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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 61501. |
line through the points (4, 7, 8), (2, 3, 4) is parallel to the lineh the points (1,-2, D, (,2, 5). |
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| 61502. |
AM ANAB AD24./1n Fig. 6.19, DE IAC and DFIAE. Prove thatBF BEFE ECFigă 6,19 |
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Answer» thankyou soo much |
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| 61503. |
14. Sides AB and AC and median AD of atriangle ABC are respectivelyproportional to sides PQ and PR andmedian PM of another triangle PQR.Show that Δ ABC-APOR.de- |
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| 61504. |
AB AD4. In Fig. 6.19, DE | AC and DF l| AE. Prove thatBF BEFE EC |
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| 61505. |
0BF and DE are the altitudes on sides AD and AB respectively of parallelogram ABCD. If the area of the/ parallelograrm is 600 cm 2, AB = 60 cm and AD-25 cm, find BF and DE. |
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| 61506. |
20. E and F are respectively the midpoints of equal sides AB and AC ofABC. Show that BF# CE. |
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| 61507. |
EXERCISE 10.2. Construct a triangle ABC in which BC-7cm, B 75o and AB +AC-13 cm2. Construct a triangle ABC in which BC 8cm, LB 45° and AB-AC-3.5 cm.3. Construct a triangle PQR in which QR-бст, <Q PR-PO 2cm.- 60° and |
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| 61508. |
AB ADy in Fig. 6. 19, DE 11 AC and DF IAE. Prove thatBF BEFE ECFig. 6.19 |
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Answer» BASIC PROPORTIONALITY THEOREM (BPT) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. That is also known as Thales theorem.SOLUTION: GIVEN:In ∆BAC, DE || ACBE/EC = BD/DA………..(1) [ By Thales theorem(BPT)] In ∆BAE, DF || AE (GIVEN)BF/FE = BD/DA………..(2) [ By Thales theorem(BPT)] (From eq 1 and 2)BF/FE = BE/ECFE/BF = EC/BE [reciprocal the terms] Hence proved. |
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| 61509. |
3Find the coordinates of the points of trisection of the linejoining the points (4, -1) and (-2, -3) |
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| 61510. |
4. 1 and m are two parallel lines intersected byanother pair of parallel lines p and q(see Fig. 7.1 9). Show that Δ ABC Δ CDA. |
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| 61511. |
Distance between the points (4, -1) and (2, 3) is |
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Answer» thanks |
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| 61512. |
Q.9 Find the values of 'a' and 'b', if the points (-4, 4) and (-16, b) lie on the locus2 |
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| 61513. |
I and m are two parallel lines intersected byanother pair of parallel lines p and q(see Fig. 7.19). Show that Δ ABC Δ CDA.4. |
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| 61514. |
A chord ofa circle of radius 10 cm subtends a right angle at the centre. Find the area。the corresponding: (i) minor segment (ii) major sector. (Use 3.14)CLE-20y*4. |
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| 61515. |
6. In △ABC, AB-AC and AD丄BC, BE-AC and CF l AB. Then, △ABC is symmetrical about(a) AD(b) BE(c) CF(d) none of these |
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| 61516. |
4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find thethe corresponding: (i) minor segment (ii) major sector. (Use n 3.14) |
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| 61517. |
ABC is a triangle in which BE and CF are perpendicular to side AC and AB if BE=CF prove that ABC is isosceles |
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| 61518. |
1. 4, 18, 48, 100, 180, 294,(A) 448(C) 416(B) 512(D) 480 |
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Answer» 1 * (2 * 2) = 42 * (3 * 3) = 183 * (4 * 4) = 484 * (5 * 5) = 1005 * (6 * 6) = 1806 * (7 * 7) = 294 So 7*(8*8) = 7*64 = 448 Like my answer if you find it useful! |
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| 61519. |
14. In a A ABC, the angles at B and C are acute. If BE and CF be drawn perpendicular on AC and ABrespectively. Prove thatBC2 = AB X BF + AC X CE |
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| 61520. |
