78° (C) 45° (D) None of these
Last Answer : (A) 102
Description : The distances AC and BC are measured from two fixed points A and B whose distance AB is known. The point C is plotted by intersection. This method is generally adopted in (A) Chain surveying (B) Traverse method of surveys (C) Triangulation (D) None of these
Last Answer : (A) Chain surveying
Description : In precision theodolite traverse if included angles are read twice and the mean reading accepted using both verniers having a least count of 30". Assuming the instrument to be in perfect adjustment, linear measurements correct to 6 mm per ... legs) (A) 50" n (B) 30" n (C) 60" n (D) None of these
Last Answer : (A) 50" n
Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th
Last Answer : answer:
Description : A traverse deflection angle is (A) Less than 90° (B) More than 90° but less than 180° (C) The difference between the included angle and 180° (D) The difference between 360° and the included angle
Last Answer : (C) The difference between the included angle and 180°
Description : The angle between the prolongation of the preceding line and the forward line of a traverse is called (A) Deflection angle (B) Included angle (C) Direct angle (D) None of the above
Last Answer : (A) Deflection angle
Description : The bearing of line AB is 152° 30' and angle ABC measured clockwise is 124° 28'. The bearing of BC is (A) 27° 52' (B) 96° 58' (C) 148° 08' (D) 186° 58'
Last Answer : (B) 96° 58'
Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th
Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF
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 : If deflection angles are measured in a closed traverse, the difference between the sum of the right-hand and that of the left hand angles should be equal to (A) 0° (B) 90° (C) 180° (D) 360°
Last Answer : (D) 360°
Description : If a linear traverse follows a sharp curve round a large lake where it is difficult to have long legs, the accuracy of the traverse may be improved by (A) Taking short legs (B) Making ... (C) Making a subsidiary traverse to determine the length of a long leg (D) All the above
Last Answer : (C) Making a subsidiary traverse to determine the length of a long leg
Description : A 4S diesel engine, when running at 2000 rpm has an injection duration of 1.5 ms.What is the corresponding of the crank angle in degrees? (a) 18° (b) 9° (c)36° (d) 15°
Last Answer : Ans :a
Description : The disc angle of a good plough varies between a) 25°-30° b) 10°-15° c) 18°-24° d) 42°-45°
Last Answer : 42°-45°
Description : An aircraft is travelling at 120 kts, what angle of bank would be required for a rate 1(one) turn? a. 30° b. 12° c. 18° d. 35°
Last Answer : c. 18°
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, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th
Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.
Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th
Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .
Description : D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th
Last Answer : Solution :-
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 : 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 : 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 : 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 : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th
Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved
Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th
Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.
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 : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th
Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.
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
Last Answer : This answer was deleted by our moderators...
Description : The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th
Description : From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th
Last Answer : (i) In Δ O C E ,D C 2 = D E 2 + E C 2 Δ O B D , D B 2 = O D 2 + B D 2 Δ O A F , O A 2 = O F 2 + A F 2 Adding we get O A 2 + O B 2 + O C 2 = O F 2 + O D 2 + O F 2 + E C 2 + B D 2 + A F 2 A F 2 + B D 2 + C E 2 = O A
Description : If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus. -Maths 9th
Last Answer : Here, we are joining A and C. In ΔABC P is the mid point of AB Q is the mid point of BC PQ∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and ... RS=PS=RQ[All sides are equal] ∴ PQRS is a parallelogram with all sides equal ∴ So PQRS is a rhombus.
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 : ABCD is a square. P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. By joining AR, BS, CP, DQ, we get a quadrilateral which is a -Maths 9th
Last Answer : According to the given statement, the figure will be a shown alongside; using mid-point theorem: In △ABC,PQ∥AC and PQ=21 AC .......(1) In △ADC,SR∥AC and SR=21 AC .... ... are perpendicular to each other) ∴PQ⊥QR(angle between two lines = angle between their parallels) Hence PQRS is a rectangle.
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
Description : In the figure, find teh perimeter of the shaded region where. ADC, AEB and BFC are semi-circles on diameters AC, AB and BC respectively. -Maths 10th
Last Answer : Given: Diameter of semicircle AEB= 2.8 cm RADIUS OF SEMICIRCLE(AEB)= 2.8/2= 1.4 cm Diameter of semicircle BFC=1.4 cm RADIUS OF SEMICIRCLE(BFC)= 1.4/2= 0.7 cm Diameter of semicircle ADC= 2.8+1.4 = ... ; .6 = 13.2 cm Hence, the perimeter of the shaded region is 13.2 cm HOPE THIS WILL HELP YOU...
Description : A simply supported uniform rectangular bar breadth b, depth d and length L carries an isolated load W at its mid-span. The same bar experiences an extension e under same tensile load. The ratio of the maximum deflection to the ... (A) L/d (B) L/2d (C) (L/2d)² (D) (L/3d)²
Last Answer : (C) (L/2d)
Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th
Last Answer : According to question prove that ar (ABCD) = ar (APQD).
Description : If is the perimete D is the closing error in departure, the correction for the departure of a traverse side of length , according to Bowditch rule, is (A) D × L/l (B) D × l²/L (C) L × l D (D) D × l/L
Last Answer : (D) D × l/L
Description : Which of the following methods of theodolite traversing is suitable for locating the details which are far away from transit stations? (A) Measuring angle and distance from one transit station (B) ... at one station and distance from other (D) Measuring distance from two points on traverse line
Last Answer : (B) Measuring angles to the point from at least two stations
Description : Consider the fractional knapsack instance n = 4, (p1, p2, p3, p4) = (10, 10, 12, 18), (w1, w2, w3, w4) = (2, 4, 6, 9) and M = 15. The maximum profit is given by (Assume p and w denotes profit and weight of objects respectively) (A) 40 (B) 38 (C) 32 (D) 30
Last Answer : (B) 38
Description : The critical angle for two media of refractive indices of medium 1 and 2 given by 2 and 1 respectively is a) 0 b) 30 c) 45 d) 60
Last Answer : b) 30
Description : The maximum deflection due to a load W at the free end of a cantilever of length L and having flexural rigidity EI, is (A) WL²/2EI (B) WL²/3EI (C) WL3 /2EI (D) WL3 /3EI
Last Answer : (D) WL3 /3EI