Lineand n are parallel lines intersected by transversal p23CD 11 EF and GH 11 KL. Find <HKL.25°600 У нiven helow, AB | | CD. Find the value of x in |
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Answer» Angle EHK = angle HKLangle EHK= (180-60 )+25thereforeangle HKL= 120° + 25° =145° |
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| 61521. |
3. Suppose ABC is a triangle in which BE and CF are respthe perpendiculars to the sides AC and AB. If BE CF, prtriangle ABC is isosceles |
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| 61522. |
ABC is a triangle in which altitudes BE and CF tosides AC and AB are equal (see Fig. 7.32). Showthat(ii) AB AC, i.e., ABC is an isosceles triangle. |
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| 61523. |
how that quadrilateral formed by the25. Two parallel lines I and m are intersected by a transversal P. Sbisectors of interior angles is a rectangle |
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| 61524. |
A quadrilateral has the vertices at the points (- 4, 2), (2, 6), (8, 5) and (9, -7Show that the mid-points of the sides of this quadrilateral are the vertices of aparallelogram.5. |
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Answer» Solutn?? Please ansr fast |
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| 61525. |
ExamnkExample 4(sec Fig. 8.15). Show that the quadrilateral formed by the bisectors of interior anglesTwo parallel lines I and m are intersected by a transversal pis a rectangle. |
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Answer» Given l||m and p is the transversalTo prove: PQRS is a rectangle Proof:RS, PS, PQ and RQ are bisectors of interior angles formed by the transversal with the parallel lines. ∠RSP = ∠RPQ (Alternate angles)Hence RS||PQSimilarly, PS||RQ (∠RPS = ∠PRQ)Therefore quadrilateral PQRS is a parallelogram as both the pairs of opposite sides are parallel. From the figure, we have ∠b + ∠b + ∠a + ∠a = 180°⇒ 2(∠b + ∠a) = 180°∴ ∠b + ∠a = 90° That is PQRS is a parallelogram and one of the angle is a right angle.Hence PQRS is a rectangle Thanks |
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| 61526. |
\frac { 7 \sqrt { 3 } - 5 \sqrt { 2 } } { \sqrt { 48 } + \sqrt { 18 } } |
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| 61527. |
Example 4 Two parallel lines 1 and m are intersected by a transversal p(see Fig. 8.15). Show that the quadrilateral formed by the bisectors of interior anglesis a rectangle.that PS II OR and transversal p intersects them at points A and C |
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| 61528. |
1. In Fig. 6.17, (i) and (ii), DE Il BC. Find EC in (i) and AD in (ii).in (111.5 cm1 cm1.8 cm7.2 cm3 cm5.4 cm |
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Answer» 1) EC/AE = DB/AD => EC = 3*1/1.5 = 2cm 3) AD/DB = AE/EC => AD = 1.8*7.2/5.4 = 2.4cm |
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| 61529. |
2. S3. Two parallel lines I and m are intersected by a transversal p. Show that the qundhow that the diagonals of a rhombus are perpendicular to each other.infoier analas is a rectangle |
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Answer» 1. |
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| 61530. |
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area ofthe corresponding:4.(i) minor segment(ii) majorsector. (Use π= 3.14) |
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| 61531. |
7. In Fig. 6.41, BA lI DE and BC Il EF. Provethat ZABC + ZDEF180°Fig. 6.41 C |
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| 61532. |
4. A chord ofa circle of radius 10 cm subtends a right ang le at the centre. Find the area ofthe comesponding: (i) minor segment(ii) major sector (Use π=3.14) |
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| 61533. |
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area ofthe corresponding: (i) minor segment (ii) major sector. (Use t 3.14)4. |
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| 61534. |
Prove that the quadrilateral formed by joining the mid points of the sides of a quadrilateral is aParallelogram. |
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| 61535. |
Show that the quadrilateral formed by joining the mid points of the consecutive sides of a rectangle is a rhombus. |
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| 61536. |
6. Name the type of quadrilateralformed, if any, by the followingpoints, and give reasons foryour answer |
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| 61537. |
[4 marks each/(1) Prove that the quadrilateral formed by joining themidpoints of sides of a quadrilateral in order is a4. Attempt the following:parallelogram.Figure :Proof |
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Answer» Let the rhombus have vertices A, B, C, D. Let the midpoints of the sides AB, BC, CD, DA be E, F, G, H. If the diagonals of EFGH are equal and bisect each other then EFGH is a rectangle. AB || HF || DC and AD || EG || BC.So HF = AB = BC = EG.ie diagonals HF and EG are equal in length. HF bisects AD and BC. EG is parallel to AD and BC. Therefore HF bisects EG also.By a similar argument, EG bisects HF. So diagonals HE and EG (i) are equal in length, and (ii) bisect each other.Therefore EFGH is a rectangle |
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| 61538. |
Two are intersected by a transversal P. Show that thequadrilateral formed by the bisectors of interior angles is a rectangle. |
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Answer» 1 2 3 Theorem:if two parallel lines are intersected by a transversal |
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| 61539. |
5 (a) In figure () given below, DEII BCtriangle0) In figure (il) given below, AB | DE and BD || EF. Prove that DC2CF x AC.(0 |
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Answer» InΔABC, we have CD/DA = CE/EB as AB || DEinΔCDB, we have CF/FD = CE/EB as EF || BD hence we get CD/DA = CF/FD=>Reciprocals: DA/CD = FD /CF=> Add 1 on bothsides: AC /CD = DC/CF => DC^2 = CF * AC thanks |
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| 61540. |
4. The perimeter of rectangle is 140 cm. If the sidesare in the ratio 3 : 4, find the lengths of the foursides and the two diagonals. |
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Answer» let the sides are 3x and 4x perimeter = 2(3x+4x) = 140 => 14x = 140=> x = 10 so, length = 4*10=40cm and breadth = 3*10 = 30cm also, Diagonal = √40²+30² = 50cm |
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| 61541. |
48 - [ 18 - \{ 16 - ( 5 - 4 - 1 ) \} ] |
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Answer» 48 - [ 18 - { 16 - ( 5 - 4 - 1 )}] = 48 - [ 18 - { 16 }] = 48 - [18 - 16 ] = 48 - 2 = 46 |
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| 61542. |
PQR is an isosceles triangle.EXERCISIn Fig. 6.17, (i) and (ii), DE Il BC. Find EC in1.5 cm1 cm7.23 cm |
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| 61543. |
(x-5)(x-18)=48 |
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| 61544. |
In Fig. 9.17, PORS and ABRS are parallelogramsand X is any point on side BR. Show that(i) ar (PQRS) = ar (ABRS)P A(ii) ar (AX S)-ar (PQRS)Fig. 9.17 |
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| 61545. |
5.In Fig. 9.17, PQRS and ABRS are parallelogramsand X is any point on side BR. Show that(i) ar (PQRS) = ar (ABRS)1801R.(ii) ar (AX S)ar (PQRS)Fig. 9.17 |
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| 61546. |
shord ofa circle of radius 10 cm subtends a right angle at the centre. Find thethe corresponding: (0) minor segment (ii) major sector.(Use t 1.140) |
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| 61547. |
1. How many rational numbers lies between 2 and3? |
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Answer» There infinite rational between 2 and 3 |
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| 61548. |
SECTION -B:MATHEMATICS. The rational number lies between 3 and 2 is253423 |
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Answer» There are infinitely many. Here are some: 1/2, 3/5, 4/7, 4/9, 5/8, 5/9, 5/11, 6/11, 7/11 The answer is option (2) |
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| 61549. |
-18- lies between consecutive integers2.Rational numberd5 |
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Answer» -18 / 5 = -3.6 It lies between - 4 and - 3 integers. rational number -18/5 lies between consecutive number = -4 & -3. |
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| 61550. |
MV = *75 l share is at discount of25 find EV of a share. (Ans. *100] |
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Answer» 100 rupees is right answer. 100 is your right answer 100 is the answer of your question |
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